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these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions. Cofunction Identities The cofunction identities are summarized in Table 9.6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 ... |
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 presented earlier may help simplify the process of verifying an identity. 1002 Chapter 9 Trigonometric Identities and Equations Given an identity, verify using ... |
9.10 Verify the identity: tan(π − θ) = − tan θ. Example 9.20 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 9.12. Show that tan θ = m2 − m1 1 + m1 m2 where m1 and m2 are the slope... |
tan α = 5 4 + 3.76 tan α = 4.7 − 5 tan α ⎞ ⎠ − 5 tan α 47 50 ⎛ ⎝ 5 tan α + 3.76 tan α = 0.7 8.76 tan 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 thi... |
sin ⎝ 11π 12 ⎞ ⎠ ⎛ tan ⎝− ⎞ ⎠ π 12 ⎛ tan ⎝ 19π 12 ⎞ ⎠ For the following exercises, rewrite in terms of sin x and cos x. 52. 53. 54. 55. ⎛ ⎝x + 11π sin 6 ⎞ ⎠ ⎛ sin ⎝x − 3π 4 ⎞ ⎠ ⎛ ⎝x − 5π cos 6 ⎞ ⎠ ⎛ ⎝x + 2π cos 3 ⎞ ⎠ For the following exercises, simplify the given expression. 56. ⎛ csc ⎝ 57. ⎛ sec ⎝ − t⎞ ⎠ − θ⎞ ⎠ π 2 ... |
��cos−1 ⎛ cos ⎝ ⎞ ⎠ + sin−sin−1 ⎛ tan ⎝ 1 2 ⎞ ⎠ − cos−1 ⎛ ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Graphical For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator. 67. ⎛ cos ⎝ π 2 − x⎞ ⎠ 68. sin(π − x) ⎛ tan ⎝ π 3 + x⎞... |
�� ⎛ sin ⎝ π 4 + x⎞ ⎠ ⎛ cos ⎝ 5π 4 + x⎞ ⎠ 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. ) 75. f (x) = sin(4x) − si... |
1 Use double-angle formulas to find exact values. 9.3.2 Use double-angle formulas to verify identities. 9.3.3 Use reduction formulas to simplify an expression. 9.3.4 Use half-angle formulas to find exact values. Figure 9.14 Bicycle ramps for advanced riders have a steeper incline than those designed for novices. Bicycl... |
Equations The second variation is: cos(2θ) = cos2 θ − sin2 θ = cos2 θ − = 2 cos2 θ − 1 ⎝1 − cos2 θ⎞ ⎛ ⎠ Similarly, to derive the double-angle formula for tangent, replacing α = β = θ in the sum formula gives tan α + tan β tan(α + β) = 1 − tan α tan β tan(θ + θ) = tan θ + tan θ 1 − tan θ tan θ 2tan θ 1 − tan2 θ tan(2θ)... |
)2 + (3)2 = c2 16 + 9 = c2 25 = c2 c = 5 Now we can draw a triangle similar to the one shown in Figure 9.15. Figure 9.15 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 9.15, we see that the hypotenuse equals 5, so sin ... |
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... |
− tan θ tan θ 1 Double-angle formula Multiply by a term that results in desired numerator. ⎞ ⎠ Use reciprocal identity for 1 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 rewri... |
formula cos(2θ) = 2 cos2 θ − 1. Solve for cos2 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: Substitute the reduction formulas. tan2 θ = sin2 θ cos2 θ 1 − cos(2θ) 2 1 + cos(2θ) 2 = ⎛ ⎞ ⎠ ⎝ 2 1 + cos(2θ) ⎞ ⎠... |
the power-reducing formulas to prove sin3 (2x) = ⎤ sin(2x) ⎦ ⎡ ⎣1 − cos(4x)⎤ ⎦ ⎡ ⎣ 1 2 Solution We will work on simplifying the left side of the equation: ⎤ ⎡ sin3(2x) = [sin(2x)] ⎣sin2(2x) ⎦ 1 − cos(4x) ⎡ = sin(2x) ⎣ 2 ⎤ ⎦ ⎛ = sin(2x) ⎝ 1 2 ⎞ ⎠[1 − cos(4x)] = 1 2 [sin(2x)][1 − cos(4x)] Substitute the power-reduction ... |
⎛ ⎝ α 2 ⎛ cos ⎝ α 2 For the tangent identity, we have cos2 θ = ⎞ ⎠ ⎞ ⎠ = 1 + cos(2θ) 2 ⎛ ⎝2 ⋅ α 1 + cos 2 2 = 1 + cos α 2 ⎠ = ± 1 + cos α ⎞ 2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1015 tan2 θ = ⎞ ⎠ = 1 − cos(2θ) 1 + cos(2θ) ⎛ ⎝2 ⋅ α ... |
. Substitute values into the formula based on the triangle. 4. Simplify. Example 9.29 Finding Exact Values Using Half-Angle Identities Given that tan α = 8 15 and α lies in quadrant III, find the exact value of the following: a. b. c. ⎛ sin ⎝ ⎞ ⎠ α 2 ⎛ cos ⎝ ⎞ ⎠ α 2 ⎛ tan ⎝ ⎞ ⎠ α 2 Solution Using the given information,... |
substitute the value of the cosine We choose the negative value of cos α 2 quadrant II. c. To find tan α 2 we found from the triangle in Figure 9.16 and simplify. tan α 2 = ± 1 − cos α 1 + cos α 1 − ( − 15 17) 1 + ( − 15 17) = ± = ± 32 17 2 17 = ± 32 2 = − 16 = −4 1018 Chapter 9 Trigonometric Identities and Equations ... |
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) 1020 Chapter 9 Trigonometric Identities and Equations 9.3 EXERCISE... |
�� ⎛ ⎠ 2 cos ⎝ ⎞ ⎠ π 4 ⎛ 4 sin ⎝ π 8 ⎛ ⎞ ⎠ cos ⎝ ⎞ ⎠ π 8 This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, find the exact value using halfangle formulas. 111. ⎛ sin ⎝ ⎞ ⎠ π 8 112. ⎛ cos ⎝− 11π 12 ⎞ ⎠ 113. ⎛ sin ⎝ 11π 12 ⎞ ⎠ 114. ⎛ cos ⎝ ⎞ ⎠ 7π 8 115. ⎛ tan ⎝ ⎞ ⎠ 5π ... |
tan4 x cos2 x 148. cos2 x sin(2x) 149. cos2 (2x)sin x 150. tan2 ⎛ ⎝ x 2 ⎞ ⎠ sin x the following exercises, For algebraically find an equivalent function, only in terms of sin x and/or cos x, and then check the answer by graphing both functions. 151. sin(4x) 152. cos(4x) Extensions For the following exercises, prove th... |
− sin(2t) 133. sin(2x) = − 2 sin(−x) cos(−x) 134. cot x − tan x = 2 cot(2x) 135. sin(2θ) 1 + cos(2θ) tan2 θ = tan θ For the following exercises, rewrite the expression with an exponent no higher than 1. 136. cos2(5x) 137. cos2(6x) 138. sin4(8x) 139. sin4(3x) 140. cos2 x sin4 x 141. cos4 x sin2 x 142. tan2 x sin2 x Tec... |
+ β) ___________________________________ 2 cos α cos β = cos(α − β) + cos(α + β) Then, we divide by 2 to isolate the product of cosines: cos α cos β = 1 2 [cos(α − β) + cos(α + β)] 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 for... |
_________________________________________ sin(α + β) + sin(α − β) = 2 sin α cos β Then, we divide by 2 to isolate the product of cosine and sine: sin α cos β = 1 2 ⎡ ⎣sin⎛ ⎝α + β⎞ ⎠ + sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ 1024 Chapter 9 Trigonometric Identities and Equations Example 9.32 Writing the Product as a Sum Containing only Sine... |
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: [cos(α − β) + cos(α + β)] cos α cos β = 1 2 sin α cos β = 1 [sin(α + β) + sin(α − β)] 2 sin α sin β = 1 2 cos α sin β = 1 2 [sin(α + β) − sin(α − β)] [cos(α − β) − co... |
�� ⎞ ⎠cos ⎝ ⎛ ⎞ ⎠cos ⎝ u − v 2 u − v 2 [sin u + sin v] ⎛ sin ⎝ ⎛ 2 sin ⎝ u + v 2 u + v 2 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 = 2sin ⎝ α + β ⎛ sin α + sin β = 2sin ⎝ 2 α − β 2 α + ... |
� 9.19 Use the sum-to-product formula to write the sum as a product: sin(3θ) + sin(θ). Example 9.35 Evaluating Using the Sum-to-Product Formula Evaluate cos(15°) − cos(75°). Check the answer with a graphing calculator. Solution We begin by writing the formula for the difference of cosines. ⎛ cos α − cos β = − 2 sin ⎝ α... |
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 9.37 Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities Verify the identity csc2 θ − 2 = cos(2θ) sin2 θ. Solution For... |
Describe a situation where we would convert an 165. equation from a product to a sum, and give an example. 184. 2 sin(100°)sin(20°) Algebraic 185. 2 sin(−100°)sin(−20°) For the following exercises, rewrite the product as a sum or difference. 186. sin(213°)cos(8°) 166. 16 sin(16x)sin(11x) 167. 20 cos(36t)cos(6t) 168. 2... |
3 x − cos x sin2 x⎞ ⎛ ⎠ = cos(3x) + cos x 2 2 tan x cos(3x) = sec x⎛ ⎝sin(4x) − sin(2x)⎞ ⎠ 199. cos(a + b) + cos(a − b) = 2 cos a cos b Chapter 9 Trigonometric Identities and Equations Extensions For the following exercises, prove the following sum-toproduct formulas. 215. ⎛ sin x − sin y = 2 sin ⎝ x − y 2 ⎛ ⎞ ⎠cos ⎝ x... |
Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 200. cos(58°) + cos(12°) 201. sin(2°) − sin(3°) 202. cos(44°) − cos(22°) 203. cos(176°)sin(9°) 204. sin(−14°)sin(85°) Technology the following exercises, algebraically determine For whether each ... |
onometric equations using a calculator. 9.5.4 Solve trigonometric equations that are quadratic in form. 9.5.5 Solve trigonometric equations using fundamental identities. 9.5.6 Solve trigonometric equations with multiple angles. 9.5.7 Solve right triangle problems. Figure 9.21 Egyptian pyramids standing near a modern ci... |
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 1032 Chapter 9 Trigonometric Identities and Equations a vari... |
θ = −2 cos θ = −1 θ = π 9.21 Solve exactly the following linear equation on the interval [0, 2π) : 2 sin 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 th... |
third and fourth quadrants. Example 9.43 Solving an Equation Involving Tangent ⎛ Solve the equation exactly: tan ⎝θ − ⎞ ⎠ = 1, 0 ≤ θ < 2π. π 2 Solution Recall that the tangent function has a period of π. On the interval [0, π), and at the angle of π 4, the tangent has a value of 1. However, the angle we want is ⎛ ⎝θ −... |
. 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... |
± 2πk and θ ≈ 4.4597 ± 2πk. 9.23 Solve cos θ = − 0.2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 9 Trigonometric Identities and Equations 1037 Solving Trigonometric Equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algeb... |
of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is θ = 2π − cos−1 ⎛ ⎝ −3 + 13 2 ⎞ ⎠ ≈ 5.02 Example 9.48 1038 Chapter 9 Trigonometric Identities and Equations Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: 2 sin2 θ − 5 sin θ... |
� = − 1 2, 11π θ = 7π 6 6, 11π 6 The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, 7π 6. 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. 2 sin2 θ + sin θ = 0 sin θ(2sin θ + 1) = 0 sin θ = 0 θ = 0, π 2 sin θ + 1 = 0 2sin θ... |
the equation. Example 9.51 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2π. cos x cos(2x) + sin x sin(2x) = 3 2 Solution Notice that the left side of the equation is the difference formula for cosine. cos x cos(2x) + sin x sin(2x) = 3 2 cos(x ... |
− 2cos2 θ 2 cos2 θ + 3 cos θ + 1 = 0 (2 cos θ + 1)(cos θ + 1) = 0 2 cos θ + 1 = 0 cos θ = − 1 2, 4π θ = 2π 3 3 cos θ + 1 = 0 cos θ = −1 θ = π Our solutions are θ = 2π 3, 4π 3, π. 1042 Chapter 9 Trigonometric Identities and Equations Solving Trigonometric Equations with Multiple Angles Sometimes it is not possible to s... |
so x = π 3 as noted. Let us revolve around the circle again: 2x = + 2π + 6π 3 π 3 π = 3 = 7π 3 so x = 7π 6. One more rotation yields 2x = + 4π + 12π 3 π 3 π = 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 = 5π 3, so x = 5π 6 as noted. Let us rev... |
Figure 9.24 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. W... |
ations 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 Trigonome... |
2 sin(x)cos(x) − sin(x) + 2 cos(x) − 1 = 0 252. cos2 θ = 1 2 253. sec2 x = 1 254. 255. tan2 (x) = −1 + 2 tan(−x) 8 sin2(x) + 6 sin(x) + 1 = 0 256. tan5(x) = tan(x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, 2π). For the following exercises, solve exactly on [0... |
− 2 = 0 For the following exercises, find exact solutions on the interval [0, 2π). Look use trigonometric identities. opportunities for to 274. 275. sin2 x − cos2 x − sin x = 0 sin2 x + cos2 x = 0 276. sin(2x) − sin x = 0 277. cos(2x) − cos x = 0 278. 2 tan x 2 − sec2 x − sin2 x = cos2 x 279. 1 − cos(2x) = 1 + cos(2x)... |
0 4 sin2 x + sin(2x)sec x − 3 = 0 Chapter 9 Trigonometric Identities and Equations 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? A 20-foot tall building has a shadow that is 55 feet 324. long. What i... |
2 x − 1 − sec3 x cos x = 0 sin(2x) sec2 x = 0 sin(2x) 2csc2 x = 0 2 cos2 x − sin2 x − cos x − 5 = 0 1 sec2 x + 2 + sin2 x + 4 cos2 x = 4 Real-World Applications An airplane has only enough gas to fly to a city 200 317. miles northeast of its current location. If the pilot knows that the city is 25 miles north, how many... |
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 function identities derived from the double-angle formulas and used to reduce the power of a trigonometric sum-to-product fo... |
tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β tan θ = cot ⎛ cos θ = sin ⎝ π ⎛ sin θ = cos ⎝ ⎛ sec θ = csc ⎝ 2 π 2 ⎛ csc θ = sec ⎝ ⎛ cot θ = tan ⎝ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ − θ⎞ ⎠ 1052 Chapter 9 Trigonometric Identities and Equations Double-angle formulas Reduction formulas sin(2θ) = 2sin θ cos θ cos(2θ)... |
�� ⎛ ⎞ ⎠cos ⎠ ⎝ 2 α − β ⎛ ⎞ ⎠sin ⎝ 2 α − β ⎛ ⎞ ⎠cos ⎝ 2 ⎞ ⎠ ⎞ ⎠ KEY CONCEPTS 9.1 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 ide... |
ines 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 9.13. • The sum and difference formulas for sine and cosine can also be used for inv... |
sines and cosines. See Example 9.31, Example 9.32, and Example 9.33. 1054 Chapter 9 Trigonometric Identities and Equations • 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, o... |
55. CHAPTER 9 REVIEW EXERCISES Solving Trigonometric Equations with Identities For the following exercises, find all solutions exactly that exist on the interval [0, 2π). 331. csc2 t = 3 332. cos2 x = 1 4 333. 2 sin θ = −1 334. tan x sin x + sin(−x) = 0 335. 9 sin ω − 2 = 4 sin2 ω 336. 1 − 2 tan(ω) = tan2(ω) 338. sin3 ... |
⎠ 8 x⎞ ⎛ x⎞ ⎛ 1 1 ⎠tan 1 − tan ⎠ ⎝ ⎝ 2 8 For the following exercises, rewrite the expression with no powers. 358. cos2 x sin4(2x) For the following exercises, find the exact value. 359. tan2 x sin3 x 348. ⎝sin−1 (0) − cos−1 ⎛ ⎛ cos ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 349. ⎝sin−1 (0) + sin−1 ⎛ ⎛ tan ⎝ ⎞ ⎞ ⎠ ⎠ 1 2 Double-Angle, Half-Angle, a... |
�� cos ⎝ ⎞ ⎠ π 3 For the following exercises, evaluate the sum by using a product formula. Write the exact answer. 363. ⎛ sin ⎝ π 12 ⎞ ⎛ ⎠ − sin ⎝ ⎞ ⎠ 7π 12 364. ⎛ 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. 365. sin(9x)cos(3x) 366... |
cos2 x + 21 cos x + 1 = 0 sec2 x − 2 sec x = 15 379. CHAPTER 9 PRACTICE TEST For the following exercises, simplify the given expression. 380. cos(−x)sin x cot x + sin2 x 381. sin(−x)cos(−2x)−sin(−x)cos(−2x) 382. ⎞ ⎛ ⎝sec2 θ − 1 csc(θ)cot(θ) ⎠ 383. ⎞ ⎞ ⎛ ⎛ ⎝1 + tan2 (θ) ⎝1 + cot2 (θ) cos2 (θ)sin2 (θ) ⎠ ⎠ For the follow... |
x) sin x − cos(2x) cos x = sec x 394. cos(2x) + sin2 x = 0 395. 2 sin2 x − sin x = 0 396. Rewrite the expression as a product instead of a sum: cos(2x) + cos(−8x). For the following exercise, rewrite the product as a sum or difference. 397. 8cos(15x)sin(3x) For the following exercise, rewrite the sum or difference as a... |
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. governed 412. Two frequencies of sound are played on an equation instrument b... |
, stands 274.9 feet tall and resides in Northern California.[1] 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 practi... |
measurements of two angles and a side that is not between the known angles. See Figure 10.4. 1. Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Fu... |
7, 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 ... |
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. Possible Outcomes for SSA Triangles Oblique triangles in the category SSA may have four different outco... |
9°) sin(35°) ≈ 2.7 c′ sin(95.1°) = c′ = 6 sin(35°) 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 10.13. 1066 Chapter 10 Further Applications of Trigonometry Figure 10.13 However, we were looking for the values... |
is impossible, and so β ≈ 48.3°. α = 180° − 85° − 131.7° ≈ − 36.7°, 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(46.7 ° ) a = sin(46.7 ° ) sin(85 ° ) 12 asin(85 ° ) 12 a = 12sin(46.7 °... |
the area formula. Observing the two triangles in Figure 10.16, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = h c or csin α = h... |
10 Further Applications of Trigonometry Figure 10.17 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... |
° 19. a = 7, c = 9, α = 43° 20. a = 7, b = 3, β = 24° In the Law of Sines, what is the relationship between the 4. angle in the numerator and the side in the denominator? 21. b = 13, c = 5, γ = 10° 5. What type of triangle results in an ambiguous case? 22. a = 2.3, c = 1.8, γ = 28° Algebraic 23. β = 119°, b = 8.2, a = ... |
for free at https://cnx.org/content/col11758/1.5 24. Find angle A when a = 24, b = 5, B = 22°. 25. Find angle A when a = 13, b = 6, B = 20°. 26. Find angle B when A = 12°, a = 2, b = 9. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a... |
.26 Figure 10.23 Find AB in the parallelogram shown in Figure 55. 10.24. In Figure 10.27, ABCD is not a parallelogram. 58. ∠ m is obtuse. Solve both triangles. Round each answer to the nearest tenth. 1076 Chapter 10 Further Applications of Trigonometry Figure 10.29 61. Figure 10.30 shows a satellite orbiting Earth. The... |
. (See Figure 10.32). How long does the vertical support holding up the back of the panel need to be? Round to the nearest tenth. Figure 10.32 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 ... |
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. A street light is mounted on a pole. A 6-foot-tall man is 71. standing on the st... |
shown in Figure 10.36. Figure 10.36 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1079 10.2 | Non-right Triangles: Law of Cosines Learning Objectives In this section, you will: 10.2.1 Use the Law of Cosines to solve oblique triangles. 10.2.2 ... |
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 ri... |
three equations. a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ (10.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1081 Figure 10.39 To solve for a missing side measurement, the corresponding opposite angle measu... |
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 α a = sin β b = sin α 10 sin α = sin(30°) 6.013 10sin(30°) 6.013 α = sin−1 ⎛ ⎝ α ≈ 56.3° 10sin(30°) 6.013 Multiply both sides of the equatio... |
(0.61) ≈ α α ≈ 52.4° Substitute the appropriate measurements. Simplify in each step. Isolate cos α. Find the inverse cosine. See Figure 10.41. Figure 10.41 Analysis Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method. 10.8 Given a = 5, b = 7... |
5050)2 − (6000)2 −2(5050)(6000) = cos θ cos θ ≈ 0.9183 θ ≈ cos−1(0.9183) θ ≈ 23.3° To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 10.43. This forms two right triangles, although we on... |
sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known. Heron’s Formula Heron’s formula finds the area of oblique triangles in which... |
= 70(70 − 62.4)(70 − 43.5)(70 − 34.1) Area = 506,118.2 Area ≈ 711.4 The developer has about 711.4 square meters. 10.10 Find the area of a triangle given a = 4.38 ft, b = 3.79 ft, and c = 5.22 ft. Access these online resources for additional instruction and practice with the Law of Cosines. • Law of Cosines (http://ope... |
67°, a = 49, b = 38 90. α = 43.1°, a = 184.2, b = 242.8 91. α = 36.6°, a = 186.2, b = 242.2 92. β = 50°, a = 105, b = 45 For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth. 93. a = 42, b = 19, c = 30; find angle A. 94. a = 14, b = 13, ... |
10.47 For the following exercises, solve for the unknown side. Round to the nearest tenth. 120. 121. 122. 123. 125. 126. 127. 128. For the following exercises, find the area of the triangle. Round to the nearest hundredth. Chapter 10 Further Applications of Trigonometry 1092 Extensions 129. A parallelogram has sides o... |
of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed. Figure 10.52 Two ships left a port at the same time. One ship 143. traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°.... |
have 152. lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral? 1094 Chapter 10 Further Applications of Trigonometry 153. The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the... |
coordinate r is the radius or length of the directed line segment from the pole. The angle θ, measured in radians, indicates the direction of r. We move counterclockwise from the polar axis by an angle of θ, and measure a directed line segment the length of r in the direction of θ. Even though we measure θ first and t... |
to the graph of the polar coordinate ⎛ ⎝2, ⎞ ⎠ shown in Figure 10.57(b). π 6 Figure 10.57 10.12 Plot the points ⎛ ⎝3, − ⎞ ⎠ π 6 and ⎛ ⎝2, 9π 4 ⎞ ⎠ on the same polar grid. Converting from Polar Coordinates to Rectangular Coordinates When given a set of polar coordinates, we may need to convert them to rectangular coord... |
ometry 1099 Figure 10.59 Example 10.16 Writing Polar Coordinates as Rectangular Coordinates Write the polar coordinates (−2, 0) as rectangular coordinates. Solution See Figure 10.60. Writing the polar coordinates as rectangular, we have x = rcos θ x = −2cos(0) = −2 y = rsin θ y = −2sin(0) = 0 The rectangular coordinate... |
. For example, the points ⎝−3 2, 5π ⎝−3 2, 5π ⎝3 2, π ⎞ ⎠ will coincide with the original solution of ⎛ ⎛ ⎞ ⎠ 4 4 4 indicates a move further counterclockwise by π, which is directly opposite π. The radius is expressed as 4 ⎝3 2, − 7π 4 ⎠. The point ⎛ ⎞ ⎠ and ⎛ ⎞ − 3 2. However, the angle 5π 4 is located in the third qu... |
x = rcos θ and y = rsin θ, we can substitute and solve for r. (rcos θ)2 + (rsin θ)2 = 9 r 2 cos2 θ + r 2 sin2 θ = 9 r 2(cos2 θ + sin2 θ) = 9 r 2(1) = 9 r = ± 3 Substitute cos2 θ + sin2 θ = 1. Use the square root property. Thus, x2 + y2 = 9, r = 3, and r = − 3 should generate the same graph. See Figure 10.63. This cont... |
b) polar form r = 6sin θ The Cartesian or rectangular equation is plotted on the rectangular grid, and the polar equation is plotted on the polar grid. Clearly, the graphs are identical. Example 10.20 Rewriting a Cartesian Equation in Polar Form Rewrite the Cartesian equation y = 3x + 2 as a polar equation. Solution We... |
r = 3 1 − 2cos θ as a Cartesian equation. Solution 1106 Chapter 10 Further Applications of Trigonometry The goal is to eliminate θ and r, and introduce x and y. We clear the fraction, and then use substitution. In order to replace r with x and y, we must use the expression x2 + y2 = r 2. 3 1 − 2cos θ r = r(1 − 2cos θ ... |
9 3x2 + 12x − y2 = − 9 3(x2 + 4x + ) − y2 = − 9 3(x2 + 4x + 4) − y2 = − 9 + 12 3(x + 2)2 − y2 = 3 (x + 2)2 − y2 3 = 1 Multiply through by −1. Organize terms to complete the square for x. 10.15 Rewrite the polar equation r = 2sin θ in Cartesian form. Example 10.23 Rewriting a Polar Equation in Cartesian Form Rewrite th... |
gebraic the following exercises, convert the given polar For to Cartesian coordinates with r > 0 and coordinates 0 ≤ θ ≤ 2π. Remember to consider the quadrant in which is located when determining θ for the the given point point. 161. ⎛ ⎝7, 7π 6 ⎞ ⎠ 162. (5, π) 163. ⎛ ⎝6, − ⎞ ⎠ π 4 164. ⎛ ⎝−3, ⎞ ⎠ π 6 165. ⎛ ⎝4, 7π 4 ⎞ ... |
Graphical For the following exercises, find the polar coordinates of the point. 195. 196. 197. 198. 199. For the following exercises, plot the points. 200. ⎛ ⎝−2, ⎞ ⎠ π 3 201. ⎛ ⎝−1, − ⎞ ⎠ π 2 202. ⎛ ⎝3.5, 7π 4 ⎞ ⎠ 203. ⎛ ⎝−4, ⎞ ⎠ π 3 204. ⎛ ⎝5, ⎞ ⎠ π 2 205. ⎛ ⎝4, −5π 4 ⎞ ⎠ 206. ⎛ ⎝3, 5π 6 ⎞ ⎠ 207. ⎛ ⎝−1.5, 7π 6 ⎞ ⎠ 2... |
exercises, convert the equation from polar to rectangular form and graph on the rectangular plane. 217. r = 6 218. r = − 4 219. θ = − 2π 3 220. θ = π 4 221. r = sec θ 222. r = −10sin θ 223. r = 3cos θ Technology 224. coordinates of ⎛ Use a graphing calculator to find the rectangular ⎞ ⎠. Round to the nearest thousandt... |
axis (positive x-axis) by an angle of θ, and extend a ray from the pole (origin) r units in the direction of θ. All points that satisfy the polar equation are on the graph. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an... |
given the equation r = 2sin(3θ). r = 2sin(3θ) −r = 2sin(3θ) The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests, the polar ... |
) yields the same result. Thus, the graph is symmetric with respect to the line θ =. π 2 2) Replacing θ with − θ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. −r = 2sin(−θ) −r = −2sin θ Even-odd identity r = 2sin θ Multiply by −1 Pa... |
the pole is 5 units. The maximum value of the cosine function is 1 when θ = 0, so our polar equation is 5cos θ, and the value θ = 0 will yield the maximum |r|. Similarly, the maximum value of the sine function is 1 when θ =, and if our polar equation is r = 5sin θ, the value π 2 will yield the maximum |r|. We may find... |
π 3 π 2 2π 3 5π 6 π r = 2sin θ r = 2sin(0) = 0 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 π 6 ⎛ r = 2sin ⎝ π 3 ⎞ ⎠ ≈ 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 2 π 2 r 0 1 1.73 2 ⎛ r = 2sin ⎝ 2π 3 ⎞ ⎠ ≈ 1.73 1.73 ⎛ r = 2sin ⎝ ⎞ ⎠ = 1 5π 6 r = 2sin(π) = 0 1 0 Table 10.1 Figure 10.70 Without converting to Cartesian coordinates, test the given equation for symm... |
circles. Figure 10.71 Example 10.26 Sketching the Graph of a Polar Equation for a Circle Sketch the graph of r = 4cos θ. Solution First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the ± kπ. zeros and maximum |r| for r = 4cos θ. First, set r = 0, and solve... |
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