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we can break it up into the sum or difference of two of the special angles. See Table 1. Sum formula for cosine Difference formula for cosine cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β Table 1 First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 3. Point P is at an angle α from the positive x-axis with coordinates (cos α, sin α) and point Q is at an angle of β from the positive x-axis with coordinates (cos β, sin β). Note the measure of angle POQ is α − β. Label two more points: A at an angle of (α − β) from the positive x-axis with coordinates (cos(α − β), sin(α − β)); and point B with coordinates (1, 0). Triangle POQ is a rotation of triangle AOB and thus the distance from P to Q is the same as the distance from A to B. y 2 π, 12 f(x) = sin x x π 2π –2π – π, 2 –π –1 –2 Figure 3 We can find the distance from P to Q using the distance formula. dPQ = √ = √ (cos α − cos β)2 + (sin α − sin β)2 cos2 α − 2 cos α cos β + cos2 β + sin2 α − 2 sin α sin β + sin2 β ———— —— Then we apply the Pythagorean Identity and simplify. = √ = √ = √ ———— —— (cos2 α + sin2 α) + (cos2 β + sin2 β) − 2 cos α cos β − 2 sin α sin β 1 + 1 − 2 cos α cos β − 2 sin α sin β 2 − 2 cos α cos β − 2 sin α sin β —— Similarly, using the distance formula we can find the distance from A to B. —— dAB = √ = √ (cos(α − β) − 1)2 + (sin(α − β) − 0)2 cos2(α − β) − 2 cos(α − β) + 1 + sin2(α − β) ——— Applying the Pythagorean Identity and simplifying we get: = √ =
√ = √ ——— —— (cos2(α − β) + sin2(α − β)) − 2 cos(α − β) + 1 1 − 2 cos(α − β) + 1 2 − 2 cos(α − β) — Because the two distances are the same, we set them equal to each other and simplify. √ —— 2 − 2 cos α cos β − 2 sin α sin β = √ 2 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos(α − β) 2 − 2 cos(α − β) — Finally we subtract 2 from both sides and divide both sides by −2. cos α cos β + sin α sin β = cos(α − β) Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles. 572 CHAPTER 7 trigonometric identities and eQuations sum and difference formulas for cosine These formulas can be used to calculate the cosine of sums and differences of angles. cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β How To… Given two angles, find the cosine of the difference between the angles. 1. Write the difference formula for cosine. 2. Substitute the values of the given angles into the formula. 3. Simplify. Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles Example 1 Using the formula for the cosine of the difference of two angles, find the exact value of cos  Solution Use the formula for the cosine of the difference of two angles. We have 5π __ 4 π __ . − 6 cos(α − β) = cos α cos β + sin α sin β cos  5π ___ 4 π __  = cos  − 6 — — — π 5π π 5π __ ___ __ ___  + sin   cos   sin   √ __ ____ ____ ____  2 2 2 2 — — 2 6 √ √ ____ ____ 4 4 — — 2 6 − √ − √ __________ 4 = − − = Try It #1 π __ π __
− . Find the exact value of cos  4 3 Example 2 Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine Find the exact value of cos(75°). Solution As 75° = 45° + 30°, we can evaluate cos(75°) as cos(45° + 30°). Thus, cos(45° + 30°) = cos(45°)cos(30°) − sin(45°)sin(30°) = = = — — — — 2 3 2  −  √ √ 1 √ __ ____ ____ ____   2 2 2 2 — 2 6 √ √ ____ ____ 4 4 — — 6 − √ 2 √ __________ 4 − Try It #2 Find the exact value of cos(105°). Using the Sum and Difference Formulas for Sine The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas. SECTION 7.2 sum and diFFerence identities 573 sum and difference formulas for sine These formulas can be used to calculate the sines of sums and differences of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β How To… Given two angles, find the sine of the difference between the angles. 1. Write the difference formula for sine. 2. Substitute the given angles into the formula. 3. Simplify. Example 3 Using Sum and Difference Identities to Evaluate the Difference of Angles Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b. a. sin(45° − 30°) b. sin(135° − 120°) Solution a. Let’s begin by writing the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(45° − 30°) = sin(45°)cos(30°) − cos(45°)sin(30°) Next, we need to find the values of the trigonometric expressions. sin(45°) =, cos(30°) =, cos(45°) = — 2 √ ____ 2 — 3 √ ____ 2 —
2, sin(30°) = 1 √ __ ____ 2 2 Now we can substitute these values into the equation and simplify. 2 3 2 √  −  1 √ √ __ ____ ____ ____   2 2 2 2 — 6 − √ 2 √ _________ 4 sin(45° − 30°) = = — — — — b. Again, we write the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(135° − 120°) = sin(135°)cos(120°) − cos(135°)sin(120°) Next, we find the values of the trigonometric expressions. sin(135°) = — 2 √, cos(120°) = − 1 __ ____, cos(135°) = − 2 2 2 √ ____ 2 —, sin(120°) = — 3 √ ____ 2 Now we can substitute these values into the equation and simplify. sin(135° − 120°) = sin(135° − 120°) = — — — — — — — ____ ____ __ ____ 2 2 2 2 2 + √ 6 − √ __________ 4 6 − √ √ _________ ____ ____ __ ____ 2 2 2 2 2 + √ 6 − √ __________ 4 6 − √ √ _________ 4 2 — — — — — — — = = = = 574 CHAPTER 7 trigonometric identities and eQuations Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function Example 4 3 1 __ __ Find the exact value of sin  cos−1 . + sin−1 5 2 3 1 __ __ and β = sin−1 Solution The pattern displayed in this problem is sin(α + β). Let α = cos−1. Then we can write 5 2 cos α = 1 __, 0 ≤ α ≤ π 2 π π sin β = 3 __ __ __ ≤ β ≤, − 2 2 5 We will use the Pythagorean identities to find sin α and cos β. sin α = √ — = √ = √ 1 − cos2 α ______ 1 − 1 __ 4 __ 3 __ 4 — 3 √ _ 2 = cos β = √ —
1 − sin2 β ______ 9 __ 1 − 25 ___ 16 __ 25 = √ = √ = 4 __ 5 Using the sum formula for sine, 3 1 __ __  = sin(α + β) sin  cos−1 + sin−1 5 2 — = = sin α cos β + cos α sin β 3 4 √ 3 + 1 ____ __ __ __ · · 2 5 2 5 — 3 + 3 4 √ ________ 10 = Using the Sum and Difference Formulas for Tangent Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern. Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, tan x =, cos x ≠ 0. sin x ____ cos x Let’s derive the sum formula for tangent. tan(α + β) = sin(α + β) _ cos(α + β) = sin α cos β + cos α sin β __________________ cos α cos β − sin α sin β = sin α cos β + cos α sin β __________________ cos α cos β cos α cos β − sin α sin β __________________ cos α cos β = sin α cos β ________ + cos α cos β cos α cos β _________ − cos α cos β cos α sin β ________ cos α cos β sin α sin β ________ cos α cos β Divide the numerator and denominator by cos α cos β = = + sin α ____ cos α 1 − sin β ____ cos β sin α sin β ________ cos α cos β tan α + tan β ____________ 1 − tan α tan β We can derive the difference formula for tangent in a similar way. SECTION 7.2 sum and diFFerence identities 575 sum and difference formulas for tangent The sum and difference formulas for tangent are: tan(α + β) = tan α + tan β ___________ 1 − tan α tan β tan(α − β) = tan α − tan β ___________ 1 + tan α tan β How To… Given two angles, find the tangent of the sum of the angles. 1. Write the sum formula for tangent. 2. Substitute the given angles into the formula. 3. Simplify. Example 5 Finding the Exact Value of
an Expression Involving Tangent π π __ __ + . Find the exact value of tan  4 6 Solution Let’s first write the sum formula for tangent and substitute the given angles into the formula. tan(α + β) = tan α + tan β __ 1 − tan α tan β π π __ __  = + tan  4 6 π π __ __  + tan   tan  4 6 ___ π π __ __ 1 −  tan     tan  4 6 Next, we determine the individual tangents within the formulas: So we have π __  = tan  6 1 _ — 3 √ π __  = 1 tan  4 π π __ __  = + tan  4 6 1 _ + 1 — 3 √ __ 1 −  1 _  (1 √ — Try It #3 Find the exact value of tan  2π __ 3 π __ . + 4 Finding Multiple Sums and Differences of Angles Example 6 π Given sin α = 3 __ __, cos. sin(α + β) b. cos(α + β), π < β < 5 __ 13 c. tan(α + β) 3π __, find 2 d. tan(α − β) 576 CHAPTER 7 trigonometric identities and eQuations Solution We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas. π a. To find sin(α + β), we begin with sin α = 3 __ __ and 0 < α <. The side opposite α has length 3, the hypotenuse 2 5 has length 5, and α is in the first quadrant. See Figure 4. Using the Pythagorean Theorem,we can find the length of side a: a2 + 32 = 52 a2 = 16 a = 4 y (4, 3) x (4, 0) 5 α Figure 4 3π __ Since cos β = − 2
quadrant. See Figure 5. Again, using the Pythagorean Theorem, we have and π < β < 5 __ 13, the side adjacent to β is −5, the hypotenuse is 13, and β is in the third (−5)2 + a2 = 132 25 + a2 = 169 a2 = 144 a = ±12 Since β is in the third quadrant, a = –12. (–5, 0) y x β 13 (–5, –12) Figure 5 The next step is finding the cosine of α and the sine of β. The cosine of α is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5: cos α = 4 __. We can also find the sine of β from the triangle in Figure 5, 5 12 __. Now we are ready to evaluate sin(α + β). as opposite side over the hypotenuse: sin β = − 13 sin(α + β) = sin αcos β + cos αsin β   − 12 4 5 __ __ __   +  13 13 5 48 __ 65 = − 3 __   − =  5 15 __ 65 63 __ 65 = − − SECTION 7.2 sum and diFFerence identities 577 b. We can find cos(α + β) in a similar manner. We substitute the values according to the formula. cos(α + β) = cos α cos β − sin α sin β   − 12 3 __ __   −  13 5 4 __   − =  5 5 __ 13 = − 20 __ 65 + 36 __ 65 = 16 __ 65 and cos α = 4 c. For tan(α + β), if sin α = 3 __ __, then 5 5 tan α = 3 __ 5 = 3 __ _ 4 4 __ 5 If sin β = − and cos β = −, then 12 __ 13 5 __ 13 Then, tan β = −12 _ 13 _ −5 _ 13 = 12 __ 5 tan(α + β) = tan α + tan β __ 1 − tan α tan β = 12 3 __ __ + 5 4 __ 12 1 − 3 __ __   5 4 = 63 __ 20 _ − 16 __ 20 = − 63 __ 16 d.
To find tan(α − β), we have the values we need. We can substitute them in and evaluate. tan(α − β) = tan α − tan β __ 1 + tan α tan β = = 12 3 __ __ − 5 4 __ 1 + 3 12 __ __   5 4 − 33 __ 20 _ 56 __ 20 = − 33 _ 56 Analysis β are angles in the same triangle, which of course, they are not. Also note that A common mistake when addressing problems such as this one is that we may be tempted to think that α and tan(α + β) = sin(α + β) ________ cos(α + β) 578 CHAPTER 7 trigonometric identities and eQuations Using Sum and Difference Formulas for Cofunctions Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two π π __ __,, those two angles are complements, and the sum of the two acute angles in a right triangle is positive angles is 2 2 so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as θ, then the other acute π __ − θ . angle must be labeled  2 π __ − θ  : opposite over hypotenuse. Thus, when two angles are complimentary, we can Notice also that sin θ = cos  2 say that the sine of θ equals the cofunction of the complement of θ. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions. π – θ 2 θ Figure 6 From these relationships, the cofunction identities are formed. cofunction identities The cofunction identities are summarized in Table 2. π __ − θ  sin θ = cos  2 π __ − θ  sec θ = csc  2 π __ − θ  cos θ = sin  2 π __ − θ  csc θ = sec  2 Table 2 π __ − θ  tan θ = cot  2
π __ − θ  cot θ = tan  2 Notice that the formulas in the table may also be justified algebraically using the sum and difference formulas. For example, using we can write cos(α − β) = cos αcos β + sin αsin β, π π π __ __ __ cos θ + sin − θ  = cos cos  sin θ 2 2 2 = (0)cos θ + (1)sin θ = sin θ Example 7 Finding a Cofunction with the Same Value as the Given Expression π __ Write tan in terms of its cofunction. 9 π __ − θ . Thus, Solution The cofunction of tan θ = cot  2 π π π __ __ __ −  = cot   tan  9 2 9 = cot  9π ___ 18 − 2π ___  18 = cot  7π ___  18 SECTION 7.2 sum and diFFerence identities 579 Try It #4 π __ in terms of its cofunction. Write sin 7 Using the Sum and Difference Formulas to Verify Identities Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules from Solving Trigonometric Equations with Identities may help simplify the process of verifying an identity. How To… Given an identity, verify using sum and difference formulas. 1. Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient. 2. Look for opportunities to use the sum and difference formulas. 3. Rewrite sums or differences of quotients as single quotients. 4. If the process becomes cumbersome, rewrite the expression in terms of sines and cosines. Example 8 Verifying an Identity Involving Sine Verify the identity sin(α + β) + sin(α − β) = 2 sin α cos β. Solution We see that the left side of the equation includes the sines of the sum and the difference of angles. sin(α + β)
= sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β We can rewrite each using the sum and difference formulas. sin(α + β) + sin(α − β) = sin α cos β + cos α sin β + sin α cos β − cos α sin β We see that the identity is verified. = 2 sin α cos β Example 9 Verifying an Identity Involving Tangent Verify the following identity. sin(α − β) _ cos α cos β = tan α − tan β Solution We can begin by rewriting the numerator on the left side of the equation. sin(α − β) _ = cos α cos β sin α cos β − cos α sin β __ cos α cos β = sin α cos β _ − cos α cos β cos α sin β _ cos α cos β = sin α _ cos α − sin β _ cos β Rewrite using a common denominator. Cancel. We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine. = tan α − tan β Rewrite in terms of tangent. 580 CHAPTER 7 trigonometric identities and eQuations Try It #5 Verify the identity: tan(π − θ) = −tan θ. Example 10 Using Sum and Difference Formulas to Solve an Application Problem Let L1 and L2 denote two non-vertical intersecting lines, and let θ denote the acute angle between L1 and L2. See Figure 7. Show that tan θ = m2 − m1 _ 1 + m1 m2 where m1 and m2 are the slopes of L1 and L2 respectively. (Hint: Use the fact that tan θ1 = m1 and tan θ2 = m2.) y L1 L2 θ θ1 θ2 x Figure 7 Solution Using the difference formula for tangent, this problem does not seem as daunting as it might. tan θ = tan(θ2 − θ1) = = tan θ2 − tan θ1 __ 1 + tan θ1 tan θ2 m2 − m1 _ 1 + m1m2 Example 11 Investigating a Guy-wire Problem For a climbing wall, a guy-wire R is attached 47 feet high on a vertical pole. Added support is provided by another guy-
wire S attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle α between the wires. See Figure 8. 47 ft 40 ft R S α β 50 ft Figure 8 Solution Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that tan β = then use difference formula for tangent. 47 __ 50, and tan( β − α) = 40 __ 50 = 4 __. We can 5 tan(β − α) = tan β − tan α __ 1 + tan β tan α SECTION 7.2 sum and diFFerence identities 581 Now, substituting the values we know into the formula, we have 4 __ = 5 47 __ − tan α 50 __ 47 ___ 1 + 50 tan α 4  1 + tan α  = 5  47 __ 50 47 __ 50 − tan α  Use the distributive property, and then simplify the functions. 4(1) + 4   tan α = 5  47 __ 50 47 __ 50  − 5tan α 4 + 3.76tan α = 4.7 − 5tan α 5tan α + 3.76tan α = 0.7 8.76tan α = 0.7 tan α ≈ 0.07991 tan−1(0.07991) ≈.079741 Now we can calculate the angle in degrees. α ≈ 0.079741  180 ___ π  ≈ 4.57° Analysis Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given. Access these online resources for additional instruction and practice with sum and difference identities. • Sum and Difference Identities for Cosine (http://openstaxcollege.org/l/sumdifcos) • Sum and Difference Identities for Sine (http://openstaxcollege.org/l/sumdifsin) • Sum and Difference Identities for Tangent (http://openstaxcollege.org/l/sumdiftan) 582 CHAPTER 7 trigonometric identities and eQuations 7.2
SeCTIOn exeRCISeS VeRBAl 1. Explain the basis for the cofunction identities and when they apply. 3. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for f(x) = sin(x) and g(x) = cos(x). (Hint: 0 − x = −x) AlGeBRAIC For the following exercises, find the exact value. 4. cos  7π ___  12 8. tan  − π __  12 5. cos  π __  12 9. tan  19π ___  12 2. Is there only one way to evaluate cos  5π ___ ? Explain 4 how to set up the solution in two different ways, and then compute to make sure they give the same answer. 6. sin  5π ___  12 7. sin  11π ___  12 For the following exercises, rewrite in terms of sin x and cos x. 10. sin  x + 11π ___  6 11. sin  x − 3π ___  4 12. cos  x − 5π ___  6 13. cos  x + 2π ___  3 For the following exercises, simplify the given expression. π __ − t  14. csc  2 π __ − θ  15. sec  2 π __ − x  16. cot  2 π __ − x  17. tan  2 18. sin(2x) cos(5x) − sin(5x) cos(2x) 19. 7 3 __ __ x  − tan  tan  x  5 2 7 3 __ __ x  tan  1 + tan  x  5 2 For the following exercises, find the requested information. and cos b = − 1 20. Given that sin a = 2 π __ __ __, π , find sin(a + b) and cos(a − b)., with a and b both in the interval 
2 4 3, and cos b = 1 21. Given that sin a = 4 π __ __ __ , find sin(a − b) and cos(a + b)., with a and b both in the interval  0, 2 3 5 For the following exercises, find the exact value of each expression. 1 __   22. sin  cos−1(0) − cos−1  2 1 1 __ __    − cos−1  24. tan  sin−1  2 2 3 2    + sin− 1  23. cos  cos−1  √ √ ____ ____ 2 2 — — SECTION 7.2 section exercises 583 GRAPHICAl For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. π __ − x  25. cos  2 π __ − x  29. tan  4 26. sin(π − x) 30. cos  7π ___ + x  6 π __ + x  27. tan  3 π __ + x  31. sin  4 π __ + x  28. sin  3 32. cos  5π ___ + x  4 For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think 2x = x + x. ) 33. f(x) = sin(4x) − sin(3x)cos x, g(x) = sin x cos(3x) 34. f(x) = cos(4x) + sin x sin(3x), g(x) = −cos x cos(3x) 35. f(x) = sin(3x)cos(6x), g(x) = −sin(3x)cos(6x) 36. f(x) = sin(4x), g(x) = sin(5x)cos x − cos(5x)sin
x 37. f(x) = sin(2x), g(x) = 2 sin x cos x 38. f(θ) = cos(2θ), g(θ) = cos2 θ − sin2 θ 39. f(θ) = tan(2θ), g(θ) = tan θ _ 1 + tan2θ 41. f(x) = tan(−x), g(x) = tan x − tan(2x) __ 1 − tan x tan(2x) TeCHnOlOGY 40. f(x) = sin(3x)sin x, g(x) = sin2(2x)cos2 x − cos2(2x)sin2 x For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. 43. sin(195°) 44. cos(165°) 45. cos(345°) 42. sin(75°) 46. tan(−15°) exTenSIOnS For the following exercises, prove the identities provided. π __  = 47. tan  x + 4 tan x + 1 ________ 1 − tan x 49. cos(a + b) ________ cos a cos b = 1 − tan a tan b 48. tan(a + b) ________ = tan(a − b) sin a cos a + sin b cos b __________________ sin a cos a − sin b cos b 50. cos(x + y)cos(x − y) = cos2 x − sin2 y 51. cos(x + h) − cos x ______________ h = cos x cos h − 1 _ h − sin x sin h _ h For the following exercises, prove or disprove the statements. 52. tan(u + v) = tan u + tan v ___________ 1 − tan u tan v 53. tan(u − v) = tan u − tan v ___________ 1 + tan u tan v 54. tan(x + y) __ = 1 + tan x tan x tan x + tan y __ 1 − tan2 x tan2 y 55. If α, β, and γ are angles in the same triangle, then prove or disprove sin(α + β) = sin γ. 56. If α, β, and γ are angles in
the same triangle, then prove or disprove: tan α + tan β + tan γ = tan α tan β tan γ. 584 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • • • Use double-angle formulas to find exact values. Use double-angle formulas to verify identities. Use reduction formulas to simplify an expression. Use half-angle formulas to find exact values. 7. 3 DOUBle-AnGle, HAlF-AnGle, AnD ReDUCTIOn FORMUlAS Figure 1 Bicycle ramps for advanced riders have a steeper incline than those designed for novices. Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. 5 __ For advanced competitors, the angle formed by the ramp and the ground should be θ such that tan θ =. The angle 3 is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one. Using Double-Angle Formulas to Find exact Values In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α = β. Deriving the double-angle formula for sine begins with the sum formula, If we let α = β = θ, then we have sin(α + β) = sin α cos β + cos α sin β sin(θ + θ) = sin θ cos θ + cos θ sin θ sin(2θ) = 2sin θ cos θ Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α + β) = cos α cos β − sin α sin β, and letting α = β = θ, we have cos(θ + θ) = cos θ cos θ − sin θ sin θ cos(2θ) = cos2 θ − sin2 θ Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is: The second interpretation is:
cos(2θ) = cos2 θ − sin2 θ = (1 − sin2 θ) − sin2 θ = 1 − 2sin2 θ cos(2θ) = cos2 θ − sin2 θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1 SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 585 Similarly, to derive the double-angle formula for tangent, replacing α = β = θ in the sum formula gives tan(α + β) = tan α + tan β ____________ 1 − tan α tan β tan(θ + θ) = tan θ + tan θ ___________ 1 − tan θ tan θ tan(2θ) = 2tan θ ________ 1 − tan2 θ double-angle formulas The double-angle formulas are summarized as follows: sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1 tan(2θ) = 2 tan θ ________ 1 − tan2 θ How To… Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value. 1. Draw a triangle to reflect the given information. 2. Determine the correct double-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify. Example 1 Using a Double-Angle Formula to Find the Exact Value Involving Tangent Given that tan θ = − 3 __ 4 and θ is in quadrant II, find the following: a. sin(2θ) b. cos(2θ) c. tan(2θ) Solution If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given tan θ = − 3 __, such that θ is in quadrant II. The tangent of an angle is equal to the opposite 4 side over the adjacent side, and because θ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
(−4)2 + (3)2 = c2 16 + 9 = c2 25 = c2 c = 5 Now we can draw a triangle similar to the one shown in Figure 2. y (–4, 3) (–4, 0) 5 θ Figure 2 x 586 CHAPTER 7 trigonometric identities and eQuations a. Let’s begin by writing the double-angle formula for sine. sin(2θ) = 2 sin θ cos θ We see that we to need to find sin θ and cos θ. Based on Figure 2, we see that the hypotenuse equals 5, so sin θ = 3 __ 5 Thus,. Substitute these values into the equation, and simplify., and cos θ = − 4 __ 5 4 3 __ __   − sin(2θ) = 2   5 5 = − 24 __ 25 b. Write the double-angle formula for cosine. Again, substitute the values of the sine and cosine into the equation, and simplify. cos(2θ) = cos2 θ − sin2 θ 2 2 3 4 __ __ −   cos(2θ) =  −  5 5 9 16 __ __ − 25 25 7 __ 25 = = c. Write the double-angle formula for tangent. In this formula, we need the tangent, which we were given as tan θ = − 3 __ 4 equation, and simplify.. Substitute this value into the tan(2θ) = 2 tan θ ________ 1 − tan2 θ tan(2θ) = 2 = 3 __  2  − 4 __ 3 __  1 −  − 4 3 __ − 2 _ 9 __ 1 − 16 16 3 __ __   = − 7 2 24 __ 7 = − Try It #1 5 __, with θ in quadrant I, find cos(2α). Given sin α = 8 Example 2 Using the Double-Angle Formula for Cosine without Exact Values Use the double-angle formula for cosine to write cos(6x) in terms of cos(3x). Solution cos(6x) = cos(2(3x)) = cos2 3x − sin2 3x = 2cos2 3x
− 1 Analysis This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function. Using Double-Angle Formulas to Verify Identities Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side. Example 3 Using the Double-Angle Formulas to Establish an Identity Establish the following identity using double-angle formulas: 1 + sin(2θ) = (sin θ + cos θ)2 SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 587 Solution We will work on the right side of the equal sign and rewrite the expression until it matches the left side. (sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ = (sin2 θ + cos2 θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ = 1 + sin(2θ) Analysis This process is not complicated, as long as we recall the perfect square formula from algebra: (a ± b)2 = a2 ± 2ab + b2 where a = sin θ and b = cos θ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent. Try It #2 Establish the identity: cos4 θ − sin4 θ = cos(2θ). Example 4 Verifying a Double-Angle Identity for Tangent Verify the identity: tan(2θ) = 2 __________ cot θ − tan θ Solution In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation. tan(2θ) = 2 tan θ ________ 1 − tan2 θ Double-angle formula Multiply by a term that results in desired numerator. = = 1 ____  2 tan θ  tan θ __ 1 ____  (1 − tan2 θ)  tan θ 2 __ tan2 θ 1 _____ ____ tan θ
tan θ − Analysis Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show = 2 __________ cot θ − tan θ Use reciprocal identity for 1 ____. tan θ Let’s work on the right side. 2tan θ ________ = 1 − tan2 θ 2 __________ cot θ − tan θ 2 __________ = cot θ − tan θ tan θ  ____ tan θ 2  __ 1 ____ − tan θ tan θ ___ (tan θ) − tan θ(tan θ) 2 tan θ = 1 _ tan θ When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier. = 2 tan θ _ 1 − tan2 θ Try It #3 Verify the identity: cos(2θ)cos θ = cos3 θ − cos θ sin2 θ. 588 CHAPTER 7 trigonometric identities and eQuations Use Reduction Formulas to Simplify an expression The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas. We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ) = 1 − 2 sin2 θ. Solve for sin2 θ: cos(2θ) = 1 − 2 sin2 θ 2 sin2 θ = 1 − cos(2θ) 1 − cos(2θ) _________ 2 Next, we use the formula
cos(2θ) = 2 cos2 θ − 1. Solve for cos2 θ: sin2 θ = cos(2θ) = 2 cos2 θ − 1 1 + cos(2θ) = 2 cos2 θ 1 + cos(2θ) _________ 2 = cos2 θ The last reduction formula is derived by writing tangent in terms of sine and cosine: tan2 θ = sin2 θ _____ cos2 θ 1 − cos(2θ) _________ 2 __ 1 + cos(2θ) _________ 2 1 − cos(2θ)   _________ 2 1 − cos(2θ) _________ 1 + cos(2θ) = =  = Substitute the reduction formulas. 2 _________ 1 + cos(2θ)  reduction formulas The reduction formulas are summarized as follows: sin2 θ = 1 − cos(2θ) _________ 2 cos2 θ = 1 + cos(2θ) _________ 2 tan2 θ = 1 − cos(2θ) _________ 1 + cos(2θ) Example 5 Writing an Equivalent Expression Not Containing Powers Greater Than 1 Write an equivalent expression for cos4 x that does not involve any powers of sine or cosine greater than 1. Solution We will apply the reduction formula for cosine twice. 1 + cos(2x) 2  _________ 2 Substitute reduction formula for cos2 x. cos4 x = (cos2 x)2 =  = 1 __ (1 + 2cos(2x) + cos2(2x)) 4 1 + cos2(2x) 1 1 = 1   __________ __ __ __ cos(2x __ __ __ __ + cos(2x) + + cos(4x) 8 2 4 8 1 1 = 3 __ __ __ cos(2x) + + cos(4x) 8 2 8 Analysis The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra. Substitute reduction formula for cos2 x. SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 589 Example 6 Using the Power-Reducing Formulas to Prove an Identity Use
the power-reducing formulas to prove sin3 (2x) =  1 __ sin(2x)  [1 − cos(4x)] 2 Solution We will work on simplifying the left side of the equation: 1 − cos(4x) _________ 2 sin3(2x) = [sin(2x)][sin2(2x)] = sin(2x)  1 __  [1 − cos(4x)] = sin(2x)  2 1 __ = [sin(2x)][1 − cos(4x)] 2  Substitute the power-reduction formula. Analysis Note that in this example, we substituted for sin2(2x). The formula states We let θ = 2x, so 2θ = 4x. 1 − cos(4x) _________ 2 sin2 θ = 1 − cos(2θ) _________ 2 Try It #4 Use the power-reducing formulas to prove that 10 cos4 x = 15 __ 4 5 __ + 5 cos(2x) + cos(4x). 4 Using Half-Angle Formulas to Find exact Values The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can α __ use when we have an angle that is half the size of a special angle. If we replace θ with, the half-angle formula for sine 2 α __ . Note that the half-angle formulas are preceded by a ± sign. is found by simplifying the equation and solving for sin  2 This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in α __ terminates. which 2 The half-angle formula for sine is derived as follows: sin2 θ = 1 − cos(2θ) _________ 2 α __  = sin2  2 α __ 1 − cos  2 ⋅  2 __ 2 = 1 − cos α ________ 2 _________ 1 − cos α ________ 2 α __ sin  2  = ± √ To derive the half-angle formula for cosine, we have cos2 θ = 1 + cos(2θ) _________ 2 α __  = cos2
 2 α __ 1 + cos  2 ⋅  2 2 __ = 1 + cos α ________ 2 _________ 1 + cos α ________ 2 α __ cos  2  = ± √ 590 CHAPTER 7 trigonometric identities and eQuations For the tangent identity, we have tan2 θ = 1 − cos(2θ) _________ 1 + cos(2θ) α __  = tan2  2 =  = ± √ α __ tan  2 α __ 1 − cos  2 ⋅  2 __ α __ 1 + cos  2 ⋅  2 1 − cos α ________ 1 + cos α _________ 1 − cos α _ 1 + cos α half-angle formulas The half-angle formulas are as follows:  = ± √ α __ sin  2  = ± √ α __ cos  2  = ± √ α __ tan  2 _________ 1 − cos α ________ 2 _________ 1 + cos α ________ 2 _________ 1 − cos α ________ 1 + cos α sin α ________ 1 + cos α 1 − cos α ________ sin α = = Example 7 Using a Half-Angle Formula to Find the Exact Value of a Sine Function Find sin(15°) using a half-angle formula. Solution Since 15° =, we use the half-angle formula for sine: 30° ___ 2 sin 30° ___ 2 = √ = √ = √ = √ √ = __________ 1 − cos30° _________ 2 ————— 3 √ _ 1 − 2 _ 2 ————— — — 3 2 − √ _______ 2 _ 2 ________ — 3 2 − √ _______ 4 _______ — 2 − √ 3 _ 2 Analysis Notice that we used only the positive root because sin(15°) is positive. How To… Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle. 1. Draw a triangle to represent the given information. 2. Determine the correct half-angle formula. 3. Substitute values into the formula based on the triangle. 4. Simplify.
SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 591 Example 8 Finding Exact Values Using Half-Angle Identities Given that tan α = and α lies in quadrant III, find the exact value of the following: 8 __ 15 α __ b. cos   2 α __ a. sin   2 α __ c. tan   2 Solution Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α = − and cos α = − 8 __ 17 15 __. 17 y α x (–15, 0) (–15, –8) 17 Figure 3 a. < α __ Before we start, we must remember that, if α is in quadrant III, then 180° < α < 270°, so 2 α __ is in quadrant II, since 90° < α __ α __ < 135°. To find sin, we begin by writing This means that the terminal side of 2 2 2 180° ____ 2 270° ____. 2 < the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3 and simplify. α _ sin 2 _________ 1 − cos α _ 2 ——————— 15 1 −  − _  17 __ 2 = ± √ = ± √ = ± √ = ± √ = ± √ ——— 32 _ 17 _ 2 ______ 32 1 _ _ · 2 17 ___ 16 _ 17 = ± 4 _ — 17 √ = — 17 4 √ _ 17 We choose the positive value of sin α __ because the angle terminates in quadrant II and sine is positive in 2 quadrant II. b. α __ To find cos, we will write the half-angle formula for cosine, substitute the value of the cosine we found from 2 the triangle in Figure 3, and simplify. 592 CHAPTER 7 trigonometric identities and eQuations __________ 1 + cos α ________ 2 ——————————— 15 ___  17 1 +  − __ 2 α __  = ± √ cos  2 = ± √ ——— 2 _ = ± √ 17 _ 2 ______ 1 2 __ __ = ± √
· 17 2 ___ 1 __ = ± √ 17 — 17 √ _ 17 = − α __ We choose the negative value of cos because the angle is in quadrant II because cosine is negative in 2 quadrant II. c. α __ To find tan, we write the half-angle formula for tangent. Again, we substitute the value of the cosine we 2 found from the triangle in Figure 3 and simplify. _________ 1 − cos α _ 1 + cos α ———————— 15 _  1 −  − 17 __ 15 _  1 +  − 17 α _ tan 2 = ± √ = ± √ = ± √ = ± √ = − √ —— 32 _ 17 _ 2 _ 17 ___ 32 _ 2 16 — α __ α __ We choose the negative value of tan lies in quadrant II, and tangent is negative in quadrant II. because 2 2 = −4 Try It #5 Given that sin α = − 4 α __ __ and α lies in quadrant IV, find the exact value of cos  . 2 5 Example 9 Finding the Measurement of a Half Angle Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tan θ = 5 __ for higher-level competition, what is the measurement of the angle for novice 3 competition? SECTION 7.3 douBle-angle, halF-angle, and reduction Formulas 593 Solution Since the angle for novice competition measures half the steepness of the angle for the high-level competition, and tan θ = 5 __ for high-level competition, we can find cos θ from the right triangle and the Pythagorean theorem so 3 that we can use the half-angle identities. See Figure 4. 32 + 52 = 34 c = √ — 34 34 5 θ 3 Figure 4 We see that cos θ = 3 _ — 34 √ = — 34 3 √ _ 34 θ _. We can use the half-angle formula for tangent: tan 2 = √ _________ 1 − cos θ _ 1 + cos θ. Since tan θ θ _ is in the first quadrant, so is tan. Thus, 2 θ
_ tan 2 —————— — 34 3 √ _ 1 − 34 __ 34 3 √ _ 1 + 34 — —————— — 34 34 − 3 √ __ 34 __ 34 + 3 √ __ 34 34 — ___________ — 34 − 3 √ 34 __________ — 34 + 3 √ 34 = √ = √ = √ ≈ 0.57 We can take the inverse tangent to find the angle: tan−1 (0.57) ≈ 29.7°. So the angle of the ramp for novice competition is ≈ 29.7°. Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas. • Double-Angle Identities (http://openstaxcollege.org/l/doubleangiden) • Half-Angle Identities (http://openstaxcollege.org/l/halfangleident) 594 CHAPTER 7 trigonometric identities and eQuations 7.3 SeCTIOn exeRCISeS VeRBAl 1. Explain how to determine the reduction identities 2. Explain how to determine the double-angle formula from the double-angle identity cos(2x) = cos2 x − sin2 x. for tan(2x) using the double-angle formulas for cos(2x) and sin(2x). 3. We can determine the half-angle formula for tan — — by dividing the formula for 1 − cos x  = ± √ x __ __  2 1 + cos x √ x x __ __  by cos  sin  . Explain how to determine two 2 2 x __ formulas for tan   that do not involve any square 2 roots. 4. For the half-angle formula given in the previous x _ exercise for tan  , explain why dividing by 0 is not a 2 concern. (Hint: examine the values of cos x necessary for the denominator to be 0.) AlGeBRAIC For the following exercises, find the exact values of a) sin(2x), b) cos(2x), and c) tan(2x) without solving for x. 5. If sin x = 1 __, and x is in quadrant I. 8 6. If cos x = 2 __, and x is in quadrant I. 3
7. If cos x = − 1 __, and x is in quadrant III. 2 8. If tan x = −8, and x is in quadrant IV. For the following exercises, find the values of the six trigonometric functions if the conditions provided hold. 9. cos(2θ) = 3 __ and 90° ≤ θ ≤ 180° 5 and 180° ≤ θ ≤ 270° 10. cos(2θ) = 1 _ — 2 √ For the following exercises, simplify to one trigonometric expression. π π __ __   cos  12. 4 sin  8 8 π π __ __   2 cos  11. 2 sin  4 4 For the following exercises, find the exact value using half-angle formulas. 14. cos  − 11π ___  12 π __  13. sin  8 15. sin  11π ___  12 16. cos  7π ___  8 17. tan  5π ___  12 18. tan  − 3π ___  12 19. tan  − 3π ___  8 x x x __ __ __ For the following exercises, find the exact values of a) sin  , and c) tan  , b) cos   without solving for x, when 2 2 2 0 ≤ x ≤ 360°. 4 __ 20. If tan x = −, and x is in quadrant IV. 3 21. If sin x = −, and x is in quadrant III. 12 __ 13 22. If csc x = 7, and x is in quadrant II. 23. If sec x = −4, and x is in quadrant II. SECTION 7.3 section exercises 595 For the following exercises, use Figure 5 to find the requested half and double angles. α 5 θ 12 Figure 5 24. Find sin(2θ), cos(2θ), and tan(2θ). θ θ θ . , and tan  , cos  26. Find sin  __ __ __ 2 2 2 25. Find sin(2α),
cos(2α), and tan(2α). α __ α __ α __ , and tan  , cos  27. Find sin  . 2 2 2 For the following exercises, simplify each expression. Do not evaluate. 28. cos2(28°) − sin2(28°) 31. cos2(9x) − sin2(9x) 29. 2 cos2(37°) − 1 32. 4 sin(8x) cos(8x) 30. 1 − 2 sin2(17°) 33. 6 sin(5x) cos(5x) For the following exercises, prove the identity given. 34. (sin t − cos t)2 = 1 − sin(2t) 35. sin(2x) = −2 sin(−x) cos(−x) 36. cot x − tan x = 2 cot(2x) 37. sin(2θ) _________ 1 + cos(2θ) tan2 θ = tan3 θ For the following exercises, rewrite the expression with an exponent no higher than 1. 38. cos2(5x) 42. cos2 x sin4 x 39. cos2(6x) 43. cos4 x sin2 x 40. sin4(8x) 44. tan2 x sin2 x 41. sin4(3x) TeCHnOlOGY For the following exercises, reduce the equations to powers of one, and then check the answer graphically. 45. tan4 x 46. sin2(2x) 49. tan4 x cos2 x 50. cos2 x sin(2x) 47. sin2 x cos2 x 51. cos2 (2x)sin x 48. tan2 x sin x x __ 52. tan2   sin x 2 For the following exercises, algebraically find an equivalent function, only in terms of sin x and/or cos x, and then check the answer by graphing both equations. 53. sin(4x) 54. cos(4x) exTenSIOnS For the following exercises, prove the identities. 55. sin(2x) = 2 tan x ________ 1 + tan2 x 57. tan(2x) = 2 sin x cos x _________ 2 cos2 x − 1 56. cos(2α)
= 1 − tan2 α ________ 1 + tan2 α 58. (sin2 x − 1)2 = cos(2x) + sin4 x 59. sin(3x) = 3 sin x cos2 x − sin3 x 60. cos(3x) = cos3 x − 3 sin2 x cos x 61. 1 + cos(2t) ___________ = sin(2t) − cos t 2 cos t ________ 2 sin t − 1 62. sin(16x) = 16 sin x cos x cos(2x)cos(4x)cos(8x) 63. cos(16x) = (cos2 (4x) − sin2 (4x) − sin(8x))(cos2 (4x) − sin2 (4x) + sin(8x)) 596 CHAPTER 7 trigonometric identities and eQuations leARnInG OBjeCTIVeS In this section, you will: • • Express products as sums. Express sums as products. 7.4 SUM-TO-PRODUCT AnD PRODUCT-TO-SUM FORMUlAS Figure 1 The UCLA marching band (credit: Eric Chan, Flickr). A band marches down the field creating an amazing sound that bolsters the crowd. That sound travels as a wave that can be interpreted using trigonometric functions. For example, Figure 2 represents a sound wave for the musical note A. In this section, we will investigate trigonometric identities that are the foundation of everyday phenomena such as sound waves. y 1 –1 0.002 0.004 0.006 0.008 0.01 x Figure 2 expressing Products as Sums We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity. Expressing Products as Sums for Cosine We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get: cos α cos β + sin α sin β = cos(α − β) + cos α cos β − sin α sin β = cos(α + β) 2 cos α cos β = cos(α − β
) + cos(α + β) Then, we divide by 2 to isolate the product of cosines: 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 SECTION 7.4 sum-to-product and product-to-sum Formulas 597 How To… Given a product of cosines, express as a sum. 1. Write the formula for the product of cosines. 2. Substitute the given angles into the formula. 3. Simplify. Example 1 Write the following product of cosines as a sum: 2 cos  Solution We begin by writing the formula for the product of cosines: Writing the Product as a Sum Using the Product-to-Sum Formula for Cosine 7x __  cos 2 3x __. 2 We can then substitute the given angles into the formula and simplify. 1 __ cos α cos β = [cos(α − β) + cos(α + β)] 2 2 cos  7x __  cos  2 1 3x __ __   cos   = (2)  2 2 7x __ 2 − 3x __  + cos  2 7x __ 2 + 3x __   2 =  cos  4x __  + cos  2 10x ___ 2   = cos 2x + cos 5x Try It #1 Use the product-to-sum formula to write the product as a sum or difference: cos(2θ)cos(4θ). Expressing the Product of Sine and Cosine as a Sum Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine. If we add the sum and difference identities, we get: sin(α + β) = sin α cos β + cos α sin β + sin(α − β) = sin α cos β − cos α sin β sin(α + β) + sin(α − β) = 2 sin α cos β Then, we divide by 2 to isolate the product of cosine and sine: 1 __ [sin(α +β) + sin(α − β)] sin α cos β = 2 Example 2 Writing the Product as a Sum Containing only Sine or Cosine Express
the following product as a sum containing only sine or cosine and no products: sin(4θ)cos(2θ). Solution Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify. 1 __ sin α cos β = [sin(α + β) + sin(α − β)] 2 1 __ [sin(4θ + 2θ) + sin(4θ − 2θ)] sin(4θ)cos(2θ) = 2 1 __ [sin(6θ) + sin(2θ)] = 2 Try It #2 Use the product-to-sum formula to write the product as a sum: sin(x + y)cos(x − y). 598 CHAPTER 7 trigonometric identities and eQuations Expressing Products of Sines in Terms of Cosine Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas: − cos(α − β) = cos α cos β + sin α sin β cos(α + β) = − (cos α cos β − sin α sin β) cos(α − β) − cos(α + β) = 2 sin α sin β Then, we divide by 2 to isolate the product of sines: 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas. the product-to-sum formulas The product-to-sum formulas are as follows: 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 1 __ [sin(α + β) − sin(α − β)] cos α sin β = 2 Example 3 Express the Product as a Sum or Difference Write cos(3θ) cos(5θ) as a sum or difference. Solution We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify. 1 __ cos α cos β =
[cos(α − β) + cos(α + β)] 2 1 __ [cos(3θ − 5θ) + cos(3θ + 5θ)] cos(3θ)cos(5θ) = 2 1 __ [cos(2θ) + cos(8θ)] = 2 Use even-odd identity. Try It #3 Use the product-to-sum formula to evaluate cos 11π ___ 12 cos π __. 12 expressing Sums as Products Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine. Let u + v _____ 2 = α and u − v _____ 2 = β. Then, α + β = + u − v _____ 2 α − β = − u − v _____ 2 u + v _____ 2 2u __ 2 = u = u + v _____ 2 2v __ 2 = v = SECTION 7.4 sum-to-product and product-to-sum Formulas 599 Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 sin  2 sin  u + v _____ 2 u + v _____ 2  cos   cos  u − v _____ 2 u − v _____ 2 1 __ [sin u + sin v]  = 2  = sin u + sin v Substitute for (α + β) and (α − β) The other sum-to-product identities are derived similarly. sum-to-product formulas The sum-to-product formulas are as follows: sin α + sin β = 2 sin  α + β _____ 2  cos  α − β  _____ 2 cos α − cos β = −2 sin  α + β _____ 2  sin  α − β  _____ 2 sin α − sin β = 2 sin  α − β _____ 2  cos �
� α + β  _____ 2 cos α + cos β = 2 cos  α + β _____ 2  cos  α − β  _____ 2 Example 4 Writing the Difference of Sines as a Product Write the following difference of sines expression as a product: sin(4θ) − sin(2θ). Solution We begin by writing the formula for the difference of sines. sin α − sin β = 2 sin  α − β _____ 2  cos  α + β  _____ 2 Substitute the values into the formula, and simplify. sin(4θ) − sin(2θ) = 2 sin  4θ − 2θ ______ 2  cos  4θ + 2θ  ______ 2 2θ  cos  __ 2 = 2 sin  = 2 sin θ cos(3θ) 6θ  __ 2 Try It #4 Use the sum-to-product formula to write the sum as a product: sin(3θ) + sin(θ). Example 5 Evaluating Using the Sum-to-Product Formula Evaluate cos(15°) − cos(75°). Solution We begin by writing the formula for the difference of cosines. α + β _____ 2 cos α − cos β = −2 sin   sin  α − β  _____ 2 Then we substitute the given angles and simplify. cos(15°) − cos(75°) = −2 sin  15° + 75° ________ 2  sin  15° − 75° ________  2 — = −2 sin(45°)sin(−30°) 2   − = −2  1 √ __ ___  2 2 2 √ ___ 2 = — 600 CHAPTER 7 trigonometric identities and eQuations Example 6 Proving an Identity Prove the identity: cos(4t) − cos(2t) _____________ sin(4t) + sin(2t) = −tan t Solution it matches the right side. We will start with the left side, the more complicated side of the equation, and rewrite the expression until cos(4t)
− cos(2t) _____________ sin(4t) + sin(2t) = −2 sin  ____  cos  2 sin  4t + 2t ______ 2 4t + 2t ______ 2 4t − 2t ______  2 4t − 2t ______  2  sin  = −2 sin(3t)sin t ___________ 2 sin(3t)cos t = −2 sin(3t)sin t ___________ 2 sin(3t)cos t = − sin t ____ cos t = −tan t Analysis Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side. Example 7 Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities Verify the identity csc2 θ − 2 = cos(2θ) ______. sin2 θ Solution For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side. cos(2θ) ______ = sin2 θ 1 − 2 sin2 θ _________ sin2 θ = 1 ____ sin2 θ − 2 sin2 θ ______ sin2 θ = csc2 θ − 2 Try It #5 Verify the identity tan θ cot θ − cos2θ = sin2θ. Access these online resources for additional instruction and practice with the product-to-sum and sum-to-product identities. • Sum to Product Identities (http://openstaxcollege.org/l/sumtoprod) • Sum to Product and Product to Sum Identities (http://openstaxcollege.org/l/sumtpptsum) SECTION 7.4 section exercises 601 7.4 SeCTIOn exeRCISeS VeRBAl 1. Starting with the product to sum formula sin α cos β = 1 __ [sin(α + β) + sin(α − β)], explain 2 how to determine the formula for cos α sin β. 3. Explain a situation where
we would convert an equation from a sum to a product and give an example. 2. Explain two different methods of calculating cos(195°)cos(105°), one of which uses the product to sum. Which method is easier? 4. Explain a situation where we would convert an equation from a product to a sum, and give an example. AlGeBRAIC For the following exercises, rewrite the product as a sum or difference. 5. 16sin(16x)sin(11x) 6. 20cos(36t)cos(6t) 7. 2sin(5x)cos(3x) 8. 10cos(5x)sin(10x) 9. sin(−x)sin(5x) 10. sin(3x)cos(5x) For the following exercises, rewrite the sum or difference as a product. 11. cos(6t) + cos(4t) 12. sin(3x) + sin(7x) 13. cos(7x) + cos(−7x) 14. sin(3x) − sin(−3x) 15. cos(3x) + cos(9x) 16. sin h − sin(3h) For the following exercises, evaluate the product using a sum or difference of two functions. Evaluate exactly. 17. cos(45°)cos(15°) 18. cos(45°)sin(15°) 19. sin(−345°)sin(−15°) 20. sin(195°)cos(15°) 21. sin(−45°)sin(−15°) For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine. 22. cos(23°)sin(17°) 23. 2sin(100°)sin(20°) 24. 2sin(−100°)sin(−20°) 25. sin(213°)cos(8°) 26. 2cos(56°)cos(47°) For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine. 27. sin(76°) + sin(14°) 28. cos(58°) − cos(12°) 29. sin(101°) − sin(32°) 30. cos(100°) + cos(200°)
31. sin(−1°) + sin(−2°) For the following exercises, prove the identity. 32. cos(a + b) ________ = cos(a − b) 1 − tan a tan b ___________ 1 + tan a tan b 34. 6cos(8x)sin(2x) ____________ sin(−6x) = −3 sin(10x)csc(6x) + 3 33. 4sin(3x)cos(4x) = 2 sin(7x) − 2 sinx 35. sin x + sin(3x) = 4sin x cos2 x 36. 2(cos3 x − cos x sin2 x)= cos(3x) + cos x 37. 2 tan x cos(3x) = sec x(sin(4x) − sin(2x)) 38. cos(a + b) + cos(a − b) = 2cos a cos b 602 CHAPTER 7 trigonometric identities and eQuations nUMeRIC For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions. Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 39. cos(58°) + cos(12°) 40. sin(2°) − sin(3°) 41. cos(44°) − cos(22°) 42. cos(176°)sin(9°) 43. sin(−14°)sin(85°) TeCHnOlOGY For the following exercises, algebraically determine whether each of the given expressions is a true identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator. 44. 2sin(2x)sin(3x) = cos x − cos(5x) 46. sin(3x) − sin(5x) _____________ cos(3x) + cos(5x) = tan x 45. cos(10θ) + cos(6θ) _______________ cos(6θ) − cos(10θ) = cot(2θ)cot(8θ) 47. 2cos(2x)cos x + sin(2x)sin x = 2 sin
x 48. sin(2x) + sin(4x) _____________ sin(2x) − sin(4x) = −tan(3x)cot x For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical. 49. sin(9t) − sin(3t) _____________ cos(9t) + cos(3t) 51. sin(3x) − sin x ____________ sin x 53. sin x cos(15x) − cos x sin(15x) exTenSIOnS 50. 2sin(8x)cos(6x) − sin(2x) 52. cos(5x) + cos(3x) ______________ sin(5x) + sin(3x) For the following exercises, prove the following sum-to-product formulas. 54. sin x − sin y = 2 sin   cos  55. cos x + cos y = 2cos   cos  x − y  _ 2 For the following exercises, prove the identity. 56. sin(6x) + sin(4x) _____________ sin(6x) − sin(4x) = tan (5x)cot x 58. cos(6y) + cos(8y) ______________ sin(6y) − sin(4y) = cot y cos(7y) sec(5y) 60. sin(10x) − sin(2x) ______________ cos(10x) + cos(2x) = tan(4x) 57. cos(3x) + cos x ____________ cos(3x) − cos x = −cot (2x)cot x 59. cos(2y) − cos(4y) ______________ sin(2y) + sin(4y) = tan y 61. cos x − cos(3x) = 4 sin2 x cos x 62. (cos(2x) − cos(4x))2 + (sin(4x) + sin(2x))2 = 4 sin2(3x) π __ − t  = 63. tan  4 1 − tan t _ 1 + tan t SECTION 7.5 solving trigonometric eQuations 603 leARnIn
G OBjeCTIVeS In this section, you will: • • • • • • • Solve linear trigonometric equations in sine and cosine. Solve equations involving a single trigonometric function. Solve trigonometric equations using a calculator. Solve trigonometric equations that are quadratic in form. Solve trigonometric equations using fundamental identities. Solve trigonometric equations with multiple angles. Solve right triangle problems. 7. 5 SOlVInG TRIGOnOMeTRIC eQUATIOnS Figure 1 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill) Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids. Solving linear Trigonometric equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2π. In other words, every 2π units, the y-values repeat. If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution. Recall the rule that gives the format
for stating all possible solutions for a function where the period is 2π: sin θ = sin(θ ± 2kπ) There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. 604 CHAPTER 7 trigonometric identities and eQuations Solving a Linear Trigonometric Equation Involving the Cosine Function Example 1 Find all possible exact solutions for the equation cos θ = 1 __. 2 Solution From the unit circle, we know that cos θ = 1 __ 2 π __ θ =, 3 5π ___ 3 These are the solutions in the interval [0, 2π]. All possible solutions are given by where k is an integer. π __ ± 2kπ and θ = θ = 3 5π ___ ± 2kπ 3 Solving a Linear Equation Involving the Sine Function Example 2 Find all possible exact solutions for the equation sin t = 1 __. 2 Solution Solving for all possible values of t means that solutions include angles beyond the period of 2π. From Section π __ and t = 7.2 Figure 2, we can see that the solutions are t = 6 5π ___. But the problem is asking for all possible values that 6 solve the equation. Therefore, the answer is where k is an integer. π __ ± 2πk and t = 6 t = 5π ___ ± 2πk 6 How To… Given a trigonometric equation, solve using algebra. 1. Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. 2. Substitute the trigonometric expression with a single variable, such as x or u. 3. Solve the equation the same way an algebraic equation would be solved. 4. Substitute the trigonometric expression back in for the variable in the resulting expressions. 5. Solve for the angle. Example 3 Solve the Trigonometric Equation in Linear Form Solve the equation exactly: 2cos θ − 3 = − 5, 0 ≤
θ < 2π. Solution Use algebraic techniques to solve the equation. 2cos θ − 3 = −5 2cos θ = −2 cos θ = −1 θ = π Try It #1 Solve exactly the following linear equation on the interval [0, 2π): 2sin x + 1 = 0. Solving equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Section 7.2 Figure 2). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π, not 2π. π __ Further, the domain of tangent is all real numbers with the exception of odd integer multiples of, unless, of course, 2 a problem places its own restrictions on the domain. SECTION 7.5 solving trigonometric eQuations 605 Example 4 Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: 2sin2 θ − 1 = 0, 0 ≤ θ < 2π. Solution As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ. Then we will find the angles. 2sin2 θ − 1 = 0 √ — 2sin2 θ = 1 sin2 θ = 1 __ 2 sin2 θ = ± √ sin θ = ± 1 _ = ± 2 √ 3π ___, 4 π __ θ =, 4 __ 1 __ 2 5π ___, 4 — — 2 √ ____ 2 7π ___ 4 Example 5 Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: csc θ = −2, 0 ≤ θ < 4π. Solution We want all values of θ for which csc θ = −2 over the interval 0 ≤ θ < 4π. csc θ = −2 = −2
1 ____ sin θ sin θ = − 1 __ 2 θ = 7π ___, 6 23π 19π 11π ___ ___ ___,, 6 6 6 Analysis As sin θ = − 1 _, notice that all four solutions are in the third and fourth quadrants. 2 Example 6 Solving an Equation Involving Tangent π __  = 1, 0 ≤ θ < 2π. Solve the equation exactly: tan  θ − 2 π __ Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of, the tangent has a 4 π π __ __  = 1, then value of 1. However, the angle we want is  θ − . Thus, if tan  4 2 π π __ __ = θ − 4 2 Over the interval [0, 2π), we have two solutions: θ = 3π ___ ± kπ 4 θ = 3π ___ and θ = 4 3π ___ 4 + π = 7π ___ 4 Try It #2 Find all solutions for tan x = √ — 3. Example 7 Identify all Solutions to the Equation Involving Tangent Identify all exact solutions to the equation 2(tan x + 3) = 5 + tan x, 0 ≤ x < 2π. 606 CHAPTER 7 trigonometric identities and eQuations Solution We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign. 2(tan x) + 2(3) = 5 + tan x 2tan x + 6 = 5 + tan x 2tan x − tan x = 5 − 6 tan x = −1 There are two angles on the unit circle that have a tangent value of −1: θ = and θ = 3π __ 4 7π __. 4 Solve Trigonometric equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example 8 Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation sin
θ = 0.8, where θ is in radians. Solution Make sure mode is set to radians. To find θ, use the inverse sine function. On most calculators, you will need to push the 2ND button and then the SIN button to bring up the sin−1 function. What is shown on the screen is sin−1(. The calculator is ready for the input within the parentheses. For this problem, we enter sin−1 (0.8), and press ENTER. Thus, to four decimals places, The solution is The angle measurement in degrees is sin−1(0.8) ≈ 0.9273 θ ≈ 0.9273 ± 2πk θ ≈ 53.1° θ ≈ 180° − 53.1° ≈ 126.9° Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ. Example 9 Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation sec θ = −4, giving your answer in radians. Solution We can begin with some algebra. sec θ = −4 = −4 1 ____ cos θ cos θ = − 1 __ 4 Check that the MODE is in radians. Now use the inverse cosine function. cos−1  − 1 __  ≈ 1.8235 4 θ ≈ 1.8235 + 2πk π __ ≈ 1.57 and π ≈ 3.14, 1.8235 is between these two numbers, thus θ ≈ 1.8235 is in quadrant II. Since 2 Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 2. SECTION 7.5 solving trigonometric eQuations 607 y θ ≈ 1.8235 x θʹ ≈ π – 1.8235 ≈ 1.3181 θʹ ≈ π + 1.3181 ≈ 4.4597 Figure 2 So, we also need to find the measure of the angle in quadrant III.
In quadrant III, the reference angle is θ´≈ π − 1.8235 ≈ 1.3181. The other solution in quadrant III is θ´ ≈ π + 1.3181 ≈ 4.4597. The solutions are θ ≈ 1.8235 ± 2πk and θ ≈ 4.4597 ± 2πk. Try It #3 Solve cos θ = −0.2. Solving Trigonometric equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example 10 Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: cos2 θ + 3 cos θ − 1 = 0, 0 ≤ θ < 2π. Solution We begin by using substitution and replacing cos θ with x. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x. We have The equation cannot be factored, so we will use the quadratic formula x = x2 + 3x − 1 = 0 — b2 − 4ac −b ± √ _____________ 2a ——. (−3)2 − 4(1)(−1) −3 ± √ ______________________ x = Replace x with cos θ, and solve. Thus, Note that only the + sign is used. This is because we get an error when we solve θ = cos−1  calculator, since the domain of the inverse cosine function is [−1, 1]. However, there is a second solution: −3 − √ _________ 2 13  on a — 2 — = −3 ± √ _________ 2 13 cos θ = — 13 −3 ± √ _________
2 θ = cos−1  — 13 −3 + √  _________ 2 — 13 −3 + √  _________ 2 θ = cos−1  ≈ 1.26 608 CHAPTER 7 trigonometric identities and eQuations This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is θ = 2π − cos−1  ≈ 5.02 — 13 −3 + √  _________ 2 Try It #4 Solve sin2 θ = 2 cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 11 Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: 2 sin2 θ − 5 sin θ + 3 = 0, 0 ≤ θ ≤ 2π. Solution Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u, or imagine it, as we factor: Now set each factor equal to zero. 2sin2 θ − 5sin θ + 3 = 0 (2sin θ − 3)(sin θ − 1) = 0 2sin θ − 3 = 0 2sin θ = 3 sin θ = 3 __ 2 sin θ − 1 = 0 sin θ = 1 Next solve for θ: sin θ ≠ 3 π __ __, as the range of the sine function is [−1, 1]. However, sin θ = 1, giving the solution θ =. 2 2 Analysis Make sure to check all solutions on the given domain as some factors have no solution. Try It #5 Solve sin2 θ = 2cos θ + 2, 0 ≤ θ ≤ 2π. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 12 Solving a Trigonometric Equation Using Algebra Solve exactly: 2sin2 θ + sin θ = 0; 0 ≤ θ < 2π Solution This problem should appear familiar as it is similar to a quadratic. Let sin θ = x. The equation becomes 2x2 + x = 0. We begin by factoring: Set each factor equal to zero.
2x2 + x = 0 x(2x + 1) = 0 x = 0 (2x + 1) = 0 1 __ x = − 2 Then, substitute back into the equation the original expression sin θ for x. Thus, sin θ = 0 The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, θ = 0, π sin θ = − 1 __ 2 11π 7π ___ ___, 6 6 11π ___. 6 7π ___, 6 θ = If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. SECTION 7.5 solving trigonometric eQuations 609 2sin2 θ + sin θ = 0 sin θ(2sin θ + 1) = 0 sin θ = 0 θ = 0, π 2sin θ + 1 = 0 2sin θ = −1 sin θ = − 1 __ 2 θ = 7π ___, 6 11π ___ 6 Analysis We can see the solutions on the graph in Figure 3. On the interval 0 ≤ θ < 2π, the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value. f(θ) = 2 sin2 θ + sin 1 π 6 π 3 π 2 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 13π 6 θ Figure 3 We can verify the solutions on the unit circle in Section 7.2 Figure 2 as well. Example 13 Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: 2sin2 θ − 3sin θ + 1 = 0, 0 ≤ θ < 2π. Solution We can factor using grouping. Solution values of θ can be found on the unit circle: (2sin θ − 1)(sin θ − 1) = 0 2sin θ − 1 = 0 sin θ = 1 __ 2 5π ___ 6 π __ θ =,
6 sin θ = 1 π __ θ = 2 Try It #6 Solve the quadratic equation 2cos2 θ + cos θ = 0. Solving Trigonometric equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation. 610 CHAPTER 7 trigonometric identities and eQuations Example 14 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2π. Solution Notice that the left side of the equation is the difference formula for cosine. cos x cos(2x) + sin x sin(2x) = cos x cos(2x) + sin x sin(2x) = — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 — 3 √ ____ 2 cos(x − 2x) = cos(−x) = cos x = Difference formula for cosine Use the negative angle identity. From the unit circle in Figure 2, we see that cos x = — 3 √ ____ 2 π __ when x =, 6 11π ___. 6 Example 15 Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: cos(2θ) = cos θ. Solution We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: cos(2θ) = cos θ 2cos2 θ − 1 = cos θ 2cos2 θ − cos θ − 1 = 0 (2cos θ + 1)(cos θ − 1) = 0 2cos θ + 1 = 0 cos θ = − 1 __ 2 cos θ − 1 = 0 cos θ = 1 So, if cos θ = − 1 __, then θ = 2 2π __ 3 ± 2πk and θ = 4π
__ ± 2πk; if cos θ = 1, then θ = 0 ± 2πk. 3 Example 16 Solving an Equation Using an Identity Solve the equation exactly using an identity: 3cos θ + 3 = 2sin2 θ, 0 ≤ θ < 2π. Solution If we rewrite the right side, we can write the equation in terms of cosine: 3cos θ + 3 = 2sin2 θ 3cos θ + 3 = 2(1 − cos2 θ) 3cos θ + 3 = 2 − 2cos2 θ 2cos2 θ + 3cos θ + 1 = 0 (2cos θ + 1)(cos θ + 1) = 0 2cos θ + 1 = 0 cos θ = − 1 __ 2 2π ___, 3 cos θ + 1 = 0 θ = 4π ___ 3 cos θ = −1 θ = π Our solutions are θ = 2π ___, 3 4π ___ 3, π SECTION 7.5 solving trigonometric eQuations 611 Solving Trigonometric equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin(2x) or cos(3x). When confronted with these equations, recall that y = sin(2x) is a horizontal compression by a factor of 2 of the function y = sin x. On an interval of 2π, we can graph two periods of y = sin(2x), as opposed to one cycle of y = sin x. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin(2x) = 0 compared to sin x = 0. This information will help us solve the equation. Example 17 Solving a Multiple Angle Trigonometric Equation Solve exactly: cos(2x) = 1 __ on [0, 2π). 2 Solution We can see that this equation is the standard equation with a multiple of an angle. If cos(α) = 1 __, we know 2 α is in quadrants I and IV. While θ = cos−1 1 __ will only yield solutions in quadrants I and II, we recognize that the 2 solutions to the equation cos θ = 1 __ will be in quadrants I and IV. 2 π 5π π __ __ __. So
, 2x = or 2x = and θ = Therefore, the possible angles are __ __ __   = cos  Does this make sense? Yes, because cos  2 . 2 3 6 5π __ 3 π __, which means that x = or x = 6 5π __. 6 Are there any other possible answers? Let us return to our first step. π π __ __ In quadrant I, 2x =, so x = as noted. Let us revolve around the circle again: 6 3 so x = 7π ___. 6 One more rotation yields π __ + 2π 2x = 3 π __ + = 3 6π ___ 3 = 7π ___ 3 π __ + 4π 2x = 3 π __ + = 3 12π ___ 3 = 13π ___ 3 x = 13π ___ 6 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). In quadrant IV, 2x =, so x = as noted. Let us revolve around the circle again: 5π __ 3 5π __ 6 2x = + 2π 5π ___ 3 = 5π ___ 3 + 6π ___ 3 so x = 11π ___. 6 = 11π ___ 3 612 CHAPTER 7 trigonometric identities and eQuations One more rotation yields 2x = + 4π 5π ___ 3 = 5π ___ 3 + 12π ___ 3 = 17π ___ 3 > 2π, so this value for x is larger than 2π, so it is not a solution on [0, 2π). x = 17π ___ 6 π __ Our solutions are x =, 6 5π __, 6 7π __ 6, and 11π ___ 6 must go around the unit circle n times.. Note that whenever we solve a problem in the form of sin(nx) = c, we Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a2 + b2 = c2, and model an equation to fit a situation. Example 18 Using the Pythagorean Theorem to Model an Equation Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits
the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4. 69.5 θ 23 Figure 4 Solution Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem. a2 + b2 = c2 (23)2 + (69.5)2 ≈ 5359 — 5359 ≈ 73.2 m √ The angle of elevation is θ, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places. tan θ = 69.5 ____ 23 The angle of elevation is approximately 71.7°, and the length of the cable is 73.2 meters. tan−1  69.5 ____ 23  ≈ 1.2522 ≈ 71.69° SECTION 7.5 solving trigonometric eQuations 613 Example 19 Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. Solution For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “a” feet from the wall, the length of the ladder will be 4a feet. See Figure 5. 4a b θ a Figure 5 The side adjacent to θ is a and the hypotenuse is 4a. Thus, cos θ = a __ 4a = 1 __ 4 The elevation of the ladder forms an angle of 75.5° with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: 1 __  ≈ 75.5° cos−1  4 a2 + b2 = (4
a)2 b2 = (4a)2 − a2 b2 = 16a2 − a2 b2 = 15a2 Thus, the ladder touches the wall at a √ 15 15 feet from the ground. b = a √ — — Access these online resources for additional instruction and practice with solving trigonometric equations. • Solving Trigonometric equations I (http://openstaxcollege.org/l/solvetrigeqI) • Solving Trigonometric equations II (http://openstaxcollege.org/l/solvetrigeqII) • Solving Trigonometric equations III (http://openstaxcollege.org/l/solvetrigeqIII) • Solving Trigonometric equations IV (http://openstaxcollege.org/l/solvetrigeqIV) • Solving Trigonometric equations V (http://openstaxcollege.org/l/solvetrigeqV) • Solving Trigonometric equations VI (http://openstaxcollege.org/l/solvetrigeqVI) 614 CHAPTER 7 trigonometric identities and eQuations 7.5 SeCTIOn exeRCISeS VeRBAl 1. Will there always be solutions to trigonometric 2. When solving a trigonometric equation involving function equations? If not, describe an equation that would not have a solution. Explain why or why not. more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not? 3. When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions? AlGeBRAIC For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2π. 4. 2sin θ = − √ 7. 2cos θ = − √ 10. cot. 2sin θ = √ 8. tan θ = −1 — 3 11. 4sin2 x − 2 = 0 For the following exercises, solve exactly on [0, 2π). 13. 2cos θ = √ — 2 16. 2sin θ = − √ — 3 19. 2cos(3θ) = − √ — 2 π __ θ  = √ 22. 2cos �
� 5 — 3 14. 2cos θ = −1 17. 2sin(3θ) = 1 3 20. cos(2θ) = − √ ____ 2 — 6. 2cos θ = 1 9. tan x = 1 12. csc2 x − 4 = 0 15. 2sin θ = −1 18. 2sin(2θ) = √ — 3 21. 2sin(πθ) = 1 For the following exercises, find all exact solutions on [0, 2π). 23. sec(x)sin(x) − 2sin(x) = 0 24. tan(x) − 2sin(x)tan(x) = 0 25. 2cos2 t + cos(t) = 1 27. 2sin(x)cos(x) − sin(x) + 2cos(x) − 1 = 0 29. sec2 x = 1 26. 2tan2(t) = 3sec(t) 28. cos2 θ = 1 __ 2 30. tan2 (x) = −1 + 2tan(−x) 31. 8sin2(x) + 6sin(x) + 1 = 0 32. tan5(x) = tan(x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π). 33. sin(3x)cos(6x) − cos(3x)sin(6x) = −0.9 34. sin(6x)cos(11x) − cos(6x)sin(11x) = −0.1 35. cos(2x)cos x + sin(2x)sin x = 1 36. 6sin(2t) + 9sin t = 0 37. 9cos(2θ) = 9cos2 θ − 4 39. cos(2t) = sin t 38. sin(2t) = cos t 40. cos(6x) − cos(3x) = 0 For the following exercises, solve exactly on the interval [0, 2π). Use the quadratic formula if the equations do not factor. 41. tan2 x − √ — 3 tan x = 0 42. sin2 x + sin x − 2 = 0 43. sin2 x − 2sin x − 4 = 0 44. 5cos2 x + 3cos x −
1 = 0 45. 3cos2 x − 2cos x − 2 = 0 46. 5sin2 x + 2sin x − 1 = 0 47. tan2 x + 5tan x − 1 = 0 48. cot2 x = −cot x 49. −tan2 x − tan x − 2 = 0 SECTION 7.5 section exercises 615 For the following exercises, find exact solutions on the interval [0, 2π). Look for opportunities to use trigonometric identities. 50. sin2 x − cos2 x − sin x = 0 51. sin2 x + cos2 x = 0 52. sin(2x) − sin x = 0 53. cos(2x) − cos x = 0 54. 2 tan x ________ 2 − sec2 x − sin2 x = cos2 x 55. 1 − cos(2x) = 1 + cos(2x) 56. sec2 x = 7 57. 10sin x cos x = 6cos x 58. −3sin t = 15cos t sin t 59. 4cos2 x − 4 = 15cos x 60. 8sin2 x + 6sin x + 1 = 0 61. 8cos2 θ = 3 − 2cos θ 62. 6cos2 x + 7sin x − 8 = 0 63. 12sin2 t + cos t − 6 = 0 64. tan x = 3sin x 65. cos3 t = cos t GRAPHICAl For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros. 66. 6sin2 x − 5sin x + 1 = 0 68. 100tan2 x + 20tan x − 3 = 0 67. 8cos2 x − 2cos x − 1 = 0 69. 2cos2 x − cos x + 15 = 0 70. 20sin2 x − 27sin x + 7 = 0 71. 2tan2 x + 7tan x + 6 = 0 72. 130tan2 x + 69tan x − 130 = 0 TeCHnOlOGY For the following exercises, use a calculator to find all solutions to four decimal places. 73. sin x = 0.27 74. sin x = −0.55 75. tan x = −0.34 76. cos x = 0.71 For the following exercises, solve the equations algebraically, and then use a calculator to find the
values on the interval [0, 2π). Round to four decimal places. 77. tan2 x + 3tan x − 3 = 0 78. 6tan2 x + 13tan x = −6 79. tan2 x − sec x = 1 80. sin2 x − 2cos2 x = 0 81. 2tan2 x + 9tan x − 6 = 0 82. 4sin2 x + sin(2x)sec x − 3 = 0 exTenSIOnS For the following exercises, find all solutions exactly to the equations on the interval [0, 2π). 83. csc2 x − 3csc x − 4 = 0 84. sin2 x − cos2 x − 1 = 0 85. sin2 x( 1 − sin2 x) + cos2 x( 1 − sin2 x) = 0 86. 3sec2 x + 2 + sin2 x − tan2 x + cos2 x = 0 87. sin2 x − 1 + 2 cos(2x) − cos2 x = 1 88. tan2 x − 1 − sec3 x cos x = 0 89. sin(2x) ______ sec2 x = 0 91. 2cos2 x − sin2 x − cos x − 5 = 0 90. sin(2x) ______ = 0 2 csc2 x 92. 1 ____ sec2 x + 2 + sin2 x + 4cos2 x = 4 616 CHAPTER 7 trigonometric identities and eQuations ReAl-WORlD APPlICATIOnS 93. An airplane has only enough gas to fly to a city 200 miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many degrees north of east should the airplane fly? 95. If a loading ramp is placed next to a truck, at a height of 2 feet, and the ramp is 20 feet long, what angle does the ramp make with the ground? 97. An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing 2 miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.) 99. A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him? 101. A 90-foot tall building has a shadow that is 2
feet long. What is the angle of elevation of the sun? 103. A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light? 94. If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground? 96. A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal? 98. A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building? 100. A 20-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun? 102. A spotlight on the ground 3 meters from a 2-meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light? For the following exercises, find a solution to the word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree. 104. A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall? 105. A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall? 106. A 23-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping? SECTION 7.6 modeling with trigonometric eQuations 617 leARnInG OBjeCTIVeS In this section, you will: • • • • Determine the amplitude and period of sinusoidal functions. Model equations and graph sinusoidal functions. Model periodic behavior. Model harmonic motion functions. 7. 6 MODelInG WITH TRIGOnOMeTRIC eQUATIOnS Figure 1 The hands on a clock are periodic: they repeat positions every twelve hours. (credit: “zoutedrop”/Flickr) Suppose we charted the average daily temperatures in
New York City over the course of one year. We would expect to find the lowest temperatures in January and February and highest in July and August. This familiar cycle repeats year after year, and if we were to extend the graph over multiple years, it would resemble a periodic function. Many other natural phenomena are also periodic. For example, the phases of the moon have a period of approximately 28 days, and birds know to fly south at about the same time each year. So how can we model an equation to reflect periodic behavior? First, we must collect and record data. We then find a function that resembles an observed pattern. Finally, we make the necessary alterations to the function to get a model that is dependable. In this section, we will take a deeper look at specific types of periodic behavior and model equations to fit data. Determining the Amplitude and Period of a Sinusoidal Function Any motion that repeats itself in a fixed time period is considered periodic motion and can be modeled by a sinusoidal function. The amplitude of a sinusoidal function is the distance from the midline to the maximum value, or from the midline to the minimum value. The midline is the average value. Sinusoidal functions oscillate above and below the midline, are periodic, and repeat values in set cycles. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is 2π. In other words, for any value of x, sin(x ± 2πk) = sin x and cos(x ± 2πk) = cos x where k is an integer standard form of sinusoidal equations The general forms of a sinusoidal equation are given as y = Asin(Bt − C) + D or y = Acos(Bt − C) + D where amplitude = ∣ A ∣, B is related to period such that the period = denotes the horizontal shift, and D represents the vertical shift from the graph’s parent graph., C is the phase shift such that 2π _ B C _ B Note that the models are sometimes written as y = a sin(ω t ± C) + D or y = a cos(ω t ± C) + D, and period is given as 2π _ ω. The difference between the sine and the cosine graphs is that the sine graph begins with the average value of the function and the cosine graph begins with the
maximum or minimum value of the function. 618 CHAPTER 7 trigonometric identities and eQuations Showing How the Properties of a Trigonometric Function Can Transform a Graph Example 1 π _ Show the transformation of the graph of y = sin x into the graph of y = 2sin  4x −  + 2. 2 Solution Consider the series of graphs in Figure 2 and the way each change to the equation changes the image. y y y – 5π 2 2π – – 3π 2 – π – y = sin x π 2 π 3π 2 2π 5π 2 4 3 2 1 π –1 2 –2 –3 –4 (a) – 5π 2 2π – – 3π 1 2 –2 –3 –4 (d) – 5π 2 2π – – 3π 2 –3 –4 (b) y = 2sin x π 2 π 3π 2 2π 5π 2 x – 5π 2 2π – – 3π 2 π – – y = 2sin (4x) π 2 π 3π 2 2π 5π 2 x 4 3 2 1 π 2 –2 –3 –4 (c) π 2 π 3π 2 2π 5π 2 y = 2sin 4x x π 2 – 5π 2 2π – – 3π 2 π – – π 2 π 3π 2 2π 5π 2 x y = 2sin 4x 1 2 –2 –3 –4 (e) Figure 2 (a) The basic graph of y = sinx (b) Changing the amplitude from 1 to 2 generates the graph of y = 2sinx. (c) The period of the sine function changes 2π ___ B. Here we have B = 4, which translates to a period of π _ 2. The graph completes one full cycle in π _ 2 units. (d) The graph. (e) Finally, the graph is shifted vertically by the value of D. In this case, the graph is shifted up by 2 units. with the value of B, such that period = C __ displays a horizontal shift equal to B π __ = π _ 2 __, or 4 8 Example 2 Finding the Amplitude and Period of a Function Find the amplitude and period of the following functions and graph one cycle. 1 _ x  a. y = 2sin
 4 π _ b. y = −3sin  2x +  2 c. y = cos x + 3 Solution We will solve these problems according to the models. 1 _ x  involves sine, so we use the form a. y = 2sin  4 y = Asin(Bt − C) + D We know that ∣ A ∣ is the amplitude, so the amplitude is 2. Period is, so the period is 2π _ B 2π _ B = 2π _ 1 __ 4 = 8π See the graph in Figure 3. y 2 1 –1 –2 y = 2 sin 4 1 x Amplitude = 2 2π 4π 6π 8π x Period = 8π Figure 3 SECTION 7.6 modeling with trigonometric eQuations 619 π _ b. y = −3sin  2x +  involves sine, so we use the form 2 y = Asin(Bt − C) + D Amplitude is ∣ A ∣, so the amplitude is ∣ −3 ∣ = 3. Since A is negative, the graph is reflected over the x-axis. Period is, so the period is 2π _ B 2π _ B = 2π _ 2 = π The graph is shifted to the left by π _ π 2 _ _ = units. See Figure 41 –2 – π 4 y = –3 sin + π 2 2x π 4 π 2 3π 4 x Figure 4 c. y = cos x + 3 involves cosine, so we use the form y = Acos(Bt − C) + D Amplitude is ∣ A ∣, so the amplitude is 1. The period is 2π. See Figure 5. This is the standard cosine function shifted up three units. y 4 2 1 Midline: y = 3 y = cos x + 3 π 2 π 3π 2 Figure 5 x 2π Try It #1 What are the amplitude and period of the function y = 3cos(3πx)? Finding equations and Graphing Sinusoidal Functions One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of 1 _ of the period. The key points will indicate the location of maximum and minimum values. equal length representing 4 If there is no vertical shift, they will
also indicate x-intercepts. For example, suppose we want to graph the function y = cos θ. We know that the period is 2π, so we find the interval between key points as follows. 2π _ 4 π _ = 2 π _ Starting with θ = 0, we calculate the first y-value, add the length of the interval to 0, and calculate the second 2 π _ repeatedly until the five key points are determined. The last value should equal the first value, y-value. We then add 2 as the calculations cover one full period. Making a table similar to Table 1, we can see these key points clearly on the graph shown in Figure 6. 620 CHAPTER 7 trigonometric identities and eQuations θ y = cos θ 0 1 π _ 2 0 Table 1 π −1 3π _ 2 0 2π 1 y 1 –1 π 2 π 3π 2 y = cos θ θ 2π Figure 6 Example 3 Graphing Sinusoidal Functions Using Key Points Graph the function y = −4cos(πx) using amplitude, period, and key points. 2π 2π _ _ Solution The amplitude is ∣ −4 ∣ = 4. The period is π = 2. (Recall that we sometimes refer to B as ω.) One ω = cycle of the graph can be drawn over the interval [0, 2]. To find the key points, we divide the period by 4. Make a table 1 _ similar to Table 2, starting with x = 0 and then adding successively to x and calculate y. See the graph in Figure 7. 2 x y = −4 cos(πx) 0 −4 1 _ 2 0 Table 2 1 4 3 _ 2 0 2 −4 y (1, 4) y = –4 cos (πx) 4 2 –2 –4 1 2 3 4 x (2, −4) Figure 7 Try It #2 Graph the function y = 3sin(3x) using the amplitude, period, and five key points. Modeling Periodic Behavior We will now apply these ideas to problems involving periodic behavior. Example 4 Modeling an Equation and Sketching a Sinusoidal Graph to Fit Criteria The average monthly temperatures for a small town in Oregon are given in Table 3. Find a sinusoidal function of the form y = Asin(Bt − C) + D that fits the
data (round to the nearest tenth) and sketch the graph. SECTION 7.6 modeling with trigonometric eQuations 621 Month January February March April May June July August September October November December Temperature, ° F 42.5 44.5 48.5 52.5 58 63 68.5 69 64.5 55.5 46.5 43.5 Solution Recall that amplitude is found using the formula Table 3 Thus, the amplitude is A = largest value − smallest value ______________________ 2 ∣ A ∣ = 69 − 42.5 ________ 2 = 13.25 The data covers a period of 12 months, so = 12 which gives B = 2π _ B 2π _ 12 π _ =. 6 The vertical shift is found using the following equation. Thus, the vertical shift is D = highest value + lowest value _____________________ 2 D = 69 + 42.5 ________ 2 = 55.8 π _ So far, we have the equation y = 13.3sin  x − C  + 55.8. 6 To find the horizontal shift, we input the x and y values for the first month and solve for C. π _ (1) − C  + 55.8 42.5 = 13.3sin  6 π _ − C  −13.3 = 13.3sin  6 π π _ _ − C  sin θ = −1 → θ = − −1 = sin  = 2π _ 3 622 CHAPTER 7 trigonometric identities and eQuations π _ We have the equation y = 13.3sin  x − 6 2π _  + 55.8. See the graph in Figure 8. 3 Average Amplitude = 13. 80 70 60 50 40 30 20 10 0 1 2 3 4 8 9 10 11 12 5 7 6 Months Figure 8 Example 5 Describing Periodic Motion The hour hand of the large clock on the wall in Union Station measures 24 inches in length. At noon, the tip of the hour hand is 30 inches from the ceiling. At 3 PM, the tip is 54 inches from the ceiling, and at 6 PM, 78 inches. At 9 PM, it is again 54 inches from the ceiling, and at midnight, the tip of the hour hand returns to its original position 30 inches from the ceiling. Let y equal the distance from the tip
of the hour hand to the ceiling x hours after noon. Find the equation that models the motion of the clock and sketch the graph. Solution Begin by making a table of values as shown in Table 4. x Noon 3 PM 6 PM 9 PM Midnight y 30 in 54 in 78 in 54 in 30 in Table 4 Points to plot (0, 30) (3, 54) (6, 78) (9, 54) (12, 30) To model an equation, we first need to find the amplitude. ∣ A ∣ =  78 − 30  ______ 2 The clock’s cycle repeats every 12 hours. Thus, The vertical shift is = 24 B = 2π _ 12 π _ = 6 D = 78 + 30 ______ 2 = 54 There is no horizontal shift, so C = 0. Since the function begins with the minimum value of y when x = 0 (as opposed to the maximum value), we will use the cosine function with the negative value for A. In the form y = A cos(Bx ± C) + D, the equation is See Figure 9. π _ y = −24cos  x  + 54 6 80 70 60 50 40 30 20 10 y (6, 78) (3, 54) (9, 54) (12, 30) (0, 30) 10 2 3 4 5 6 7 8 9 10 11 12 x Figure 9 SECTION 7.6 modeling with trigonometric eQuations 623 Example 6 Determining a Model for Tides The height of the tide in a small beach town is measured along a seawall. Water levels oscillate between 7 feet at low tide and 15 feet at high tide. On a particular day, low tide occurred at 6 AM and high tide occurred at noon. Approximately every 12 hours, the cycle repeats. Find an equation to model the water levels. Solution As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as The cycle repeats every 12 hours; therefore, B is ∣ A ∣ =  (15 − 7)  _______ 2 = 4 2π _ 12 π _ = 6 There is a vertical translation of = 11. Since the value of the function is at a maximum at t = 0, we will use the cosine function, with the positive value for A. (15 + 7) _______ 2 See Figure 10. π _ y = 4cos
  t + 11 6 (0, 15) (12, 15) Midline: y = 11 (6, 7) 16 14 12 10 10 11 12 Time Figure 10 Try It #3 The daily temperature in the month of March in a certain city varies from a low of 24°F to a high of 40°F. Find a sinusoidal function to model daily temperature and sketch the graph. Approximate the time when the temperature reaches the freezing point 32°F. Let t = 0 correspond to noon. Example 7 Interpreting the Periodic Behavior Equation The average person’s blood pressure is modeled by the function f(t) = 20 sin(160πt) + 100, where f(t) represents the blood pressure at time t, measured in minutes. Interpret the function in terms of period and frequency. Sketch the graph and find the blood pressure reading. Solution The period is given by 2π ____ 160π 1 __ 80 In a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by 2π ___ ω = = ω __ 2π = 160π ____ 2π = 80 See the graph in Figure 11. f(t) = 20sin(160πt) + 100 f(t) 120 100 80 60 1 40 1 20 3 40 t 1 10 Figure 11 The blood pressure reading on the graph is 120  ____ 80 maximum . _________ minimum 624 CHAPTER 7 trigonometric identities and eQuations 120 _ 80 Analysis Blood pressure of measures the pressure in the arteries when the heart contracts. The bottom number is the minimum or diastolic reading, which measures the pressure in the arteries as the heart relaxes between beats, refilling with blood. Thus, normal blood pressure can be modeled by a periodic function with a maximum of 120 and a minimum of 80. is considered to be normal. The top number is the maximum or systolic reading, which Modeling Harmonic Motion Functions Harmonic motion is a form of periodic motion, but there are factors to consider that differentiate the two types. While general periodic motion applications cycle through their periods with no outside interference, harmonic motion requires a restoring force. Examples of harmonic motion include springs, gravitational force, and magnetic force. Simple Harmonic Motion A type of motion described as simple harmonic motion involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object
hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When t = 0, d = 0. simple harmonic motion We see that simple harmonic motion equations are given in terms of displacement: d = acos(ωt) or d = asin(ωt) where \|a\| is the amplitude, 2π _ ω is the period, and ω _ 2π is the frequency, or the number of cycles per unit of time. Example 8 Finding the Displacement, Period, and Frequency, and Graphing a Function For the given functions, 1. Find the maximum displacement of an object. 2. Find the period or the time required for one vibration. 3. Find the frequency. 4. Sketch the graph. a. y = 5sin(3t) b. y = 6cos(πt) π _ c. y = 5cos  t  2 Solution a. y = 5sin(3t) 1. The maximum displacement is equal to the amplitude, ∣ a ∣, which is 5. 2. The period is 2π _ ω = 2π _. 3 3. The frequency is given as ω _ 2π = 3 _. 2π 4. See Figure 12. The graph indicates the five key points. b. y = 6cos(πt) 2. The period is 1. The maximum displacement is 6. 2π _ ω = ω _ 2π 2π _ π = 2. π 1 _ _ = =. 2π 2 3. The frequency is 4. See Figure 13. y 5 3 1 –1 –3 –5 y 6 4 2 –2 –4 –6 y = 5sin(3t) π 6 π 3 π 2 t 2π 3 Figure 12 y = 6cos(πt) 1 2 3 4 t Figure 13 SECTION 7.6 modeling with trigonometric eQuations 625 π _ c. y = 5cos   t 2 1. The maximum displacement is 5. 2. The period is = 4. 2π _ ω = 2π _ π _ 2 1 _ 3. The frequency
is. 4 4. See Figure 14. Damped Harmonic Motion y 5 3 1 –1 –3 –5 π t y = 5cos 2 1 2 3 4 t Figure 14 In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion. Friction is typically the damping factor. In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models. damped harmonic motion In damped harmonic motion, the displacement of an oscillating object from its rest position at time t is given as where c is a damping factor, ∣ a ∣ is the initial displacement and 2π _ ω is the period. f(t) = ae−ctsin(ωt) or f(t) = ae−ctcos(ωt) Example 9 Modeling Damped Harmonic Motion Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of 0.5 cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of 0.5 and the second has a damping factor of 0.1. Solution At time t = 0, the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models. ω _ 2π We are given the frequency f = of 0.5 cycles per second. Thus, = 0.5 ω _ 2π ω = (0.5)2π The first spring system has a damping factor of c = 0.5. Following the general model for damped harmonic motion, we have f(t) = 10e−0.5t cos(πt) Figure 15 models the motion of the first spring system. = π f(t) 12 8 4 f(t) = 10e–0.5tcos(πt) –15 –9 –3 9 15 t –4 –8 Figure 15 626 CHAPTER
7 trigonometric identities and eQuations The second spring system has a damping factor of c = 0.1 and can be modeled as f (t) = 10e−0.1tcos(πt) Figure 16 models the motion of the second spring system. f(t) f(t) = 10e–0.1tcos (πt) 12 8 4 –12 –8 –4 4 8 12 t –4 –8 –12 Figure 16 Analysis Notice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor c = 0.5 decreases much more rapidly than that of the function with c = 0.1. Example 10 Finding a Cosine Function that Models Damped Harmonic Motion Find and graph a function of the form y = ae−ctcos(ωt) that models the information given. a. a = 20, c = 0.05, p = 4 b. a = 2, c = 1.5, f = 3 Solution Substitute the given values into the model. Recall that period is 2π _ ω and frequency is ω _. 2π π _ a. y = 20e−0.05tcos  t . See Figure 17. 2 b. y = 2e−1.5tcos(6πt). See Figure 18. y 30 20 10 π t y = 20e–0.05t cos 2 –24 –16 –8 8 16 24 t –10 –20 –30 Figure 17 y 6 4 2 y = 2e–1.5t cos(6πt) –6 –4 –2 2 4 6 t –2 –4 –6 Figure 18 Try It #4 The following equation represents a damped harmonic motion model: f(t) = 5e−6tcos(4t) Find the initial displacement, the damping constant, and the frequency. SECTION 7.6 modeling with trigonometric eQuations 627 Example 11 Finding a Sine Function that Models Damped Harmonic Motion Find and graph a function of the form y = ae−ct sin(ωt) that models the information given. π _ a. a = 7, c = 10, p = 6 b. a = 0.3, c = 0.2, f = 20 Solution Calculate the value of ω and substitute the known values into the model.
a. As period is 2π _ ω, we have π 2π _ _ = ω 6 ωπ = 6(2π) ω = 12 The damping factor is given as 10 and the amplitude is 7. Thus, the model is y = 7e−10tsin(12t). See Figure 19. y 3 2 1 –1 y = 7e–10t sin(12t) 1 2 3 4 t Figure 19 20 = ω _ 2π 40π = ω b. As frequency is, we have ω _ 2π The damping factor is given as 0.2 and the amplitude is 0.3. The model is y = 0.3e−0.2tsin(40πt). See Figure 20. y 1 –1 5 10 15 20 25 t –1 Figure 20 Analysis A comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when t = 0, and the sine function π _ is at the equilibrium point when t = 0. For example, consider the equation y = 20e−0.05tcos  t  from Example 10. We 2 can see from the graph that when t = 0, y = 20, which is the initial amplitude. Check this by setting t = 0 in the cosine equation: π _ y = 20e−0.05(0)cos   (0) 2 = 20(1)(1) Using the sine function yields Thus, cosine is the correct function. = 20 π _ y = 20e−0.05(0)sin   (0) 2 = 20(1)(0) = 0 628 CHAPTER 7 trigonometric identities and eQuations Try It #5 Write the equation for damped harmonic motion given a = 10, c = 0.5, and p = 2. Example 12 Modeling the Oscillation of a Spring A spring measuring 10 inches in natural length is compressed by 5 inches and released. It oscillates once every 3 seconds, and its amplitude decreases by 30% every second. Find an equation that models the position of the spring t seconds after being released. Solution The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we
will write A as a negative value. We can write the amplitude portion of the function as We put (1 − 0.30)t in the form ect as follows: A(t) = 5(1 − 0.30)t 0.7 = ec c = ln 0.7 c = −0.357 Now let’s address the period. The spring cycles through its positions every 3 seconds, this is the period, and we can use the formula to find omega. 3 = ω = 2π _ ω 2π _ 3 The natural length of 10 inches is the midline. We will use the cosine function, since the spring starts out at its maximum displacement. This portion of the equation is represented as Finally, we put both functions together. Our the model for the position of the spring at t seconds is given as y = cos  2π _ t  + 10 3 See the graph in Figure 21. y = −5e−0.357t cos  2π _ t  + 10 3 y 24 16 8 –8 –24 –16 –8 y = –5e–0.357t cos 3 2πt + 10 8 16 24 t Figure 21 Try It #6 A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time 1 _ t = 0 and allowed to oscillate. After second, it is observed that the mass returns to its highest position. Find a 3 function to model this motion relative to its initial resting position. Example 13 Finding the Value of the Damping Constant c According to the Given Criteria A guitar string is plucked and vibrates in damped harmonic motion. The string is pulled and displaced 2 cm from its resting position. After 3 seconds, the displacement of the string measures 1 cm. Find the damping constant. Solution The displacement factor represents the amplitude and is determined by the coefficient ae−ct in the model for damped harmonic motion. The damping constant is included in the term e−ct. It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation ae−c(t + 3) = 1 _ ae−ct 2 SECTION 7.6 modeling with trigonometric eQuations 629 Use algebra and the laws of exponents to solve for c. Divide out a. Divide out e−ct. Take reciproc
als. ae−c(t + 3) = 1 _ ae−ct 2 e−ct · e−3c = 1 _ e−ct 2 e−3c = 1 _ 2 e3c = 2 e3c = 2 3c = ln(2) c = ln(2) _ 3 Then use the laws of logarithms. The damping constant is ln(2) ___. 3 Bounding Curves in Harmonic Motion Harmonic motion graphs may be enclosed by bounding curves. When a function has a varying amplitude, such that the amplitude rises and falls multiple times within a period, we can determine the bounding curves from part of the function. Example 14 Graphing an Oscillating Cosine Curve Graph the function f(x) = cos(2πx) cos(16πx). Solution The graph produced by this function will be shown in two parts. The first graph will be the exact function f(x) (see Figure 22), and the second graph is the exact function f(x) plus a bounding function (see Figure 23). The graphs look quite different. f(x) 1 –1 f(x) = cos (2πx) cos (16πx) 1 x 2 Figure 22 y 2 1 –1 –2 f(x) = cos (2πx) cos (16πx) y = cos (2πx) 1 y = –cos (2πx) x 2 Figure 23 Analysis The curves y = cos(2πx) and y = −cos(2πx) are bounding curves: they bound the function from above and below, tracing out the high and low points. The harmonic motion graph sits inside the bounding curves. This is an example of a function whose amplitude not only decreases with time, but actually increases and decreases multiple times within a period. Access these online resources for additional instruction and practice with trigonometric applications. • Solving Problems Using Trigonometry (http://openstaxcollege.org/l/solvetrigprob) • Ferris Wheel Trigonometry (http://openstaxcollege.org/l/ferriswheel) • Daily Temperatures and Trigonometry (http://openstaxcollege.org/l/dailytemp) • Simple Harmonic Motion (http://openstaxcollege.org/l/simpleharm) 630 CHAPTER 7 trigonometric identities and eQu
ations 7.6 SeCTIOn exeRCISeS VeRBAl 1. Explain what types of physical phenomena are best modeled by sinusoidal functions. What are the characteristics necessary? 3. If we want to model cumulative rainfall over the course of a year, would a sinusoidal function be a good model? Why or why not? 2. What information is necessary to construct a trigonometric model of daily temperature? Give examples of two different sets of information that would enable modeling with an equation. 4. Explain the effect of a damping factor on the graphs of harmonic motion functions. AlGeBRAIC For the following exercises, find a possible formula for the trigonometric function represented by the given table of values. 5. x 3 0 y −4 −1 7. x y 0 2 9. x 0 y −2 π _ 4 7 1 4 9 12 15 6 2 −1 −4 −1 18 2 π 3π π _ _ 4 2 2 −3 2 5π _ 4 7 3π _ 2 2 2 10 4 3 4 −2 5 4 6 10 6. x y 8. x y 10. x y 0 5 4 2 1 −3 6 1 8 5 12 10 1 −3 2 1 3 0 1 −3 −7 −3 6 5 4 1 −3 −7 1 0 5 −3 2 5 3 13 5 4 5 −3 6 5 11. x y 12. x −3 −1 − √ — 2 −1 y — 3 − 2 √ −2 −1 0 0 − √ ____ GRAPHICAl For the following exercises, graph the given function, and then find a possible physical process that the equation could model. 13. f (x) = −30cos  xπ _  − 20cos2  6 xπ _  + 80 [0, 12] 6 14. f (x) = −18cos   − 5sin   + 100 on the interval [0, 24] xπ _ 12 xπ _ 12 15. f (x) = 10 − sin  xπ _  + 24tan  6 xπ _ 240  on the interval [0, 80] TeCHnOlOGY For the following exercise, construct a function modeling behavior and use a calculator to find desired results. 16. A city�
�s average yearly rainfall is currently 20 inches and varies seasonally by 5 inches. Due to unforeseen circumstances, rainfall appears to be decreasing by 15% each year. How many years from now would we expect rainfall to initially reach 0 inches? Note, the model is invalid once it predicts negative rainfall, so choose the first point at which it goes below 0. SECTION 7.6 section exercises 631 ReAl-WORlD APPlICATIOnS For the following exercises, construct a sinusoidal function with the provided information, and then solve the equation for the requested values. 17. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of 105°F occurs at 5PM and the average temperature for the day is 85°F. Find the temperature, to the nearest degree, at 9AM. 18. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of 84°F occurs at 6PM and the average temperature for the day is 70°F. Find the temperature, to the nearest degree, at 7AM. 19. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between 47°F and 63°F during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight does the temperature first reach 51°F? 20. Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the temperature varies between 64°F and 86°F during the day and the average daily temperature first occurs at 12 AM. How many hours after midnight does the temperature first reach 70°F? 21. A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How much of the ride, in minutes and seconds, is spent higher than 13 meters above the ground? 22. A Ferris wheel is 45 meters in diameter and boarded from a platform that is 1 meter above the ground. The six o’clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 27 meters above the ground? Round to the nearest second 23. The sea ice area around the North Pole
fluctuates between about 6 million square kilometers on September 1 to 14 million square kilometers on March 1. Assuming a sinusoidal fluctuation, when are there less than 9 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day. 24. The sea ice area around the South Pole fluctuates between about 18 million square kilometers in September to 3 million square kilometers in March. Assuming a sinusoidal fluctuation, when are there more than 15 million square kilometers of sea ice? Give your answer as a range of dates, to the nearest day. 25. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. If the rainfall fluctuates between a low of 2 inches on day 10 and 12 inches on day 55, during what period is daily rainfall more than 10 inches? 26. During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorded on day 30, and overall the average daily rainfall was 8 inches. During what period was daily rainfall less than 5 inches? 27. In a certain region, monthly precipitation peaks at 8 inches on June 1 and falls to a low of 1 inch on December 1. Identify the periods when the region is under flood conditions (greater than 7 inches) and drought conditions (less than 2 inches). Give your answer in terms of the nearest day. 28. In a certain region, monthly precipitation peaks at 24 inches in September and falls to a low of 4 inches in March. Identify the periods when the region is under flood conditions (greater than 22 inches) and drought conditions (less than 5 inches). Give your answer in terms of the nearest day. For the following exercises, find the amplitude, period, and frequency of the given function. 29. The displacement h(t) in centimeters of a mass 30. The displacement h(t) in centimeters of a mass suspended by a spring is modeled by the function h(t) = 8sin(6πt),where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. suspended by a spring is modeled by the function h(t) = 11sin(12πt), where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. 632 CHAPTER 7 trigonometric identities and eQuations 31. The displacement h(t) in centimeters of a mass suspended by a spring is
modeled by the function π _ h(t) = 4cos  t , where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. 2 For the following exercises, construct an equation that models the described behavior. 32. The displacement h(t), in centimeters, of a mass suspended by a spring is modeled by the function h(t) = −5 cos(60πt), where t is measured in seconds. Find the amplitude, period, and frequency of this displacement. For the following exercises, construct an equation that models the described behavior. 33. A deer population oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 160 each year. Find a function that models the population, P, in terms of months since January, t. 34. A rabbit population oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 110 each year. Find a function that models the population, P, in terms of months since January, t. 35. A muskrat population oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, P, in terms of months since January, t. 36. A fish population oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, P, in terms of months since January, t. 37. A spring attached to the ceiling is pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. 38. A spring attached to the ceiling is pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. 39. A spring attached to the ceiling is pulled 17 cm down 40. A spring attached to the ceiling is
pulled 19 cm down from equilibrium and released. After 3 seconds, the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. from equilibrium and released. After 4 seconds, the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, D, the end of the spring is from equilibrium in terms of seconds, t, since the spring was released. For the following exercises, create a function modeling the described behavior. Then, calculate the desired result using a calculator. 41. A certain lake currently has an average trout population of 20,000. The population naturally oscillates above and below average by 2,000 every year. This year, the lake was opened to fishermen. If fishermen catch 3,000 fish every year, how long will it take for the lake to have no more trout? 42. Whitefish populations are currently at 500 in a lake. The population naturally oscillates above and below by 25 each year. If humans overfish, taking 4% of the population every year, in how many years will the lake first have fewer than 200 whitefish? 43. A spring attached to a ceiling is pulled down 11 cm from equilibrium and released. After 2 seconds, the amplitude has decreased to 6 cm. The spring oscillates 8 times each second. Find when the spring first comes between −0.1 and 0.1 cm, effectively at rest. 44. A spring attached to a ceiling is pulled down 21 cm from equilibrium and released. After 6 seconds, the amplitude has decreased to 4 cm. The spring oscillates 20 times each second. Find when the spring first comes between −0.1 and 0.1 cm, effectively at rest. SECTION 7.6 section exercises 633 45. Two springs are pulled down from the ceiling and released at the same time. The first spring, which oscillates 8 times per second, was initially pulled down 32 cm from equilibrium, and the amplitude decreases by 50% each second. The second spring, oscillating 18 times per second, was initially pulled down 15 cm from equilibrium and after 4 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than 0.1 cm. 46. Two springs are pulled down from the ceiling and released at the same
time. The first spring, which oscillates 14 times per second, was initially pulled down 2 cm from equilibrium, and the amplitude decreases by 8% each second. The second spring, oscillating 22 times per second, was initially pulled down 10 cm from equilibrium and after 3 seconds has an amplitude of 2 cm. Which spring comes to rest first, and at what time? Consider “rest” as an amplitude less than 0.1 cm. exTenSIOnS 47. A plane flies 1 hour at 150 mph at 22° east of north, then continues to fly for 1.5 hours at 120 mph, this time at a bearing of 112° east of north. Find the total distance from the starting point and the direct angle flown north of east. 48. A plane flies 2 hours at 200 mph at a bearing of 60°, then continues to fly for 1.5 hours at the same speed, this time at a bearing of 150°. Find the distance from the starting point and the bearing from the starting point. Hint: bearing is measured counterclockwise from north. π _ For the following exercises, find a function of the form y = abx + csin  x  that fits the given data. 2 49. x y 0 6 1 29 2 96 3 379 50. x y 0 6 1 34 2 150 3 746 51. x y 0 4 1 0 3 2 16 −40 π _ x  + c that fits the given data. For the following exercises, find a function of the form y = abx cos  2 52. x y 53. x y 1 2 1 −11 0 11 634 CHAPTER 7 trigonometric identities and eQuations CHAPTeR 7 ReVIeW Key Terms damped harmonic motion oscillating motion that resembles periodic motion and simple harmonic motion, except that the graph is affected by a damping factor, an energy dissipating influence on the motion, such as friction double-angle formulas identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal even-odd identities set of equations involving trigonometric functions such that if f (−x) = −f (x), the identity is odd, and if f (−x) = f (x), the identity is even half-angle formulas identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions product-to-sum formula a
trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions Pythagorean identities set of equations involving trigonometric functions based on the right triangle properties quotient identities pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine reciprocal identities set of equations involving the reciprocals of basic trigonometric definitions reduction formulas identities derived from the double-angle formulas and used to reduce the power of a trigonometric function simple harmonic motion a repetitive motion that can be modeled by periodic sinusoidal oscillation sum-to-product formula a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions Key equations Pythagorean identities Even-odd identities Reciprocal identities sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ 1 + tan2 θ = sec2 θ tan(−θ) = −tan θ cot(−θ) = −cot θ sin(−θ) = −sin θ csc(−θ) = −csc θ cos(−θ) = cos θ sec(−θ) = sec θ sin θ = 1 _ csc θ cos θ = 1 _ sec θ tan θ = 1 _ cot θ csc θ = 1 _ sin θ sec θ = 1 _ cos θ cot θ = 1 _ tan θ CHAPTER 7 review 635 Quotient identities tan θ = sin θ _ cos θ cot θ = cos θ _ sin θ Sum Formula for Cosine cos(α + β) = cos α cos β − sin α sin β Difference Formula for Cosine cos(α − β) = cos α cos β + sin α sin β Sum Formula for Sine sin(α + β)= sin α cos β + cos α sin β Difference Formula for Sine sin(α − β) = sin α cos β − cos α sin β Sum Formula for Tangent Difference Formula for Tangent Cofunction identities Double-angle formulas Reduction formulas tan(α + β) = tan α + tan β ___________ 1 − tan α tan β tan(α − β) = tan α
− tan β ___________ 1 + tan α tan β π _ sin θ = cos  − θ  2 π _ − θ  cos θ = sin  2 π _ − θ  tan θ = cot  2 π _ − θ  cot θ = tan  2 π _ − θ  sec θ = csc  2 π _ − θ  csc θ = sec  2 sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 1 − 2sin2 θ = 2cos2 θ − 1 tan(2θ) = 2tan θ ________ 1 − tan2 θ sin2θ = 1 − cos(2θ) _________ 2 cos2θ = 1 + cos(2θ) _________ 2 tan2θ = 1 − cos(2θ) _________ 1 + cos(2θ) 636 CHAPTER 7 trigonometric identities and eQuations Half-angle formulas Product-to-sum Formulas Sum-to-product Formulas α _ sin 2 α _ cos 2 α _ tan 2 = ± √ = ± √ = ± √ _________ 1 − cos α _______ 2 _________ 1 + cos α _______ 2 _________ 1 − cos α _______ 1 + cos α = = sin α _______ 1 + cos α 1 − cos α _______ sin α 1 __ [cos(α − β) + cos(α + β)] cos α cos β = 2 1 __ [sin(α + β) + sin(α − β)] sin α cos β = 2 1 __ [cos(α − β) − cos(α + β)] sin α sin β = 2 1 __ [sin(α + β) − sin(α − β)] cos α sin β = 2 sin α + sin β = 2 sin  sin α − sin β = 2 sin  cos α − cos β = −2 sin  cos α + cos β = 2cos   cos   cos  α + β _____ 2 α − β _____
2 α + β _____ 2 α + β _____ 2  cos  α − β  _____ 2 α + β  _____ 2 α − β  _____ 2 α − β  _____ 2  sin  Standard form of sinusoidal equation y = A sin(Bt − C) + D or y = A cos(Bt − C) + D Simple harmonic motion Damped harmonic motion Key Concepts d = a cos(ωt) or d = a sin(ωt) f (t) = ae−ct sin(ωt) or f (t) = ae−ct cos(ωt) 7.1 Solving Trigonometric Equations with Identities • There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem. • Graphing both sides of an identity will verify it. See Example 1. • Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example 2 and Example 3. • The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example 4. • We can create an identity by simplifying an expression and then verifying it. See Example 5. • Verifying an identity may involve algebra with the fundamental identities. See Example 6 and Example 7. CHAPTER 7 review 637 • Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example 8, Example 9, and Example 10. 7.2 Sum and Difference Identities • The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles. • The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See Example 1 and Example 2. • The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle
and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See Example 3. • The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See Example 4. • The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See Example 5. • The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See Example 6. • The cofunction identities apply to complementary angles and pairs of reciprocal functions. See Example 7. • Sum and difference formulas are useful in verifying identities. See Example 8 and Example 9. • Application problems are often easier to solve by using sum and difference formulas. See Example 10 and Example 11. 7.3 Double-Angle, Half-Angle, and Reduction Formulas • Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example 1, Example 2, Example 3, and Example 4. • Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example 5 and Example 6. • Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example 7, Example 8, and Example 9. 7.4 Sum-to-Product and Product-to-Sum Formulas • From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine. • We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines
. See Example 1, Example 2, and Example 3. • We can also derive the sum-to-product identities from the product-to-sum identities using substitution. • We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines. See Example 4. 638 CHAPTER 7 trigonometric identities and eQuations • Trigonometric expressions are often simpler to evaluate using the formulas. See Example 5. • The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side. See Example 6 and Example 7. 7.5 Solving Trigonometric Equations • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example 1, Example 2, and Example 3. • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example 4, Example 5, and Example 6, and Example 7. • We can also solve trigonometric equations using a graphing calculator. See Example 8 and Example 9. • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example 10, Example 11, Example 12, and Example 13. • We can also use the identities to solve trigonometric equation. See Example 14, Example 15, and Example 16. • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example 17. • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example 18. 7.6 Modeling with Trigonometric Equations • Sinusoidal functions are represented by the sine and cosine graphs. In standard form, we can find the amplitude, period, and horizontal and vertical shifts. See Example 1 and Example
2. • Use key points to graph a sinusoidal function. The five key points include the minimum and maximum values and the midline values. See Example 3. • Periodic functions can model events that reoccur in set cycles, like the phases of the moon, the hands on a clock, and the seasons in a year. See Example 4, Example 5, Example 6 and Example 7. • Harmonic motion functions are modeled from given data. Similar to periodic motion applications, harmonic motion requires a restoring force. Examples include gravitational force and spring motion activated by weight. See Example 8. • Damped harmonic motion is a form of periodic behavior affected by a damping factor. Energy dissipating factors, like friction, cause the displacement of the object to shrink. See Example 9, Example 10, Example 11, Example 12, and Example 13. • Bounding curves delineate the graph of harmonic motion with variable maximum and minimum values. See Example 14. CHAPTER 7 review 639 CHAPTeR 7 ReVIeW exeRCISeS SOlVInG TRIGOnOMeTRIC eQUATIOnS WITH IDenTITIeS For the following exercises, find all solutions exactly that exist on the interval [0, 2π). 1. csc2 t = 3 4. tan x sin x + sin(−x) = 0 2. cos2 x = 1 __ 4 5. 9sin ω − 2 = 4 sin2 ω 3. 2sin θ = −1 6. 1 − 2tan(ω) = tan2(ω) For the following exercises, use basic identities to simplify the expression. 7. sec x cos x + cos x − 1 ____ sec x 8. sin3 x + cos2 x sin x For the following exercises, determine if the given identities are equivalent. 9. sin2 x + sec2 x − 1 = (1 − cos2 x)(1 + cos2 x) __ cos2 x 10. tan3 x csc2 x cot2 x cos x sin x = 1 SUM AnD DIFFeRenCe IDenTITIeS For the following exercises, find the exact value. 11. tan  7π ___  12 12. cos  25π ___  12 13. sin(70°)cos(25°) − cos(70°)sin(25°) 14. cos(83°)
cos(23°) + sin(83°)sin(23°) For the following exercises, prove the identity. 15. cos(4x) − cos(3x)cosx = sin2 x − 4cos2 x sin2 x 16. cos(3x) − cos3 x = − cos x sin2 x − sin x sin(2x) For the following exercise, simplify the expression. 1 1 __ __ x  + tan  x  tan  8 2 ___ 1 1 __ __ x  x  tan  1 − tan  2 8 17. For the following exercises, find the exact value. 1 __   18. cos  sin−1 (0) − cos−1  2 1 __   19. tan  sin−1 (0) + sin−1  2 DOUBle-AnGle, HAlF-AnGle, AnD ReDUCTIOn FORMUlAS For the following exercises, find the exact value. 20. Find sin(2θ), cos(2θ), and tan(2θ) given π cos θ = − 1 __ __, π  and θ is in the interval  3 2 21. Find sin(2θ), cos(2θ), and tan(2θ) given π sec θ = − 5 __ __, π  and θ is in the interval  2 3 22. sin  7π ___  8 23. sec  3π ___  8 640 CHAPTER 7 trigonometric identities and eQuations For the following exercises, use Figure 1 to find the desired quantities. α 25 β 24 Figure 1 24. sin(2β), cos(2β), tan(2β), sin(2α), cos(2α), and tan(2α) α α β β β , sin  , tan  , cos  25. sin  __ __ __ __ __ , cos  , 2 2 2 2 2 α __ and tan   2 For the following exercises, prove the identity. 26
. 2cos(2x) _______ sin(2x) = cot x − tan x 27. cot x cos(2x) = − sin(2x) + cot x For the following exercises, rewrite the expression with no powers. 28. cos2 x sin4(2x) 29. tan2 x sin3 x SUM-TO-PRODUCT AnD PRODUCT-TO-SUM FORMUlAS For the following exercises, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. π π _ __   sin  30. cos  3 4 π π __ __   cos  32. 2cos  3 5 31. 2sin  2π __  sin  3 5π __  6 For the following exercises, evaluate the sum by using a product formula. Write the exact answer. 33. sin  π __ 12  − sin  7π __  12 34. cos  5π __ 12  + cos  7π __  12 For the following exercises, change the functions from a product to a sum or a sum to a product. 35. sin(9x)cos(3x) 37. sin(11x) + sin(2x) 36. cos(7x)cos(12x) 38. cos(6x) + cos(5x) SOlVInG TRIGOnOMeTRIC eQUATIOnS For the following exercises, find all exact solutions on the interval [0, 2π). 39. tan x + 1 = 0 40. 2sin(2x) + √ — 2 = 0 For the following exercises, find all exact solutions on the interval [0, 2π). 41. 2sin2 x − sin x = 0 43. 2sin2 x + 5 sin x + 3 = 0 42. cos2 x − cos x − 1 = 0 44. cos x − 5sin(2x) = 0 45. 1 ____ sec2 x + 2 + sin2 x + 4cos2 x = 0 CHAPTER 7 review 641 For the following exercises, simplify the equation algebraically as much as possible. Then use a calculator to find the solutions on the interval
[0, 2π). Round to four decimal places. 46. √ — 3 cot2 x + cot x = 1 47. csc2 x − 3csc x − 4 = 0 For the following exercises, graph each side of the equation to find the zeroes on the interval [0, 2π). 48. 20cos2 x + 21cos x + 1 = 0 49. sec2 x − 2sec x = 15 MODelInG WITH TRIGOnOMeTRIC eQUATIOnS For the following exercises, graph the points and find a possible formula for the trigonometric values in the given table. 50. x y 52. x y 0 1 1 6 2 11 −2 3 4 1 5 6 51. x y 0 −2 −2 −5 −1 3 −1 − 2 √ — 2 53. A man with his eye level 6 feet above the ground is standing 3 feet away from the base of a 15-foot vertical ladder. If he looks to the top of the ladder, at what angle above horizontal is he looking? 54. Using the ladder from the previous exercise, if a 6-foot-tall construction worker standing at the top of the ladder looks down at the feet of the man standing at the bottom, what angle from the horizontal is he looking? For the following exercises, construct functions that model the described behavior. 55. A population of lemmings varies with a yearly low of 500 in March. If the average yearly population of lemmings is 950, write a function that models the population with respect to t, the month. 56. Daily temperatures in the desert can be very extreme. If the temperature varies from 90°F to 30°F and the average daily temperature first occurs at 10 AM, write a function modeling this behavior. For the following exercises, find the amplitude, frequency, and period of the given equations. 57. y = 3cos(xπ) 58. y = −2sin(16xπ) For the following exercises, model the described behavior and find requested values. 59. An invasive species of carp is introduced to Lake Freshwater. Initially there are 100 carp in the lake and the population varies by 20 fish seasonally. If by year 5, there are 625 carp, find a function modeling the population of carp with respect to t, the number of years from now. 60. The native fish population of Lake Freshwater averages 2500 fish, varying by 100 fish seasonally
. Due to competition for resources from the invasive carp, the native fish population is expected to decrease by 5% each year. Find a function modeling the population of native fish with respect to t, the number of years from now. Also determine how many years it will take for the carp to overtake the native fish population. 642 CHAPTER 7 trigonometric identities and eQuations CHAPTeR 7 PRACTICe TeST For the following exercises, simplify the given expression. 1. cos(−x)sin x cot x + sin2 x 2. sin(−x)cos(−2x)−sin(−x)cos(−2x) For the following exercises, find the exact value. 3. cos  7π __  12 4. tan  3π ___  8 — 2  + tan−1 √ 5. tan  sin−1  √ ____ 2 — 3  π π __ __   sin  6. 2sin  6 4 For the following exercises, find all exact solutions to the equation on [0, 2π). 7. cos2 x − sin2 x − 1 = 0 9. cos(2x) + sin2 x = 0 11. Rewrite the expression as a product instead of a sum: cos(2x) + cos(−8x). 13. Find the solutions of sec2 x − 2sec x = 15 on the interval [0, 2π) algebraically; then graph both sides of the equation to determine the answer. θ θ θ  given , and tan  , cos  15. Find sin  __ __ __ 2 2 2 7 __ cos θ = 25 and θ is in quadrant IV. For the following exercises, prove the identity. 8. cos2 x = cos x 4sin2 x + 2sin x − 3 = 0 10. 2sin2 x − sin x = 0 12. Find all solutions of tan(x) − √ — 3 = 0. 14. Find sin(2θ), cos(2θ), and tan(2θ) given cot θ = − 3 π __ __, π . and θ is on the interval  2 4 16
. Rewrite the expression sin4 x with no powers greater than 1. 17. tan3 x − tan x sec2 x = tan(−x) 18. sin(3x) − cos x sin(2x) = cos2 x sin x − sin3 x 19. sin(2x) ______ − sin x cos(2x) ______ cos x = sec x 20. Plot the points and find a function of the form y = Acos(Bx + C) + D that fits the given data. x y 0 −2 1 2 2 −2 3 2 4 −2 5 2 21. The displacement h(t) in centimeters of a mass 22. A woman is standing 300 feet away from a 2,000- suspended by a spring is modeled by the function 1 __ h(t) = sin(120πt), where t is measured in seconds. 4 Find the amplitude, period, and frequency of this displacement. foot building. If she looks to the top of the building, at what angle above horizontal is she looking? A bored worker looks down at her from the 15th floor (1500 feet above her). At what angle is he looking down at her? Round to the nearest tenth of a degree. 23. Two frequencies of sound are played on an instrument governed by the equation n(t) = 8 cos(20πt)cos(1,000πt). What are the period and frequency of the “fast” and “slow” oscillations? What is the amplitude? 24. The average monthly snowfall in a small village in the Himalayas is 6 inches, with the low of 1 inch occurring in July. Construct a function that models this behavior. During what period is there more than 10 inches of snowfall? 25. A spring attached to a ceiling is pulled down 20 cm. After 3 seconds, wherein it completes 6 full periods, the amplitude is only 15 cm. Find the function modeling the position of the spring t seconds after being released. At what time will the spring come to rest? In this case, use 1 cm amplitude as rest. 26. Water levels near a glacier currently average 9 feet, varying seasonally by 2 inches above and below the average and reaching their highest point in January. Due to global warming, the glacier has begun melting faster than normal. Every year, the water levels rise by a steady 3 inches. Find a function modeling the depth of the water t months from now. If
the docks are 2 feet above current water levels, at what point will the water first rise above the docks? 8 Further Applications of Trigonometry Figure 1 General Sherman, the world’s largest living tree. (credit: Mike Baird, Flickr) CHAPTeR OUTlIne 8.1 non-right Triangles: law of Sines 8.2 non-right Triangles: law of Cosines 8.3 Polar Coordinates 8.4 Polar Coordinates: Graphs 8.5 Polar Form of Complex numbers 8.6 Parametric equations 8.7 Parametric equations: Graphs 8.8 Vectors Introduction The world’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California.[27] Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully. 27 Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. 643 644 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Use the Law of Sines to solve oblique triangles. • Find the area of an oblique triangle using the sine function. • Solve applied problems using the Law of Sines. 8.1 nOn-RIGHT TRIAnGleS: lAW OF SIneS Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. 15° 35
° 20 miles Figure 1 Using the law of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2. β α γ Figure 2 2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 3. β α γ Figure 3 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4. β α γ Figure 4 SECTION 8.1 non-right triangles: law oF sines 645 Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5. γ b h a α c Figure 5 β h h _ _ and sin β = Using the right triangle relationships, we know that sin α = a. Solving both equations for h gives two b different expressions for h. We then set the expressions equal to each other. h = bsin α and h = asin β bsin α = asin β Similarly, we can compare the other ratios. 1 _  ab 1 _  (bsin α) = (asin β)  ab = sin α _ a sin β _ b 1 _  Multiply both sides by. ab sin α _ a = sin γ _ c and sin β _ b = sin γ _ c Collectively, these relationships are called the
Law of Sines. sin α _ a = sin β _ b = sin γ _ c Note the standard way of labeling triangles: angle α (alpha) is opposite side a; angle β (beta) is opposite side b; and angle γ (gamma) is opposite side c. See Figure 6. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. β c a α b Figure 6 γ Law of Sines Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. = = sin α _ a a _ sin α = sin β _ b b _ sin β = sin γ _ c c _ sin γ To solve an oblique triangle, use any pair of applicable ratios. 646 CHAPTER 8 Further applications oF trigonometry Example 1 Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in Figure 7 to the nearest tenth. β c 50° α 10 b Figure 7 30° γ Solution The three angles must add up to 180 degrees. From this, we can determine that β = 180° − 50° − 30° = 100° To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c. = sin(30°) _ c sin(50°) _ 10 sin(50°) _ c 10 = sin(30°) c = sin(30°) c ≈ 6.5 10 _ sin(50°) Multiply both sides by c. Multiply by the reciprocal to isolate c. Similarly, to solve for b, we set up another proportion. sin(50°) _ 10 = sin(100°) _ b bsin(50°) = 10sin(100°) Multiply both sides by b. b = 10sin(100°) _ sin(50°) b ≈ 12.9 Multiply by the reciprocal to isolate b. Therefore, the complete
set of angles and sides is α = 50° β = 100° γ = 30° a = 10 b ≈ 12.9 c ≈ 6.5 Try It #1 Solve the triangle shown in Figure 8 to the nearest tenth. β c a 43° 98° α 22 Figure 8 Using The law of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. SECTION 8.1 non-right triangles: law oF sines 647 possible outcomes for SSA triangles Oblique triangles in the category SSA may have four different outcomes. Figure 9 illustrates the solutions with the known sides a and b and known angle α. No triangle, a < h γ γ Right triangle, a = h Two triangles, a > h, a < b One trianglea) β α (b) β α (c) Figure 9 β α (d) β Example 2 Solving an Oblique SSA Triangle Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth. γ 8 6 35° α β Figure 10 Solution Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we have the proportion = = sin β _ b sin β _ 8 sin α _ a sin(35°) _ 6 8sin(35°) _ 6 0.7648 ≈ sin β sin−1(0.7648) ≈ 49.9° β ≈ 49.9° = sin β However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria. γ' 8 6 6 α' 35° β Figure 11 β' φ The angle supplementary to β is approximately equal to
49.9°, which means that β = 180° − 49.9° = 130.1°. (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ, we have γ = 180° − 35° − 130.1° ≈ 14.9° 648 CHAPTER 8 Further applications oF trigonometry We can then use these measurements to solve the other triangle. Since γ' is supplementary to α and β, we have γ' = 180° − 35° − 49.9° ≈ 95.1° Now we need to find c and c'. We have Finally, c _ = sin(14.9°) 6 _ sin(35°) c = 6sin(14.9°) _ sin(35°) ≈ 2.7 c' _ = sin(95.1°) 6 _ sin(35°) c' = 6sin(95.1°) _ sin(35°) ≈ 10.4 To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12. γ 14.9° a = 6 130.1° β 35° α' Figure 12 γ' 95.1° a' = 6 49.9° β' b' = 8 c' ≈ 10.4 (b) b = 8 35° α c ≈ 2.7 (a) However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first triangle (a) in Figure 12. Try It #2 Given α = 80°, a = 120, and b = 121, find the missing side and angles. If there is more than one possible solution, show both. Example 3 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth. 12 β a α 85° 9 Figure 13 Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85°, and its corresponding side c = 12, and we know side b = 9. We will use this proportion to solve for β. sin(85°) _ 12 = sin β _
9 9sin(85°) _ 12 = sin β Isolate the unknown. SECTION 8.1 non-right triangles: law oF sines 649 To find β, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. 9sin(85°)  β = sin−1  _ 12 β ≈ sin−1(0.7471) β ≈ 48.3° In this case, if we subtract β from 180°, we find that there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7°. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives α = 180° − 85° − 131.7° ≈ − 36.7°, which is impossible, and so β ≈ 48.3°. To find the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7°. Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions. sin(85°) _ 12 sin(85°) _ 12 a = sin(46.7°) _ a = sin(46.7°) a = 12sin(46.7°) __ sin(85°) ≈ 8.8 The complete set of solutions for the given triangle is α ≈ 46.7° β ≈ 48.3° γ = 85° a ≈ 8.8 b = 9 c = 12 Try It #3 Given α = 80°, a = 100, b = 10, find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. Example 4 Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Solution Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14. = sin α _ 10 sin(50°) _ 4 10sin(50°) _________ 4 sin α ≈ 1.
915 sin α = α 4 50° 10 Figure 14 We can stop here without finding the value of α. Because the range of the sine function is [−1, 1], it is impossible for the sine value to be 1.915. In fact, inputting sin−1 (1.915) in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions. Try It #4 Determine the number of triangles possible given a = 31, b = 26, β = 48°. 650 CHAPTER 8 Further applications oF trigonometry Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the 1 _ area of an oblique triangle. Recall that the area formula for a triangle is given as Area = bh, where b is base and h is 2 property sin α = height. For oblique triangles, we must find h before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric h _ to write an equation for area in oblique triangles. In the acute triangle, we have sin α = c or csin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α', or 180 − α. opposite _ hypotenuse'α γ Figure 15 a b γ Thus, Similarly, 1 1 _ _ Area = (base)(height) = b(csin α) 2 2 1 1 _ _ a(bsin γ) = Area = a(csin β) 2 2 area of an oblique triangle The formula for the area of an oblique triangle is given by 1 _ Area = bcsin α 2 1 _ = acsin β 2 1 _ = absin γ 2 This is equivalent to one-half of the product of two sides and the sine of their included angle. Example 5 Finding the Area of an Oblique Triangle Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102°. Round the area to the nearest integer. Solution Using the formula, we have 1 _ Area = absin γ 2 1 _ Area = (
90)(52)sin(102°) 2 Area ≈ 2289 square units Try It #5 Find the area of the triangle given β = 42°, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth. Solving Applied Problems Using the law of Sines The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. SECTION 8.1 non-right triangles: law oF sines 651 Example 6 Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile. a 15° 20 miles Figure 16 35° Solution To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a, and then use right triangle relationships to find the height of the aircraft, h. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180° − 15° − 35° = 130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. sin(130°) _ = 20 sin(35°) _ a asin(130°) = 20sin(35°) a = 20sin(35°) _ sin(130°) a ≈ 14.98 The distance from one station to the aircraft is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for h. sin(15°) = opposite _ hypotenuse h _ sin(15°) = a h _ 14.98 sin(15°) = h = 14.98sin(15°) h ≈ 3.88 The aircraft is at an altitude of approximately 3.9 miles. Try It #6 The diagram shown in Figure 17 represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B, is 62°, and the distance between the viewing points of the two end zones is 145 yards. C 70° A 145 yards Figure 17 62° B Access the following online resources for additional instruction and practice with trig
onometric applications. • law of Sines: The Basics (http://openstaxcollege.org/l/sinesbasic) • law of Sines: The Ambiguous Case (http://openstaxcollege.org/l/sinesambiguous) 652 CHAPTER 8 Further applications oF trigonometry 8.1 SeCTIOn exeRCISeS VeRBAl 1. Describe the altitude of a triangle. 3. When can you use the Law of Sines to find a missing angle? 2. Compare right triangles and oblique triangles. 4. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator? 5. What type of triangle results in an ambiguous case? AlGeBRAIC For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 6. α = 43°, γ = 69°, a = 20 9. a = 4, α = 60°, β = 100° 7. α = 35°, γ = 73°, c = 20 10. b = 10, β = 95°, γ = 30° 8. α = 60°, β = 60°, γ = 60° For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. 11. Find side b when A = 37°, B = 49°, c = 5. 13. Find side c when B = 37°, C = 21, b = 23. 12. Find side a when A = 132°, C = 23°, b = 10. For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. 14. α = 119°, a = 14, b = 26 17. a = 12, c = 17, α = 35° 20. a = 7, b = 3, β = 24° 23. β = 119°, b =
8.2, a = 11.3 15. γ = 113°, b = 10, c = 32 18. a = 20.5, b = 35.0, β = 25° 21. b = 13, c = 5, γ = 10° 16. b = 3.5, c = 5.3, γ = 80° 19. a = 7, c = 9, α = 43° 22. a = 2.3, c = 1.8, γ = 28° For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 24. Find angle A when a = 24, b = 5, B = 22°. 26. Find angle B when A = 12°, a = 2, b = 9. 25. Find angle A when a = 13, b = 6, B = 20°. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a = 5, c = 6, β = 35° 30. a = 7.2, b = 4.5, γ = 43° GRAPHICAl 28. b = 11, c = 8, α = 28° 29. a = 32, b = 24, γ = 75° For the following exercises, find the length of side x. Round to the nearest tenth. 31. 32. 33. 10 x 70° 50° 25° 6 120° x 45° 15 x 75° SECTION 8.1 section exercises 653 34. 35. 18 x 40° 110° 14 50° x 42° 36. 8.6 111° 22° x For the following exercises, find the measure of angle x, if possible. Round to the nearest tenth. 98° 5 38. 37° 8 39. 5 x 13 22° 37. 40. x 5.7 x 10 5.3 59° x 11 41. Notice that x is an obtuse angle. 42. 65° 12 x 10 21 24 x 55° For the following exercises, solve the triangle. Round each answer to the nearest tenth. 43. A 24.1 32.6 93° B C For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. 44. 45. 18 46. 16 30° 10 25
° 15 4.5 51° 2.9 47. 58° 9 11 51° 48. 49. 25 18 40° 30° 3.5 115° 30 50 654 CHAPTER 8 Further applications oF trigonometry exTenSIOnS 50. Find the radius of the circle in Figure 18. Round to 51. Find the diameter of the circle in Figure 19. Round the nearest tenth. to the nearest tenth. 145° 3 8.3 110° Figure 18 Figure 19 52. Find m ∠ADC in Figure 20. Round to the nearest 53. Find AD in Figure 21. Round to the nearest tenth. tenth. A 10 9 8 60° B C D Figure 20 A 13 Figure 21 12 53° B 44° D C 54. Solve both triangles in Figure 22. Round each 55. Find AB in the parallelogram shown in Figure 23. answer to the nearest tenth. A 48° 4.2 B 46° E Figure 22 48° C 2 D B A 130° 10 12 130° D C Figure 23 56. Solve the triangle in Figure 24. (Hint: Draw a 57. Solve the triangle in Figure 25. (Hint: Draw a perpendicular from H to JK). Round each answer to the nearest tenth. perpendicular from N to LM). Round each answer to the nearest tenth. H 7 10 Figure 24 20° J K L 5 74° M 4.6 Figure 25 N SECTION 8.1 section exercises 655 58. In Figure 26, ABCD is not a parallelogram. ∠m is obtuse. Solve both triangles. Round each answer to the nearest tenth. A m x 35° n 29 45 h k D B y 40 Figure 26 65° C ReAl-WORlD APPlICATIOnS 59. A pole leans away from the sun at an angle of 7° to the vertical, as shown in Figure 27. When the elevation of the sun is 55°, the pole casts a shadow 42 feet long on the level ground. How long is the pole? Round the answer to the nearest tenth. 60. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in Figure 28. Determine the distance of the boat from station A and the distance of the boat from shore. Round your answers to the nearest whole foot. Figure 27 70° 60° A B Figure 28 61. Figure 29
shows a satellite orbiting Earth. The satellite passes directly over two tracking stations A and B, which are 69 miles apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 86.2° and 83.9°, respectively. How far is the satellite from station A and how high is the satellite above the ground? Round answers to the nearest whole mile. 62. A communications tower is located at the top of a steep hill, as shown in Figure 30. The angle of inclination of the hill is 67°. A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. The angle formed by the guy wire and the hill is 16°. Find the length of the cable required for the guy wire to the nearest whole meter. 83.9° A 86.2° B Figure 29 16° 165m 67° Figure 30 656 CHAPTER 8 Further applications oF trigonometry 63. The roof of a house is at a 20° angle. An 8-foot solar panel is to be mounted on the roof and should be angled 38° relative to the horizontal for optimal results. (See Figure 31). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. 64. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 6.6 km apart, to be 37° and 44°, as shown in Figure 32. Find the distance of the plane from point A to the nearest tenth of a kilometer. 8 ft 38° 20° Figure 31 65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33. Find the distance of the plane from point A to the nearest tenth of a kilometer. 32° 56° A Figure 33 B 37° 44° A B Figure 32 66. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move
300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot. 67. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 35°. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. 1 _ miles apart spot 69. A man and a woman standing 3 2 a hot air balloon at the same time. If the angle of elevation from the man to the balloon is 27°, and the angle of elevation from the woman to the balloon is 41°, find the altitude of the balloon to the nearest foot. 68. Points A and B are on opposite sides of a lake. Point C is 97 meters from A. The measure of angle BAC is determined to be 101°, and the measure of angle ACB is determined to be 53°. What is the distance from A to B, rounded to the nearest whole meter? 70. Two search teams spot a stranded climber on a mountain. The first search team is 0.5 miles from the second search team, and both teams are at an altitude of 1 mile. The angle of elevation from the first search team to the stranded climber is 15°. The angle of elevation from the second search team to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. SECTION 8.1 section exercises 657 71. A street light is mounted on a pole. A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. A 6-foot-tall woman is standing on the same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. 73. Two streets meet at an 80° angle. At the corner, a park is
being built in the shape of a triangle. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. 72. Three cities, A, B, and C, are located so that city A is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 miles from city B, how far is city A from city B? Round the distance to the nearest tenth of a mile. 74. Brian’s house is on a corner lot. Find the area of the front yard if the edges measure 40 and 56 feet, as shown in Figure 34. House 56 ft 135° 40 ft Figure 34 75. The Bermuda triangle is a region of the Atlantic 76. A yield sign measures 30 inches on all three sides. What is the area of the sign? Ocean that connects Bermuda, Florida, and Puerto Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. 77. Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35. 4.5 feet 32° Figure 35 4 feet 42° 658 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Use the Law of Cosines to solve oblique triangles. • Solve applied problems using the Law of Cosines. Use Heron’s formula to find the area of a triangle. • 8.2 nOn-RIGHT TRIAnGleS: lAW OF COSIneS Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1. How far from port is the boat? 8 mi 20° 10 mi Port Figure 1 Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side
-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases. Using the law of Cosines to Solve Oblique Triangles The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level. Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted. y C (b cosθ, b sinθ) A (0, 0 Figure 2 SECTION 8.2 non-right triangles: law oF cosines 659 We can drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that cos θ = x(adjacent) __ b(hypotenuse) and sin θ = y(opposite) __ b(hypotenuse) In terms of θ, x = bcos θ and y = bsin θ. The (x, y) point located at C has coordinates (bcos θ, bsin θ). Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagorean Theorem. Thus, a2 = (x − c)2 + y2 = (bcos θ − c)2 + (bsin θ)2 Substitute (bcos θ) for x and (bsin θ) for y. = (b2 cos2 θ − 2bccos θ + c
2) + b2 sin2 θ Expand the perfect square. = b2 cos2 θ + b2 sin2 θ + c2 − 2bccos θ Group terms noting that cos2 θ + sin2 θ = 1. = b2(cos2 θ + sin2 θ) + c2 − 2bccos θ Factor out b2. a2 = b2 + c2 − 2bccos θ The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve. Law of Cosines The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 3, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations. a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ β a To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle. c α b Figure 3 γ cos α = b2 + c2 − a2 _ 2bc cos β = a2 + c2 − b2 _ 2ac cos γ = a2 + b2 − c2 _ 2ab How To… Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle. 1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles. 2. Apply the Law of Cosines to find the length of the unknown side or angle. 3. Apply the Law of Sines or Cosines
to find the measure of a second angle. 4. Compute the measure of the remaining angle. 660 CHAPTER 8 Further applications oF trigonometry Example 1 Finding the Unknown Side and Angles of a SAS Triangle Find the unknown side and angles of the triangle in Figure 4. γ b a 5 10 α c 5 12 Figure 4 30° β Solution First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines. Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b, as we know the measurement of the opposite angle β. b2 = a2 + c2−2accos β b2 = 102 + 122 − 2(10)(12)cos(30°) — 3 √  b2 = 100 + 144 − 240  _ 2 b2 = 244 − 120 √ b = √ b ≈ 6.013 — — 3 244 − 120 √ 3 — Substitute the measurements for the known quantities. Evaluate the cosine and begin to simplify. Use the square root property. Because we are solving for a length, we use only the positive square root. Now that we know the length b, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α, we have sin β _ b = sin α _ a sin α _ 10 = sin(30°) _ 6.013 sin α = 10sin(30°) _ 6.013 10sin(30°)  _ 6.013) α = sin−1  α ≈ 56.3° Multiply both sides of the equation by 10. Find the inverse sine of 10sin(30°) _ 6.013. The other possibility for α would be α = 180° − 56.3° ≈ 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0° and 180°. Proceeding with α ≈ 56.3

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