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�sin2cos2tan27sin2532228cos530212tan532csc423cos5024sin53212cos133225sin132sec5322tan224tan312sin13csc7sec4 197 30. Without using your calculator, show that. 31. Let θ be a Quadrant III angle with. Sho...
32sin2 198 4.5 Product to Sum and Sum to Product Formulas Learning Objectives In this section you will:  Learn and apply the Product to Sum Formulas.  Learn and apply the Sum to Product Formulas. This section begins with an example that uses identities from Sections 4.2 and 4.3 to write a trigonometric exp...
�from sum identity for cosine from double angle identities2sin21cos3232333cos32coscos2sincos2coscos21coscos2coscos2cos2cos4cos3cos.cos334cos3coscos3coscos4cos5cos 199 Product to Sum Formulas Our next batch of identities,...
left as exercises. 1 These are also known as the Prosthaphaeresis Formulas and have a rich history. Conduct some research on them as your schedule allows. 1sinsincoscos21coscoscoscos21sincossinsin2cos2cos626coscos1cos2cos6c...
reminded that all of the identities presented in this chapter which regard the circular functions as functions of angles in radian measure apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. sinsin2sincos22coscos2coscos22coscos2...
/odd or cofunction identity. 7. 10. 8. 11. 9. 12. In Exercises 13 – 20, verify the identity. Assume all quantities are defined. 13. 15. 17. 18. 19. 20. 14. 16. (HINT: Use result for 16.) 21. In Example 4.5.1, we wrote as a polynomial in terms of. In Exercise 16, we had you verify an identity which expresses as a polyno...
�48coscos44cos233sin33sin4sin42cos48cos8cos133sin44sincos4sincos2432sincos2cos22cos4cos64232sincos2cos22cos4cos68642cos8128cos256cos160cos32cos1cos3coscos4coscoscos5cos6cosnsin3
sinsin5sin4 202 4.6 Using Sum Identities in Determining Sinusoidal Formulas Learning Objectives In this section you will:  Write a trigonometric function of sines and cosines in general sinusoidal format: or. The motivation for this section lies in the occasions when a trigonometric function is defined in term...
sincoscossinSxAxAxBCxSxcos23sin2fxxxfxcosCxAxB0fxsinSxAxB0cos23sin2fxxxcosCxAxBcos23sin2coscossinsinxxAxAxB 203. By matching up corresponding coefficients and constants, we get and. To determine A and ϕ, a bit more work is involved. Re...
��cos3sincoscossinsinAA+20+2ωBωxxxx20Bcos2sin2cos2sin2xxxx13cossinAAcos2xcosAcos1Asin2xsin3A22222222222cos1cossinsi113113132n3AAAAAAAAAAAA Pythagorean Identity from and multiplying through by =2A2cos12sin31cos23sin23...
33132cos2sin222cos23sin2fxxxxxxxx2cos23fxxcos23sin2fxxx 204 2. Proceeding as before, we equate with the expanded form of to get or Once again, we may take and. To determine A and ϕ, we begin by rewriting the equation. We equate the coefficients of, then, on either s...
�cos2sin2cos2sin2xxxx1si-3cosnAAcos2xsin2xsin1Acos3A22cossin12A2sin11sin22cos33cos25652sin26fxx52sin26552sin2coscos2sin66312sin2cos2223sin2cos2cos23sin2fxxxxxxxxxx52sin
26fxxcos23sin2fxxx 205 Graphing the three formulas for on a graphing calculator of computer graphing program will result in three identical graphs, verifying our analytical work. It is worth mentioning that, had we chosen instead of as we worked through the preceding example, our final answers would have loo...
(with ) can be written as. cosCxAxBsinSxAxB0022sin2cos1fxxx33sin33cos3fxxxsincos2fxxx13sin2cos222fxxx23cos2sinfxxx333cos2sin2622fxxx13cos5sin522fxxx63cos36sin33fxxx5252sincos22fxxx3sin33cos66xxfxCxSx...
�,PabsincosfxaxbxB022sinfxabxB 207 CHAPTER 5 THE INVERSE TRIGONOMETRIC FUNCTIONS Chapter Outline 5.1 Properties of the Inverse Cosine and Sine Functions 5.2 Properties of the Inverse Tangent and Cotangent Functions 5.3 Properties of the Inverse Secant and Cosecant Functions 5.4 Calculators a...
sine functions to algebraic expressions. In this chapter, we concern ourselves with finding inverses of the (circular) trigonometric functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domain of each circular functi...
point on the Unit Circle of an oriented arc of length whose initial point is. Hence, we may view the inputs to as oriented arcs and the outputs as x- coordinates on the Unit Circle. The function f –1, then, would take x-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arccosine. Below are th...
��22x1arcsingxxarccos()Fxx1,10,arccosxt0tcostxcosarccosxx11xarccoscosxx0xarcsinGxx1,1,22arcsinxt22tsintxsinarcsinxx11xarcsinsinxx22xxyxy 211 Everything in Theorem 5.1 is a direct consequence of the fact that for a...
is a real number t between and with. The number we seek is. Hence,. 3. We begin by observing that is equivalent to. The number lies in the interval with. Our answer is. cosfxx0xarccosFxxsingxx22xarcsinGxx1arccos22arcsin212cos211sin2arccoscos6111coscos6...
sin2t4t2arcsin2412cos22arccos212cos2t0,2cos2t123cos24 212 4. To find, we seek the number t in the interval with. The answer is so that. 5. Since, we could simply invoke Theorem 5.1 to get. However, in order to make sure we understand why this is the case, we choose ...
arccoscosarccos623arccos20t3cos2t6t3arccoscosarccos62.611611113coscoscos6213cos2611113coscoscos62.613coscos535133coscos55 213 However, as before, to really understand why this cancellation occu...
move on to finding. A geometric approach is to sketch the angle t, along with its corresponding reference angle. We then introduce a ‘reference triangle’ in Quadrant II. Since, the reference triangle will have. We label the adjacent side with length 3 and the hypotenuse with length 5. The Pythagorean Theorem can be us...
of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. We begin this solution by letting, so that. We sketch a right triangle representing. To find the length of the opposite side, y, in terms of x, the Pythagorean Theorem yields 222235164yyy As a length, y is positive. 4sin55yt4si...
(t) in terms of x. 2 tanarccostan1xtxxoppositesince tangent is adjacentarccostxarccosx1,111x0x21tanarccosxxx1,00,11,0arccos2x21xx10xarcsintx,22sintxcos2arcsincos2xtcos2t22cossintt22cos1t212sintsinxt22cos2arcsincos212sin12xttx...
��arcsintxarcsinx11x2cos2arcsin12xx1,1212x2cos2arcsin12xx11x2xx0x 217 5.1 Exercises In Exercises 1 – 18, find the exact value. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. 17. In Exercises 19 – 44, find the exact value or state that it is undefined. 19. 22. 25. 28. 31. 34. 37. 20. 23. 26. 29. 32....
11cos2arccos01arccos212cos213cos2arccos11sinarcsin212sinsin213sinsin5sinarcsin0.425sinarcsin412coscos21cosarccos215coscos13cosarccos0.998cosarccosarcsinsin6arcsinsin3�
��13sinsin4111sinsin64arcsinsin31coscos42arccoscos313coscos21coscos65arccoscos41sinarccos2 40. 43. 218 41. 44. 42. In Exercises 45 – 54, rewrite the quantities as algebraic expressions of x and state the domain on which the...
�15sinsin1344sin2arcsin53cos2arcsin5sinarccosx1tansinx1sin2cosxsinarccos2xsinarccos5x1cossin2xsin2arcsin7x13sin2sin3xcos2arcsin4x11sinsincosxxsin2x22sin2arcsinarccos2xx11x11sin302
1sinfxx1cosgxxcosyx1cosyx1coscos66 219 5.2 Properties of the Inverse Tangent and Cotangent Functions In this section you will: Learning Objectives  Learn and apply properties of the inverse tangent and cotangent functions, including domain and range.  Find exact values of inverse tangent...
�22x1arctanfxxxyxy 220 The Inverse Cotangent Function Next, we restrict to its fundamental cycle on to obtain the arccotangent: or. Once again, the vertical asymptotes and of the graph of become the horizontal asymptotes and of the graph of when the graph of the cotangen...
��xy 221 Theorem 5.2. Properties of the Arctangent and Arccotangent Functions  Properties of – Domain: – Range: – as, ; as, – – – – if and only if and for for all real numbers x provided – arctangent is odd  Properties of – Domain: – Range – as, ; as, – – – if and only if and for for all real numbers x – – ...
��arccot0xarccotxt0tcottx1arccotarctanxx0xcotarccotxxarccotcotxx0xarctan3arccot3cotarccot513sintan4 Solution. 222 1. We know is the real number t between and with. We find, so. 2. The real number lies in the interval with. We get. 3. We can apply Theorem 5.2 directl...
5arccot36cotarccot55arccot5tcot5t0tcotarccot5cot5.t13sintan413tan4t3tan4t22ttan0t02t221cotcsctttantsint4cot3t3tan4tcsct 223 With, we choose so that. Hence, Example 5.2.2. Rewrite the following as algebraic expressions of x and state t...
�02t5csc3t3sin5t13sintansin43.5ttan2arctanx1coscot2xarctantx22ttantxtan2arctantan2xttan2t2, or2.42tktk,224ttan2t,,,244442 224 Returning to, we note the double angle identity is valid for all of the values of t u...
. Our goal is to then express in terms of x. Using the identity, which is valid for t in, we can write. Thus With and, the Pythagorean identity provides a path for expressing in terms of x. arctan2t22tantan21tanttt22tan2arctantan22tan1tan2.1xtttxxarctantxarctanx4ttantan41.xt...
1cscsintt221cotcscttsint 225 Since t is between 0 and π,. Thus, and. Finally, This is true for all real numbers x since is defined for all real numbers x and we encountered no additional restrictions on t. 222221cotcsc12csccsc41ttxttxcsc0t2csc41tx21sin41tx12coscot22sin2.41xx...
��3arccot31cot013cot3arccot1arccot31tantan11tantan35tanarctan12tanarctan0.9651tantan3cotarccot11cotcot37cotarccot241cotcot0.00117cotarccot4arctantan31tantan41tantanarctantan212tantan3arccotcot31co...
��1cotcot22arccotcot3sinarctan21sincot5cosarctan71coscot3 227 40. 43. 46. 41. 44. 47. 39. 42. 45. 48. In Exercises 49 – 55, rewrite the quantities as algebraic expressions of x and state the domain on which the equivalence is valid. 49. 52. 55. 50. 53. 51. 54. 56. If for, find an expres...
�1costanx1sin2tanxcos2arctanx1costan3xcosarcsinarctanxx1tan2sinx1sinarctan2xtan7x2211sin222 228 5.3 Properties of the Inverse Secant and Cosecant Functions In this section you will: Learning Objectives  Learn and apply properties of the inverse secant and cosecant...
�cscyx,11,1,,1xyxyxyxy 229 The Inverse Secant Function For, we restrict the domain to, corresponding to that of cosine, and reflect about the line to obtain the graph of. on The Inverse Cosecant Function We restrict to, in correspondence with the sine, and reflect about the line to obtain. on secf...
��xyxyxyxy 230 Note that and may also be written as and, respectively. For both arcsecant and arccosecant, the domain is, which can be written as. Theorem 5.3. Properties of the Arcsecant and Arccosecant Functions  Properties of – Domain: – Range: – as, ; as, – –...
��xarcsec2xxarcsec2xarcsecxt02t2tsectx1arcsecarccosxx1xsecarcsecxx1xarcsecsecxx02x2xarccscGxx:1,11,xx,00,22xarccsc0xxarccsc0xarccscxt02t02tcsctx1arccscarcsinxx1xcscarccscxx1xarccs
ccscxx02x02xarcsec21csc215secsec4cotarccsc3 231 Solution. 1. Since, we can use Theorem 5.3 and have 2. Once again, with, Theorem 5.3 comes to our aid giving 3. Since doesn’t fall between 0 and or between and π, we cannot use the inverse property stated in Theorem 5.3. We can, nevertheless, begi...
221cotcsctt22221cotcsc1cot3cot8cot22ttttt 232 Since,, so we get. Example 5.3.2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. We begin simplifying by letting. Then, for t in, and we seek a formula for. Since is d...
222221tansec1tantan1tttxtx0,2tan0t,2tan0ttant22 1, if 0t<2tan1, if 2xtxtsecxt02t1x2t1xtanarcsecx 233 2. To simplify, we start by letting. Then for t in, and we now set about finding an expression for Since is defined for all t, there are no additio...
�csc4tx,00,221coscsc4cosxtcostcsc4tx1sin4txcost22cossin1tt2222222cossin11cos14161cos16161cos4tttxxtxxtx,00,22cos0t2161cos4xtx1csc4tx1cscx1x1csc4x41x4141 or 4111 or 44xxxxx21161coscsc44xxx11,,44...
find the exact value. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. In Exercises 17 – 52, find the exact value or state that it is undefined. 17. 20. 23. 26. 29. 32. 35. 18. 21. 24. 27. 30. 33. 36. 3. 6. 9. 12. 15. 19. 22. 25. 28. 31. 34. 37. arcsec21csc21sec21csc223arcsec323arccsc31sec1arccsc1...
12csccsc2cscarccsc1.0001cscarccsc41secsec44arcsecsec315secsec61secsec25arcsecsec3arccsccsc615csccsc412csccsc3arccsccsc211arccsccsc6111secsec12 38. 41. 44. 47. 50. 235 39. 42. 45. 48. 51. 40. 43. 46. ...
terms of x. 57. Show that for as long as we use as the range of 58. Show that.. for as long as we use as the range of 9arccsccsc81sincsc31cossec55tanarcsec31cotcsc53secarccos2112secsin13secarctan10110seccot10cscarccot93cscarcsin512csc...
024tan41arcsecarccosxx1x0,,22arcsecfxx111cscsinxx1x,00,221cscfxx 236 5.4 Calculators and the Inverse Circular Functions In this section you will: Learning Objectives  Use technology to evaluate approximate values of the inverse trigonometric functions.  ...
1cot23arccsc220arccot21arctan21arccot2arctan20.4636 radians.51111sec5cos51.3694 radians. 237 3. Since the argument –2 is negative, we cannot directly apply Theorem 5.2 to help us find. We will, however, be able to use by first establishing a relationship between and.  By definitio...
1cot22.67793arccsc231232arccscarcsin230.7297 radians.xy 1cot(2) radians   238 Domain and Range of Inverse Trigonometric Functions Example 5.4.2. Find the domain and range of the following functions. Check your answers using graphing technology. 1. 2. 3. Solution. 1. Since ...
�arccos25xfx5,5arccosFxx1,0,21,05,20,05,2arccos25xfx,225,5,22 239 2. To find the domain of, we note the domain of is all real numbers. The only restrictions, if any, on the domain of come from the argument, 4x, and since 4x is define...
tanFxx13tan4fxx0,02y2y1tanyFxx13tan4yfxx33,22,33,22 240 3. To find the domain of, we proceed as above. Since the domain of is, and is defined for all x, we get the domain of g is as well. As for the range, we note that the range of, like that of, is limited by a pair ...
�1tanFxx0yyarccot2xygxarccotyGxx,2arccot2xgx2arccotarctan2xx02x0x0x2arctangxx 241  When, we can use the same argument in Example 5.4.1, part 3, that gave us to give us. Thus, for,  What about? We know, and neither of the formulas for g involving arctang...
�0x0arccot0gygx2arctan2 when 0arccot when 022arctan when 0xxxygxxxxarccot2xygx,,2 242 Applications of Inverse Trigonometric Functions The inverse trigonometric functions are typically found in applications where the measure of an angle is requi...
 6 feet 12 feet  243 1arctan226.57. 244 5.4 Exercises In Exercises 1 – 6, use a calculator to evaluate each expression. Express answers to the nearest hundredth. 1. 4. 2. 5. 3. 6. In Exercises 7 – 18, find the domain of the given function. Write your answers in interval notation. 7. 10. 13. 16. 8. 11. 14. 17...
fxxarctan4fxx122cot9xfxx1tanln21fxxarccot21fxx1sec12fxxarccsc5fxx3arcsec8xfx12cscxfxe 245 22. A parasailor is being pulled by a boat on Lake Powell. The cable is 300 feet long and the parasailor is 100 feet above the surface of the water. What is the angle of elev...
igonometric Functions/Arguments Introduction Chapter 6 focuses on solving trigonometric equations through the implementation of tools and formulas accumulated throughout the preceding chapters. In Section 6.1, inverse trigonometric functions are used to find acute angles within right triangles, as well as to find exact...
each case, we appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of common angles. If, on the other hand, we had been asked to find all angles with or to solve for real numbers t, we would have been hard pressed to do so. With the introduction of the inverse trigonometric functio...
position, intersects the Unit Circle at. Geometrically, we see that this happens at two places: in Quadrant I and in Quadrant II. If we let α denote the acute solution to the equation, then all of the solutions to this equation in Quadrant I are coterminal with α, and α serves as the reference angle for all solutions ...
the solutions from Quadrant II differ by exactly π units from the solutions in Quadrant IV, so all the solutions to are of the form Switching back to the variable t, we record our final answer to as for integers k. 21arcsin2, for integers.3kkk1sin31arcsin23k1arcsin23ktan2t...
not x represents an angle or a real number. We assume the latter, but note that we will use angles and the Unit Circle to solve the equation regardless. Adopting an angle approach, we consider and note that our solutions lie in Quadrants II and III. Since isn’t the secant of any of the common angles, we’ll need to exp...
�5arcsec23xk 252 Note that, with, it follows from Theorem 5.3 that. Another way to write our solution to is or for integers k. The reader is encouraged to check the answers found in Example 6.1.2, both analytically and with technology. 51353arcsecarccos355sec3x3arccos25xk3...
9. 12. 15. 18. 21. 24. 7. 10. 13. 16. 19. 22. 25. 8. 11. 14. 17. 20. 23. 0,27sin11x2cos9xsin0.569xcos0.117xsin0.008x359cos360xtan117xcot12x3sec2x90csc17xtan10x3sin8x7cos16xtan0.03xsin0.3502xsin0.721xcos0.9824xcos0.5637x1cot117xtan0.6109x 254 6.2 S...
respectively, and solve as above. If, convert to cosine or sine, or, there are no real solutions.  To solve for any real number c, first solve for u in the interval and add integer multiples of the period π.  To solve for, convert to tangent and solve as above. If, the solution to is for integers k. Using the above ...
. The solutions provide practice with, and extensions of, the technique applied in solving. We will add to the general solution of each equation specific solutions that fall in the interval. Equations Involving Cosines or Sines Example 6.2.1. Solve the equation, giving the exact solutions which lie in. Solution. The so...
uting them into the left hand side of the original equation,.  Starting with, we have  Similarly, we find, for, This confirms the solution of or for integers k. Equations Involving Tangents or Cotangents Next, we look at an example of a cotangent function. Example 6.2.2. Solve, giving the exact solutions which lie in...
, and, corresponding to through. To check the solution of, we start with the left side of. This confirms our solution of for integers k. Equations Involving Secants or Cosecants We look next at an equation involving a cosecant function which we will solve through conversion to its reciprocal, a sine function. Example 6...
together, we have or for integers k. Despite the infinitely many solutions of, we find that none of them lie in. This can be verified graphically by plotting and, and observing that the two functions do not intersect at all over the interval. csc2u2sin2u24uk324uk13x1234xk13234xk123412...
the profile of equations thus far in this section. Example 6.2.4. Solve. List the solutions which lie in the interval. Solution. The complication in solving an equation like comes not from the argument of secant, which is just x, but rather from the fact that secant is being squared. Thus, we begin by extracting squar...
 2y 260 k … –1 0 1 2 … … … … … The solutions in the interval are,, and. Solutions that Include Inverse Trigonometric Functions Our remaining two examples require inverse trigonometric functions in their solutions. Otherwise, the solution process is similar to that of the first four examples. Example 6.2.5. Solve an...
��2arctan302arctan32xk0,20k2arctan30x2arctan32xk0k 261  Next, we turn our attention to and get. Starting with the inequality, we add 2π and get. This means lies in.  Advancing k to 2 produces. Once again, we get from that. Since this is outside the interval, we discard and all solutions...
��0,22arctan34x2arctan32xk2ktan32x0,22arctan323.7851xsin20.87x0,2sin20.87xsin0.87uarcsin0.872ukarcsin0.872uk2arcsin0.872xk2arcsin0.872xk1arcsin0.872xk1arcsin0.8722xk0,21arcsin0.872x0arcsin0.87210arcsin012.8724 by definitio...
and All told, we have found four solutions to in :, and.., We will continue solving equations containing trigonometric functions in Section 6.3, with the added complexity of multiple trigonometric functions and/or arguments. 1arcsin0.872xk0k1k0,21arcsin0.872x1arcsin0.872x1arcsin0.8722xk1arcs...
list those solutions which are in the interval. 1. 4. 7. 10. 13. 16. 2. 5. 8. 11. 14. 17. In Exercises 19 – 26, solve the equation. 19. 22. 25. 20. 23. 26. 3. 6. 9. 12. 15. 18. 21. 24. 27. With the help of your classmates, determine the number of solutions to in. Then find the number of solutions to, and in. A pattern...
��12arccsc23x229arcsin0x229arccos0x1sin2x0,21sin22x1sin32x1sin42x0,21sin112x1sin22x31sin22x51sin22x12 264 6.3 Solving Equations of Multiple Trigonometric Functions/Arguments In this section you will: Learning Objectives  Solve equations containing multiple trigonome...
�323sinsinxx2sinx323223sinsin3sinsin0sin3sin10xxxxxxFactor out from both terms.2sinx2sin0x3sin10xsinxsin0x1sin3xsin0xxk0xx0,21sin3x1arcsin23xk1arcsin23xk0,21arcsin3x1arcsin3x 265 We can visualize the solutions by graphing and. It may be necess...
2sinyx0xx0,2323sinsinxx0,20x1arcsin3x1arcsin3xx2sectan3xx0,22sectan3xx22sec1tanxx2x0,22x0,2 33sinyx 2sinyxxy 33sinyx 2sinyx 33sinyx 2sinyx 266 This gives or. Since, we have or.  From, we get for integers k. ...
interval. As discussed above, these solutions include,, and. 222222sec1tansectan31tantan3tantan20201t0an2xxxxxxxxuuuuux since for 1u2utanuxtan1xtan2xtan1x4xk0,234x74xtan2xarctan2xkarctan2xarctan2x0,22secyxtan3yx2sectan3xx...
��xy 2secyx tan3yx 267 Equations Containing Multiple Arguments of the Same Function Some trigonometric equations can be solved by treating them as quadratic equations. Before proceeding in this manner with the following example, we will need to apply a double angle identity. Example 6.3.3. Solve the equation. ...
for 12u1ucosux1cos2xcos1x1cos2x23xk523xkcos1x2xk0,20x353cos2yx3cos2yx0,2cos313cosxx3cos34cos3cosxxxcosx 268 We get, or and. Since, our solutions would result from,. Both 2 and –2 are outside the range of cosine. Thus, the only real solution to in is, so that fo...
� for 0u2u2ucosuxcos0xcos2xcos2xcos313cosxxcos0x2xk0,22x32xsin23cosxxsin23cosxxsin22sincosxxxsin23cos2sincos3cos2sincos3cos0cos2sin30xxxxxxxxxx 269 Then or. From, we obtain for integers k. From, we get or for integers k. The answers...
cos0x2xk3sin2x23xk223xk0,22x32323sincoscossin122xxxxsincoscossin122xxxxsin2xxsincoscossin122sin123sin12xxxxxxx3sin12x3sin123222433xxkxk since sine is equal to 1 after multiplying both sides by 23 270 The solution of for integers k has the fol...
. 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 0,2sincosxxsin2sinxxsin2cosxxcos2sinxxcos2cosxxcos225cosxx3cos2cos20xxcos25sin2xx3cos2sin2xx22sec3tanxx2tan1secxx2cot3csc3xxsec2cscxx2coscsccot6cotxxxxsin2tanxx...
�tan22cos0xx32csccsc4csc4xxx22tan1tanxxtansecxxsin6coscos6sinxxxxsin3coscos3sinxxxxcos2cossin2sin1xxxx3cos5cos3sin5sin32xxxxsincos1xxsin3cos1xx2cos2sin1xx3sin2cos21xxcos23sin22xx33sin33cos333xxcos3cos5xxcos4cos2xx...
 273 CHAPTER 7 BEYOND RIGHT TRIANGLES Chapter Outline 7.1 Solving Triangles with the Law of Sines 7.2 Applications of the Law of Sines 7.3 The Law of Cosines Introduction Chapter 7 introduces some new tools that will help in solving obtuse triangles, and in solving real-life applications. In Section 7.1, triangles are...
the angles in decimal degrees, rounded to the nearest hundredth of a degree. Solution. For definitiveness, we label the triangle below.  To find the length of the missing side a, we use the Pythagorean Theorem to get, which then yields units. Now that all three sides of the triangle are known, there are several ways ...
this. 4 Don’t worry! Radians will be back before you know it! 4cos74arccos755.154sin74arcsin734.8590,a,b,c 276 Theorem 7.1. The Law of Sines: Given a triangle with angle-side opposite pairs, and, the following ratios hold Or, equivalently, The proof of the Law of Sines can be broken in...
sinsinacABQACQsinsinbcbcaABC   caABChQ  bcABCQh'   277 2. For our next case, consider the triangle below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: and. Proceeding as before, we get and so that. Dropping an altitude from vertex B also...
�'sinhcBCQ'sinha'sinhasinsincaABC1207a45 278 Solution. Knowing an angle-side opposite pair, namely and, we may proceed in using the Law of Sines. Now that we have two angle-side pairs, it is time to find the third. To find γ, we use the fact that the sum of the measures of the angles in a triang...
triangle. Solution. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since. As in the previous example, we are forced to use a derived value in our computations since the only angle-side pair available is. The Law of Sines gives To fi...
. Using the Law of Sines, we get Now γ is an angle in a triangle which also contains the angle. This means that γ must measure between 0° and 150° in order to fit inside the triangle with α. The only angle that satisfies this requirement and has is. In other words, we have a right triangle. We find the measure of β to ...
sin30sin434sin30sin32sin33001502sin32arcsin32arcsin3 282  In the case radians ≈ 41.81°, we find5. Using the Law of Sines with the angle-side opposite pair and β, we find units.  In the case radians ≈ 138.19°, we repeat the exact same steps and find β ≈ 11.81° and b ≈ 1.23 units.6 B...
180252arcsinarcsin63632arcsin36 283 Then γ must inhabit a triangle with, so we must have. Since the measure of γ must be strictly less than 150°, there is just one angle which satisfies both required conditions, namely. So and, using the Law of Sines one last time, Some remarks are in order. 1. If...
to be angle-side pairs in a triangle where α, a and c are given. Let.     If If If If, then no triangle exists which satisfies the given criteria., then so exactly one (right) triangle exists which satisfies the criteria., then two distinct triangles exist which satisfy the given criteria., then γ is acute and exa...
to argue that each of these angles ‘fit’ into a triangle with α. o Since and are angle-side opposite pairs, the assumption in this case gives. Since is acute, we must have that α is acute as well. This means that one triangle can contain both α and, giving us one of the triangles promised in the theorem. o If we manip...
6. 7. 9. 288 8. 10. 11. Notice that x is an obtuse angle. 12. In Exercises 13 – 32, solve for the remaining side(s) and angle(s) if possible. As in the text,, and are angle-side opposite pairs. 13. 15. 17.,,,,,, 14. 16. 18.,,,,,, ,a,b,c13175a73.254.1117a958533.33a956233.33a11735a42b117...
�3.05a10216.75b13c10216.75b18c1023516.75b29.1383.95314.15b120614c5025a12.5bABC 290 37. Discuss with your classmates why the Law of Sines cannot be used to find the angles in a triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them...
angles and all three sides. To minimize propagated error, we choose, from Theorem 7.3, because it uses the most pieces of given information. We are given and, and we calculated in Example 7.1.2. Using these values, we find ,a,b,c111sinsinsin222Abcacab1207a451sin2Aac7a457sin15sin120...

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