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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. Show that this is not enough information to determine the sign of by first assuming and then assuming. Compute in both cases. 2362241cos5sin2732
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 expression as the sum of trigonometric expressions. Example 4.5.1. Express as a polynomial in terms of. Solution. The double angle identity expresses as a polynomial in terms of. We are now asked to find such an identity for. Finally, we exchange for, courtesy of a Pythagorean identity, and get Thus, can be expressed as the polynomial. Having just shown how we could rewrite as the sum of powers of, it might occur to you that similar operations could be applied to or to rewrite the expressions as sums of powers of. This will be of use in calculus, as will the formulas yet to be presented in this section. cos3cos2cos22cos1cos2coscos3232cos3cos2cos2cossin2sin2cos1cos2sincossin2coscos2sincos�
�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, the Product to Sum Formulas1, are easily verified by expanding each of the right hand sides in accordance with the Sum and Difference Identities. The details are left as exercises. Theorem 4.9. Product to Sum Formulas. For all angles α and β,    Example 4.5.1. Write as a sum. Solution. Identifying and, we use the Product to Sum Formula for. Sum to Product Formulas Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Chapter 6. These are easily verified using the Product to Sum Formulas and, as such, their proofs are
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1cos2cos6cos26cos26211cos4cos82211cos4cos822 even property of cosine 200 Theorem 4.10. Sum to Product Formulas. For all angles α and β,    Example 4.5.2. Write as a product. Solution. Using the Sum to Product Formula for, with and, yields the following. Where the last equality is courtesy of the odd identity for sine,. The reader is
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coscos2sinsin22sinsin3sinsin333sinsin32sincos222sincos22sincos2sinsin 201 4.5 Exercises In Exercises 1 – 6, write the given product as a sum. You may need to use an even/odd identity. 1. 4. 2. 5. 3. 6. In Exercises 7 – 12, write the given sum as a product. You may need to use an even
/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 polynomial in terms of. Can you find a polynomial in terms of for?? Can you find a pattern so that could be written as a polynomial in cosine for any natural number n? 22. In Exercise 15, we had you verify an identity which expresses as a polynomial in terms of. Can you do the same for? What about for? If not, what goes wrong? 23. Verify the Product to Sum Identities. 24. Verify the Sum to Product Identities. cos3cos5sin2sin7sin9coscos2cos6sin3sin2cossin3cos3cos5sin2sin7sin5cos6sin9sinsincoscossin48sincos44cos23�
�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 terms of both sines and cosines. By rewriting the function as a sine function or as a cosine function, properties of the function, such as amplitude and period, will become more apparent. This will be a tool used in future courses such as differential equations. To get started, if we use the sum identity for cosine, we can expand to yield Similarly, using the sum identity for sine, is equivalent to.. Making these observations allows us to recognize (and graph) functions as sinusoids which, at first glance, don’t appear to fit the forms of either or. Example 4.6.1. Consider the function. 1. Find a formula for in the form 2. Find a formula for in the form Check your answers analytically using identities. Solution. for for.. 1. The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. Equating with the expanded form of, we get or cosCxAxBsinSxAxBcosCxAxBcoscossinsinCxAxAxBsinSxAxB
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. Rewriting the equation will help. On the left hand side, the coefficient of is 1, while on the right hand side it is. Since this equation is to hold for all real numbers, we must have that. Similarly, we find by equating the coefficients of that. We now have a system of nonlinear equations that will allow us to determine values for A and ϕ. Choosing, we have and or, after some rearrangement, and. One such angle which satisfies this criteria is. Hence, one way to write as a sinusoid is. We can easily check our answer using the sum formula for cosine. This verifies that is equivalent to. �
��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fx2cos23fxx2cos232cos2cossin2sin
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 side and get and. Using the Pythagorean identity as before, we get. Then, or, and, which means. One such angle which meets these criteria is. Hence, we have. We check our work analytically, using the sum formula for sine. Thus, is equivalent to. cos23sin2fxxxsinSxAxBcos23sin2sincoscossinxxAxAxBcos3sincossinsincosAA+20+2ωBωxxxx20B�
�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 looked different. The reader is encouraged to rework the example using and to then use identities to show that the formulas are all equivalent. It is important to note that in order for the technique presented in the example to fit a function into the form of one of the general equations, or arguments of the cosine and sine function must match. That is, while, the is a sinusoid, is not.1 The general equations of sinusoids will be explored further in the exercises. 1 This graph does, however, exhibit sinusoid-like characteristics. Check it out! fx2A2A2AcosCxAxBsinSxAxBcos23sin2fxxxcos23sin3gxxx 206 4.6 Exercises In Exercises 1 – 10, use Example 4.6.1 as a guide to show that the function is a sinusoid by rewriting it in the forms and for and. 1. 3. 5. 7. 9. 2. 4. 6. 8. 10. 11. In Exercises 1 – 10, you should have noticed a relationship between the phases ϕ for and. Show that if, then where. 12. Let ϕ be an angle measured in radians and let be a point on the terminal side of ϕ when it is drawn in standard position. Use Theorem 2.6 and the sum identity for sine to show that
(with ) can be written as. cosCxAxBsinSxAxB0022sin2cos1fxxx33sin33cos3fxxxsincos2fxxx13sin2cos222fxxx23cos2sinfxxx333cos2sin2622fxxx13cos5sin522fxxx63cos36sin33fxxx5252sincos22fxxx3sin33cos66xxfxCxSxsinfxAxBcosfxAxB2�
�,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 and the Inverse Circular Functions Introduction Chapter 5 introduces the valuable inverse (circular) trigonometric functions. The first three sections are devoted to defining these inverse functions and identifying properties of each. Emphasis is on determining function values and rewriting expressions containing inverse trigonometric functions. Section 5.1 focuses on the inverse cosine and sine functions. In Section 5.2, the inverse tangent and cotangent functions are added, followed by the inverse secant and cosecant in Section 5.3. Throughout the first three sections, relationships between the inverse functions are developed. Because of the necessity for using inverse trigonometric functions in solving real-world applications, the calculation of degree or radian measure is often desired. Section 5.4 includes techniques for determining approximate values of inverse trigonometric functions through technology. In addition to real-world applications of inverse trigonometric functions, Section 5.4 includes finding domains and ranges, and verifying domains and ranges through graphing technology. Chapter 5 is essential to the understanding of trigonometric functions and their uses. The inverse trigonometric functions will be needed in solving trigonometric equations and solving triangles, the focus of the next two chapters. 208 5.1 Properties of the Inverse Cosine and Sine Functions In this section you will: Learning Objectives  Learn and be able to apply properties of the inverse cosine and sine functions, including domain and range.  Find exact values of inverse cosine and sine functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse cosine or
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 function to obtain a one-to-one function. The Inverse Cosine Function We first consider. Choosing the interval allows us to keep the range as along with the properties of being smooth and continuous. Restricting the domain of to. Recall that the inverse of a function f is typically denoted f –1. The notation for the inverse of is denoted as either or, read ‘arc-cosine of x’.1 To understand the ‘arc’ in arccosine, recall that an inverse function, by definition, reverses the process of the original function. The function takes a real number input t, associates it with the angle radians, and returns the value. Digging deeper, we have that is 1 The obvious pitfall here is our convention of writing (cos(x))2 as cos2(x), (cos(x))3 as cos3(x) and so on. It is easy to confuse cos–1(x) as (cos(x))–1, which is equivalent to sec(x), not the inverse of cos(x). Always be careful to check the context of cos–1(x)! cosfxx0,1,1cosfxx0,cosfxx11cosfxx1arccosfxxcosftttcoscoscostxy 209 the x-coordinate of the terminal
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 the graphs of and, where we obtain the latter from the former by reflecting it across the line. This is achieved by switching the x and y coordinates., The Inverse Sine Function We restrict in a similar manner, although the interval of choice is. The inverse of, denoted, is read ‘arcsine of x’. t1,0cosfttcosfxx1arccosfxxyxcosfxx0x1arccosfxxsingxx,22singxx1arcsingxxxyxyxy 210, We list some important facts about the arccosine and arcsine functions in the following theorem. Theorem 5.1. Properties of the Arccosine and Arcsine Functions  Properties of – Domain: – Range: – – if and only if and provided – – arccosine is neither even nor odd provided  Properties of – Domain: – Range: – – – – arcsine is odd if and only if and provided provided singxx�
��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 and are inverses of each other, as are for and. It’s about time for an example. Example 5.1.1. Find the exact values of the following. 2. 3. 4. 6. 7. 8. 1. 5. Solution. 1. To find, we need to find the real number t (or, equivalently, an angle measuring t radians) with and. We know that meets these criteria, so. 2. The value of
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13coscos53sinarccos51arccos21cos2t0t3t1arccos232arcsin2222
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 to work the example through using the definition of arccosine. Working from the inside out,. Now, is the real number t with and. We find, so that 6. Since does not fall between 0 and π, Theorem 5.1 does not apply. We are forced to work through from the inside out starting with. From the previous problem, we know. Hence, 7. One way to simplify is to use Theorem 5.1 directly. Since is between –1 and 1, we have and we are done. 11sin2,221sin2t6t11sin2606arccoscos663
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 occurs, we let. Then, by definition,. Hence, 8. To evaluate, as in the previous example, we let so that for some t where. Since, we can narrow this down a bit and conclude that, so that t corresponds to an angle in Quadrant II. We
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 used to find the length y of the opposite side. 13cos5t3cos5t13coscoscos53.5t3sinarccos53arccos5t3cos5t0tcos0t2t3sinarccossin5tt3cos5t3cos5txy5y3t t 3,y 214 Then. And, since sine is positive in Quadrant II,. Finally, Most of the common errors encountered in dealing with the inverse circular functions come from the need to restrict the domains of the original functions so that they are one-to-one. One instance of this phenomenon is the fact that as opposed to, as observed in part 6 of the previous example. Example 5.1.2. Rewrite the following as algebraic expressions
of x and state the domain on which the equivalence is valid. 1. Solution. 2. 1. We begin 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4sin5t3sinarccossin54.5t11arccoscos66116tanarccosxcos2arcsinxarccostxcostxadjacenthypotenusecos1xt222222111xyyxyx1xty 215 This results in To determine the values of x for which this equivalence is valid, we consider our substitution. The domain of is, or. Additionally, since the tangent is not defined when the cosine is 0, we need to discard. Hence, is valid for x in. Note that for x in, we have, resulting in values of the tangent being less than or equal to zero. This corresponds correctly in sign with the values obtained for since. 2. We proceed as in the previous problem by writing so that so that t lies in the interval with. We aim to express in terms of x. We have three choices for rewriting :, or. Since we know, it is easiest to use the last form.2 2 To use the first or second form, we could use a right triangle, like we did in part 1, to evaluate sin
(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1xt 21x 216 To find the restrictions on x, we revisit our substitution. Since is defined only for, the equivalence is valid only on. Even though the expression we arrived at in part 2 of the last example, namely, is defined for all real numbers, the equivalence is valid for only. This is similar to the fact that while the expression x is defined for all real numbers, the equivalence is valid only for. For this reason, it pays to be careful when we determine the intervals where such equivalences are valid. �
��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. 35. 38. 3. 6. 9. 12. 15. 18. 21. 24. 27. 30. 33. 36. 39. arcsin113sin212sin21arcsin2arcsin01arcsin212sin23arcsin2arcsin11cos113cos22arccos2
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 equivalence is valid. 45. 48. 51. 54. 46. 49. 52. 47. 50. 53. 55. If for, find an expression for in terms of x. 56. Show that for. 57. Discuss with your classmates why. 58. Why do the functions and have different ranges? 59. Since the functions and are inverse functions, why is not equal to? 13sincos55cosarcsin13�
�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 and cotangent functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse tangent or cotangent functions to algebraic expressions. The Inverse Tangent Function We restrict to its fundamental cycle on to obtain the inverse of tangent. The inverse, reflect, named arctangent, is also denoted. In the following graphs, we about the line to obtain. Note that the vertical asymptotes and from the graph of become the horizontal asymptotes and on the graph of., tanfxx,221arctanfxx1tanxtanfxxyx1arctanfxx2x2xtanfxx2y2y1arctanfxxtanfxx�
�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 cotangent is reflected about the line to obtain the graph of the arccotangent., cotgxx0,1arccotgxx11cotgxx0xxcotgxx0yy1arccotgxxyxcotgxx0x1arccotgxxxy�
��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 – – arccotangent is neither even nor odd provided Example 5.2.1. Find the exact values of the following. 1. 2. 3. 4. arctanFxx,,22xarctan2xxarctan2xarctanxt22ttantx1arctanarccotxx0xtanarctanxxarctantanxx22xarccotGxx,0,xarccotx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 directly and obtain. However, working it through provides us with yet another opportunity to understand why this is the case. If we let, then for some t,. Hence, 4. We start simplifying by letting. Then for some t,. Since, we know, in fact,. One way to proceed is to use the Pythagorean identity, since this relates the reciprocals of and and is valid for all t under consideration.1 Along with this identity, we use, from, to solve for. 1 It’s always a good idea to make sure the identities used in these situations are valid for all values t under consideration. Check our work back in Example 5.1.1. Were the identities we used there valid for all t under consideration? arctan322tan3t3tarctan33arccot3t0,cot3t
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 the domain on which the equivalence is valid. 1. Solution. 2. 1. If we let, then and. We look for a way to express in terms of x. Before we get started using identities, we note that is undefined, for any integer k, when The only members of this family which lie in are, which means the values of t under consideration for are. 222221cotcsc41csc325csc95csc3ttttt�
�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 under consideration. Hence, we get To find where this equivalence is valid, we check back with our substitution. Since the domain of is all real numbers, the only exclusions come from, the values of t we discarded earlier. We exclude corresponding values of x: Hence, the equivalence holds for all x in. 2. To get started, we let so that where. Then and, since is always defined, there are no additional restrictions on t
. 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22tan2arctan1xxx,11,11,1cot2txcot2tx0t1coscot2cosxtcostcostcoscotsinttt0,coscotsinttt1coscot2coscotsin2sin.xtttxtcot2tx
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12coscot22sin2.41xxtxx1cot2x 226 5.2 Exercises In Exercises 1 – 14, find the exact value. 1. 4. 7. 10. 13. 2. 5. 8. 11. 14. In Exercises 15 – 48, find the exact value or state that it is undefined. 15. 18. 21. 24. 27. 30. 33. 36. 16. 19. 22. 25. 28. 31. 34. 37. 3. 6. 9. 12. 17. 20. 23. 26. 29. 32. 35. 38. arctan31tan113tan3arctan03arctan31tan11tan31cot3arccot1�
��3arccot31cot013cot3arccot1arccot31tantan11tantan35tanarctan12tanarctan0.9651tantan3cotarccot11cotcot37cotarccot241cotcot0.00117cotarccot4arctantan31tantan41tantanarctantan212tantan3arccotcot31cotcot4arccotcot�
��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 expression for in terms of x. 125tansin51tanarccos2tanarccot1212cotarcsin1313cotcos21cottan0.253tanarctan3arccos51sin2tan2cos2arccot5arctan2sin2�
�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 functions, including domain and range.  Find exact values of inverse secant and cosecant functions, and of their composition with other trigonometric functions.  Convert compositions of trigonometric and inverse secant or cosecant functions to algebraic expressions. The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Section 3.4, are given below with the fundamental cycles highlighted. The graph of Fundamental cycle of The graph of Fundamental cycle of It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of and restricts the domain of the function so that it is one-to- one. The same is true for cosecant. Thus, in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely, and another piece to cover the bottom, namely. secyxsecyxcscyx�
�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 secfxx0,,22secfxxyx1arcsecfxxsecfxx0,,221arcsecfxxcscgxx,00,22yx1arccscgxxcscgxx,00,221arccscgxx�
��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, – – – – if and only if or and provided provided provided or – arcsecant is neither even nor odd  Properties of – Domain: – Range: – as, ; as – – – – if and only if provided provided, or and provided or – arccosecant is odd Example 5.3.1. Find the exact values of the following. 1. 2. 3. 4. arcsecxarccscx1secx1cscx,11,:1xxarcsecFxx:1,11,xx0,,22�
��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, begin by working inside out which yields 4. One way to begin to simplify is to let. Then and, since this is negative, we have that t lies in the interval. We are after, knowing, so we use the Pythagorean identity. 211arcsec2arccos2.321111csc2sin2.65422111215secsecsec241cos23.4 Note: from Theorem 5.3cotarccsc3arccsc3tcsc3t,02cottarccsc3t
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 defined for all t values under consideration, there are no additional restrictions on t. To relate to, we use the identity, which is valid for all t under consideration. If t belongs to then ; if, on the other hand, t belongs to then. As a result, we get a piecewise defined function for. Now we need to determine what these conditions on t mean for. When, it follows that, and when, we have. With no further restrictions on t, we can express as an algebraic expression of x. 02tcot0tcotarccsc322tanarcsecx1coscsc4xtanarcsecxarcsectxsectx0,,22tanttantsecttant221tansectt
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 additional restrictions on t. From get. To find, we can make use of the identity.., we Since t belongs to, we know, so we choose. (The absolute value here is necessary since x could be negative.) To find the values of x for which this equivalence is valid, we look back at our original substitution. The domain of requires its argument x to satisfy, and so the domain of will require. With no additional restrictions on t, the equivalence holds for all x in. 22 1, if 1tanarcsec1, if 1xxxxx1coscsc4x1csc4tx�
�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 234 5.3 Exercises In Exercises 1 – 16,
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arcsec21sec2123sec31sec11csc2arccsc223arccsc3arccsc11secsec2secarcsec111secsec21secsec0.75secarcsec1171csccsc223cscarccsc3
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. 49. 52. In Exercises 53 – 55, rewrite the quantities as algebraic expressions of x and state the domain on which the equivalence is valid. 53. 54. 55. 56. If for, find an expression for in
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12csctan3cosarcsec3arctan2113sin2csc5125cos2sec7secarctanxcscarccosxsecarctan2tanarctan2xxsec4x
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.  Find domains and ranges of inverse trigonometric functions.  Use inverse trigonometric functions to solve real-world applications. In the sections to come, we will have need to approximate values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as sin-1, cos-1 and tan-1, respectively. If we are asked for an arccotangent, arcsecant or arccosecant, we often need to employ some ingenuity, as the next example illustrates. Using a Calculator to Find Values Example 5.4.1. Use a calculator to approximate the following values to four decimal places. 1. 2. 3. 4. Solution. 1. Since, we can use a property from Theorem 5.2 to rewrite as. After verifying that our calculator or other graphing tool is in radian mode, we find 2. Noting that, we can use a property from Theorem 5.3 to write arccot21sec5
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 definition, the real number satisfies with. Since, we know more specifically that, so t corresponds to an angle β in Quadrant IV with.  We next visualize the angle radians and note that θ is a Quadrant II angle with. This means θ is exactly π units away from β, and we get Hence, radians. 4. To approximate, noting that, we can use Theorem 5.3: 1cot211tan21cot211tan211tan2t1tan2t22ttan0t02t1tan21cot21tan211tan22.6779.
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 the domain of is, we can find the domain of by setting the argument of the arccosine, in this case, between –1 and 1. So the domain of is. To determine the range of f, we select three key points on the graph of, and. We use transformations to track these points to :, and on the graph of. Plotting these values tells us that the range of f is. The following graphing calculator screen confirms a domain of and range of. arccos25xfx13tan4fxxarccot2xgxarccosFxx11xarccos25xfx5x11515515555xxx�
�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 defined for all real numbers, we have established that the domain of f is all real numbers. To determine the range of f, we can choose the key point along with horizontal asymptotes and from the graph of. We find that the graph of differs by a horizontal compression with a factor of 4 and a vertical stretch with a factor of 3. It is the latter which affects the range, producing a range of. We confirm the domain of and range of using graphing technology. arccos25xyfx13tan4fxx1
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 of horizontal asymptotes, in this case and. We graph starting with and first performing a horizontal expansion by a factor of 2, followed by a vertical shift upwards by π. This latter transformation is the one which affects the range, making it. To check this graphically using technology, it may be necessary to create a piecewise defined function for that makes use of the arctangent function.  Using Theorem 5.2, we have that when, or, in this case, when. Hence, for, we have. 13tan4yfxxarccot2xgxarccotGxx,2x,arccotGxx�
�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 arctangent will produce this result. Finally, we have a piecewise function to use when graphing with technology such as graphing calculators. The graph confirms a domain of and range of. 02x111cot2tan22arccotarctan2xx0xarccot22arctan2arctan2.xgxxx�
�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 required. One such scenario is presented in the following example. Example 5.4.3.1 The roof on the house below has a 6/12 pitch. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to the nearest hundredth of a degree. Front View Side View Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using trigonometric functions of right triangles, we find the angle of inclination, labeled θ below, satisfies. Since θ is an acute angle, we can use the arctangent function and we find (using a calculator in degree mode) 1 Thanks to Dan Stitz for this problem. 61tan122
 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. 9. 12. 15. 18. 19. A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it touches level ground 360 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place. 20. At Cliffs of Insanity Point, the Great Sasquatch Canyon is 7117 feet deep. From that point, a fire is seen at a location known to be 10 miles away from the base of the sheer canyon wall. What angle of depression is made by the line of sight from the canyon edge to the fire? Express your answer using degree measure rounded to one decimal place. 21. Shelving that is being built at the college library is to be 14 inches deep. An 18-inch rod will be attached to the wall and to the underside of the shelf, at its edge away from the wall, forming a right triangle under the shelf to support it. What angle, to the nearest degree, will the rod make with the wall? 1cos0.4arcsin0.233arccos51cos0.81tan6arctan61sin5fxx131cos2xfx2arcsinsin2fxx21arccos4
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 elevation from the boat to the parasailor? Express your answer using degree measure rounded to one decimal place. 23. A tag-and-release program to study the Sasquatch population of the eponymous Sasquatch National Park is begun. From a 200 foot tall tower, a ranger spots a Sasquatch lumbering through the wilderness directly towards the tower. Let θ denote the angle of depression from the top of the tower to a point on the ground. If the range of the rifle with a tranquilizer dart is 300 feet, find the smallest value of θ for which the corresponding point on the ground is in range of the rifle. Round your answer to the nearest hundredth of a degree. 24. Suppose a 13-foot ladder leans against the side of a house, reaching to the bottom of a second-floor window 12 feet above the ground. What angle does the ladder make with the house? Round your answer to the nearest tenth of a degree. 246 CHAPTER 6 TRIGONOMETRIC EQUATIONS Chapter Outline 6.1 Solving Equations Using the Inverse Trigonometric Functions 6.2 Solving Equations Involving a Single Trigonometric Function 6.3 Solving Equations of Multiple Tr
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 solutions of trigonometric equations. Section 6.2 applies identities and inverse functions in solving equations of a single trigonometric function. The process continues in Section 6.3 for equations that include multiple trigonometric functions and/or differing arguments. This essential chapter is primarily used to develop algebraic skills necessary for solving equations. In Chapter 7, these skills will be put to use in the many applications of triangles made possible through the Law of Sines and the Law of Cosines. 247 6.1 Solving Equations Using the Inverse Trigonometric Functions In this section you will: Learning Objectives  Use inverse trigonometric functions to solve right triangles  Use inverse trigonometric functions to solve for angles in trigonometric equations. Solving for Angles in Right Triangles In Section 5.4, we used inverse trigonometric functions to solve real-world applications. Here, we apply the same technique in solving for acute angles within right triangles. Using inverse trigonometric functions, we can determine an angle, being given only the value of a trigonometric function, such as cosine in the following example. Thus, inverse trigonometric functions truly serve as inverses with the angle representing the inverse of the trigonometric function. Example 6.1.1. Solve the following triangle for the angle θ. Solution. Because we know the lengths of the hypotenuse and the side adjacent to the angle θ, it makes sense for us to use the cosine function. 19cos129cos120.7227 radians or 41.4096 from properties of the arccosine 248 Knowing the measure of one acute angle in a right triangle, we can easily determine the measure of the second acute angle. In the example above, the measure of the angle opposite the side of length 9 would be, approximately,. Note that the exact measure is. Solving for Angles in Trigonometric Equations In Section 2.3, we learned how to solve equations like and. In
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 functions, we are now in a position to solve these equations. A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation is a lot like in that it has friendly ‘common value’ answers. The equation, on the other hand, is a lot like. We know there are answers but we can’t express them using ‘friendly’ numbers.1 To solve, we make use of the square root function and write. We can certainly approximate these answers using a calculator, but as far as exact answers go, we leave them as. In the same way, we will use the arcsine function to solve, as seen in the following example. Example 6.1.2. Solve the following equations. 1. Find all angles θ for which. 2. Find all real numbers t for which. 1 This is all, of course, a matter of opinion. Many find just as nice as. 1809041.409648.590419sin121sin2tan1t1sin3tan2t24x1sin22x27x1sin327x7x7x1sin31sin3tan2t72 249 3. Solve for x. Solution. 1. If, then the terminal side of θ, when plotted in standard
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 in Quadrant II. Noting that is not the sine of any of the common angles, we use the arcsine function to express our answers. The real number is defined so it satisfies with. Hence, radians. Since the solutions in Quadrant I are all coterminal with α, we get part of our solution to be Turning our attention to Quadrant II, we get one solution to be. Hence, the Quadrant II solutions are 5sec3x1sin313y131arcsin3t02t1sin3t1arcsin321arcsin2, for integers.3kkkxy 13arcsin radians 13xy 13  250 Our final answer is that the solution to is or for all integers k. 2. We may visualize the solutions to as angles θ with. Since tangent is negative only in Quadrants II and IV, we focus our efforts there. We note that none of the common angles have tangent –2, so we need to use the arctangent, or inverse tangent, function to express our answers. The real number satisfies and. If we let radians, we see that all of the Quadrant IV solutions to are coterminal with β. Moreover,
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tan21tan2ttan2t02t1tan2tan2ttan2t1tan2 for some integer.kkktan2t1tan2tkxy 1tan2 radiansxy   251 3. The last equation we are asked to solve,, poses an immediate problem. We are not told whether or
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 express our solutions in terms of the arcsecant function. The real number is defined so that with. If we let, we see that β is a Quadrant II angle. To obtain a Quadrant III angle solution, we may simply use. Since all angle solutions are coterminal with β or –β, we get our solutions for to be, for integers k, or or Switching back to the variable x, we record our final answer to as or for integers k. 5sec3x5sec3535arcsec3x2x5sec3x5arcsec35arcsec35sec32k2k5arcsec23k5arcsec2.3k5sec3x5arcsec23xk�
�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3arccos25xkxy 35arccos radiansxy 35arccos radians 35arccos radians 253 6.1 Exercises In Exercises 1 – 2, find the angle θ in the given right triangle. Express your answers using degree measure rounded to two decimal places. 1. 2. In Exercises 3 – 5, find the two acute angles in the right triangle whose sides have the given lengths. Express your answers using degree measure rounded to two decimal places. 3. lengths 3, 4 and 5 4. lengths 5, 12 and 13 5. lengths 336, 527 and 625 In Exercises 6 – 25, solve the equation using the techniques discussed in Example 6.1.2. Then approximate the solutions which lie in the interval to four decimal places. 6.
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 Solving Equations Involving a Single Trigonometric Function In this section you will: Learning Objectives  Write complete solutions to equations containing a single trigonometric function.  Evaluate exact solutions in the interval [0,2π). In this section, we continue solving some basic equations involving trigonometric functions. Below we summarize the techniques that were first introduced in Section 2.3. Note that we use the letter u as the argument of each circular function for generality. Strategies for Solving Basic Equations Involving Trigonometric Functions  To solve or for, first solve for u in the interval and add integer multiples of the period 2π. If or, there are no real solutions.  To solve or for
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 guidelines, we can comfortably solve and find the solution or for integers k. How do we solve something like? This equation has the form, so we know the solutions take the form or for integers k. Then, since the argument of sine is 3x, we have or for integers k. To cosucsinuc11c0,21c1csecuccscuc1c1c11ctanuc,22cotuc0c0ccot0u2uk1sin2x26xk526xk1sin32x1sin2u26uk526uk326xk5326xk 255 solve for x, we divide both sides1 of the equations by 3, and obtain our solution of or for integers k. In the remainder of this section, we look at examples of equations which contain a single trigonometric function
. 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 solutions to are or for integers k. Since the argument of cosine here is 2x, this means or for integers k. Solving for x gives or for integers k. To determine which of our solutions lie in, we substitute the integer values into and : k … –2 – Don’t forget to divide the 2πk by 3 as well! 2183xk52183xk1sin32x0,23cos22x0,23cos2u526uk726uk5226xk7226xk512xk712xk0,20,1,2,k512xk712xk512xk191271251217122912712xk171251271219123112 256 The solutions in the interval correspond to and. They are,, and. In the preceding example, the solutions and can be checked analytically by substit
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. 0,20k1k512x71217121912512xk712xk3cos22x512xk55cos2cos21265cos63.2kk since the period of cosine is 2712xk77cos2cos21267cos63.2kk512xk712xkcot30x0,2 257 Solution. Since has the form, we know. So, in this case, for integers k. Solving for x yields. We next determine which of our solutions lie in. k … –1 0 1 2 3 4 5 6 … … … The solutions in are,,,
, 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 6.2.3. Solve, giving the exact solutions which lie in. cot30xcot0u2uk32xk63xk0,263xk6625676321161360,26x25676321160k5k63xkcot30xcot3cot632cot20.kk since the period of cotangent is 63xk1csc23x0,2 258 Solution. Noting that this equation has the form, we rewrite it as and find or for integers k. Since the argument of cosecant here is, or. We solve the first of these equations for x. Solving the other equation,, produces. Putting these two solutions
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123412341523453241564xkxkxkxkxk add to both sides combine the like terms and multiply both sides by 3413234xk2164xk1564xk2164xk1csc23x0,21csc3yx2y0,2 259 The reader is encouraged to check the solutions of Example 6.2.3 as we did following the first two examples in this section. We next solve an equation that at first glance does not fit
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 square roots to get. Converting to cosines, we have.  For  For, we get or for integers k., we get or for integers k. If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of. As a result, these solutions can be combined and we may write our solutions as and for integers k. The solutions in the interval come from the values and as indicated in the following table. 2sec4x0,22sec4xsec2x1cos2x1cos2x23xk523xk1cos2x223xk423xk33xk23xk0,20k1kxy 1csc3yx
 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 and determine solutions that lie in the interval. Solution. The equation has the form, whose solution is. Hence,, so for integers k. To determine which of our answers lie in the interval, we first need to get an idea of the value of. While we could easily find an approximation using a calculator,2 we proceed analytically. Since, it follows that. Multiplying through by 2 gives. We are now in a position to argue which of the solutions lie in.  For, we get, so we discard this answer and all answers where. 2 Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be? 3xk233437323xk32353830,23x234353tan32x0,2tan32xtan3uarctan3ukarctan32xk2arctan32xk0,22arctan330arctan302�
��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 of the form for. Thus, the only solution of in the interval is. A similar process determines the solutions of the following equation involving a sine function. Example 6.2.6. Solve and find solutions in the interval. Solution. To solve, we first note that the equation has the form, which has the family of solutions or for integers k. Since the argument of sine here is 2x, we get or which gives or for integers k. To determine which of these solutions lie in, we first need to get an idea of the value of. Once again, we could use a calculator but we adopt an analytic route here. 1k2arctan32x2arctan302arctan3222arctan32x0,22arctan34x2arctan3032arctan344�
��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 definition after multiplying through by 262  Starting with the family of solutions, we use the same kind of arguments as in our solution to Example 6.2.5 and find only the solutions corresponding to and lie in : and.  Next, we move to the family. Here we need to get a better estimate of. Proceeding with the usual arguments, we find the only solutions which lie in correspond to and, namely
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1arcsin0.872210arcsin0.872410arcsin0.87241arcsin0.8722241arcsin0.8742222 from above multiply through by -1 add 0,20k1k1arcsin0.8722x31arcsin0.8722xsin20.87x0,21arcsin0.872x1arcsin0.872x1arcsin0.8722x31arcsin0.8722x 263 6.2 Exercises In Exercises 1 –18, find the exact solutions of the equation and then
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 should emerge. Explain how this pattern would help you solve equations like. Now consider, and. What do you find? Replace with –1 and repeat the whole exploration. 0,2sin50x1cos32x3sin22xtan61xcsc41xsec32x3cot23xcos99x2sin32x5cos06x1sin232x72cos34xcsc0xtan21x2tan3x24sec3x21cos2x23sin4xarccos2x2arcsin2x4arctan310x6arccot250x4arcsec2x�
��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 trigonometric functions and/or arguments.  Evaluate exact solutions in the interval [0,2π). Each of the examples in Section 6.2 featured one trigonometric function. If an equation involves two different trigonometric functions, or if the equation contains the same trigonometric function but with different arguments, we will need to use identities and algebra to reduce the equation to a form similar to the equations in Section 6.2. Equations with Different Powers of the Same Function Example 6.3.1. Solve the equation and list the solutions which lie in the interval. Solution. We resist the temptation to divide both sides of by (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor. We get or. Solving for, we find or.  The solution to is, with and being two solutions which lie in.  To solve, we use the arcsine function to get or for integers k. We find the two solutions here which lie in to be and. 323sinsinxx0,2�
�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 necessary to ‘zoom in’ close to and to verify that the graphs do in fact intersect four times. interval close to 0 close to π To summarize, the solutions to in the interval are,, and.1 Equations Containing Multiple Trigonometric Functions In the next example, we make use of a Pythagorean identity. Example 6.3.2. Solve the equation and list the solutions in. Solution. Analysis of reveals two different trigonometric functions, so an identity is in order. Since, we have 1 Note that we are not counting as a solution since it is not in the interval. In the forthcoming solutions, remember that while may be a solution to the equation, it isn’t counted among the solutions in. 33sinyx
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. The solutions which lie in are and.  To solve, we employ the arctangent function and get for integers k. Using the same sort of argument we saw in Example 6.2.5, we get and as solutions in. The points of intersection on the following graph of and correspond to the four solutions of on the
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0,2arctan2x34xarctan2x74x�
��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. Solution. In the equation we have the same trigonometric function, namely cosine, on both sides, but the arguments differ. Using the identity, we obtain an equation quadratic in form and can proceed as we have done in the past. This gives or. Since, we have or.  Solving, we get or for integers k.  From, we get for integers k. The answers which lie in are, and. Try graphing and to verify that the curves intersect in three places on, and that the x-coordinates of these points confirm our results. Next, we look at an example that uses a technique similar to Example 6.3.3, but relies on an identity established in Example 4.5.1. Example 6.3.4. Solve the equation. Solution. From Example 4.5.1, we know that. This transforms our equation into a polynomial in terms of. cos23cos2xxcos23cos2xx2cos22cos1xx222cos22cos1coscos23cos22cos13cos22cos3cos1023102110xxuxxxxxxxuuuu since
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 for integers k. The only solutions which lie are and. Equations Containing Different Functions and Different Arguments The following example includes different trigonometric functions and different arguments. Example 6.3.5. Solve for x. Solution. In examining the equation, not only do we have different functions involved, namely sine and cosine, we also have different arguments to contend with, namely 2x and x. Using the identity makes all of the arguments the same and we proceed as we would solving any nonlinear equation, We gather all of the nonzero terms on one side of the equation and factor. 3332coscos313cos4cos3cos13cos4cos16cos041604404220uxxxxxxxxuuuuuuu�
� 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 which lie in are,, and. The last example in this section, which also includes different functions and arguments, tests our memory a bit and introduces another solution technique. Example 6.3.6. Solve the following equation for x:. Solution. Unlike the previous problem, there seems to be no quick way to get the circular functions or their argument to match in the equation. If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for. Hence, our original equation is equivalent to. We proceed in solving for x. cos0x3sin2x
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 following two solutions in the interval : and. When solving trigonometric equations, try something! Practice will help, but trying different solution techniques will improve your problem solving skills. After working through the examples in this section, spend some time with the problems in the Exercises. Try checking your solutions through viewing the intersection of graphs, as in Example 6.3.1 and Example 6.3.2. 433xk0,23x53x 271 6.3 Exercises In Exercises 1 –40, solve the equation, giving the exact solutions which lie in. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35
. 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42cot4csc7xx2cos2csc0xx3tan3tanxx23tansec2xx3coscosxx�
�tan22cos0xx32csccsc4csc4xxx22tan1tanxxtansecxxsin6coscos6sinxxxxsin3coscos3sinxxxxcos2cossin2sin1xxxx3cos5cos3sin5sin32xxxxsincos1xxsin3cos1xx2cos2sin1xx3sin2cos21xxcos23sin22xx33sin33cos333xxcos3cos5xxcos4cos2xx 37. 39. 272 38. 40. sin5sin3xxcos5cos2xxsin6sin0xxtancosxx
 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 defined as Angle-Side-Angle, Angle-Angle-Side and Side-Side- Angle. The Law of Sines is introduced to assist with solving triangles of these types. Section 7.2 further promotes the Law of Sines as a tool in finding the area of a triangle and in applying the Law of Sines to various applications. In Section 7.3, the Law of Cosines is used to solve triangles of the types Side- Angle-Side and Side-Side-Side. Solutions to many applications are made possible through the Law of Cosines. Heron’s Formula, which is proved through the Law of Cosines, provides a simple method for finding the area of a triangle in which the lengths of all three sides are known. This is a critical chapter in developing aptitude for solving real-life trigonometric applications. It is followed by Chapter 8, in which we delve into the more theoretical, but still very application-oriented, polar coordinates and complex numbers. 274 7.1 Solving Triangles with the Law of Sines Learning Objectives In this section you will:  Use the Law of Sines to solve oblique triangles.  Distinguish between ASA, AAS and ASS triangles.  Determine the existence of, and values for, multiple solutions of oblique triangles.  Determine when given criteria will not result in a triangle. Trigonometry literally means ‘measuring triangles’ and with Chapters 1 – 6 under our belts, we are more than prepared to do just that. The main goal of this chapter is to develop theorems which allow us to solve triangles – that is, find the length of each side of a triangle and the measure of each of its angles. Solving Right Triangles We have had some experience solving right triangles. The following example reviews what we know. Example 7.1.1. Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express
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 we can find α and β using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to the data given to us in the problem. In this case, the lengths 4 and 7 were given. 22247a33ab = 4c = 7a   275  We want to relate the lengths 4 and 7 to α. Since and α is an acute angle, radians. Converting to degrees, we find.  We see that, so radians and we have. Note that we could have used the measure of angle α to find the measure of angle β, using the fact that α and β are complements, from which. A few remarks about Example 7.1.1 are in order. 1. First, we adhere to the convention that a lower case Greek letter denotes an angle1 and the corresponding lowercase English letter represents the side2 opposite that angle. Thus, a is the side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs, and are called angle-side opposite pairs. 2. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error.3 3. Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Law of Sines The Pythagorean Theorem along with the definitions of the trigonometric functions in Section 2.1 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? Any triangle that is not a right triangle is an oblique triangle. In certain cases, we can use the Law of Sines to solve oblique triangles. 1 as well as the measure of said angle 2 as well as the length of said side 3 Your Science teachers should thank us for
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 into three cases. 1. For our first case, consider the triangle below, all of whose angles are acute, with angle- side pairs, and. If we drop an altitude from vertex B, we divide the triangle into two right triangles: and. If we call the length of the altitude h (for height), we get that and so that. After some rearrangement of the last equation, we get. If we drop an altitude from vertex A, we can proceed as above using the triangles and to get, completing the proof for this case. ,a,b,csinsinsinabcsinsinsinabcABC,a,b,cABQBCQsinhcsinhasinsinhca
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 generates two right triangles, and. We know that so that. Since, so in fact we have. Proceeding to, we get so,. Putting this together with the previous equation, we get, and we are finished with this case. 3. The remaining case is when is a right triangle. In this case, the definitions of trigonometric functions from Section 2.1 can be used to verify the Law of Sines and this verification is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least three measurements of angles and/or sides, including at least one of the sides. Also, note that we need to be given, or be able to find, at least one angle-side opposite pair. We will investigate three possible oblique triangle problem situations. AAS (Angle-Angle-Side) Here, we know the measurements of two angles and a side that is not between the known angles. Example 7.1.2. Solve the triangle:, units,. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. ABCABQACQsinhbsinhcsinsinbcABQBCQ'sin'hc'sin'hc'180sin'sin�
�'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 triangle is 180°. Hence,. To find c, we have no choice but to use the derived value, yet we can minimize the propagation of error here by using the given angle-side opposite pair. The Law of Sines gives us The exact value of could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus ‘exact’ here means. 1207a7sin45sin1207sin45sin12072232725.72 u45nits3bbbb since 180120451515,a7sin15sin1207sin15sin1202.09 unitscccsin157sin15sin120c 279 ASA (Angle-Side-Angle) In this case, we know the measurements of two angles and the included side. Example 7.1.3. Solve the triangle:,, units. Give exact answers and decimal approximations (rounded to hundredths) and sketch the
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 find b we use the angle-side pair which yields ASS (Angle-Side-Side) Knowing the measurement of two sides and an angle that is not between the known sides proves to be more complex than our first two scenarios. While we can use the Law of Sines to solve any oblique triangle, 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 ASS, 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 solutions. 85305.25c180853065,c5.25sin85sin655.25sin85sin655.77 unitsaaa,c5.25sin30sin655.25sin30sin652.90 unitsbbb 280 Example 7.1.4. Solve the triangle:, unit, units. Solution. Since we are given and c, we use the Law of Sines to find the measure of γ. Since the range of the sine function is, there is no real number with. Geometrically, we see that side a is just too short to make a triangle. The following examples keep the same value for the measure of α and the length of c while varying the length of a. We will discuss the preceding case in more detail after we see what happens in the next three examples. Example 7.1.5. Solve the triangle:, units, units. Solution
. 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 be and then determine b using the Law of Sines. 301a4c,asin30sin41sin4sin30sin21,1sin2302a4csin30sin42sin2sin30sin130sin190180309060 281 In this case, the side a is precisely long enough to form a unique right triangle. Example 7.1.6. Solve the triangle:, units, units. Solution. Proceeding as we have in the previous two examples, we use the Law of Sines to find γ. In this case, we have Since γ lies in a triangle with, we must have. There are two angles γ that fall in this range and have : radians, approximately 41.81°, and radians, approximately 138.19°. sin30sin6022sin60sin3023212233.46 unitsbbbb303a4c
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 Both triangles are drawn below. Example 7.1.7. Solve the triangle:, units, units. Solution. For this last problem, we repeat the usual Law of Sines routine to find that 5 To find an exact expression for β, we convert everything back to radians: radians, radians and radians. Hence, radians ≈ 108.19°. 6 An exact answer for β in this case is radians ≈ 11.81°. 2arcsin31803041.81108.19,a3sin108.195.70sin30b2arcsin3304a4csin30sin44sinsin301sin23062arcsin3
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 we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely determined using the equation. Knowing the measures of all three angles of a triangle completely determines its shape. 2. If, in addition to being given the measures of two angles, we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides in order to determine the size of the triangle. This is true for the previously described AAS and ASA cases. 3. If we are given the measure of just one of the angles along with the lengths of two sides, only one of which is adjacent to the given angle, we have the ASS case.7 As we saw in Examples 7.14 – 7.17, this information may describe one right triangle, one oblique triangle, two oblique triangles, or no triangle. 7 In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 300150301803030120sin30sin12044sin120sin3043212436.93 unitsbbbb180 The four possibilities in the ASS case are summarized in the following theorem. 284 Theorem 7.2. Suppose and are intended
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 exactly one triangle exists which satisfies the given criteria. Theorem 7.2 is proved on a case-by-case basis.  If then. If a triangle were to exist, the Law of Sines would have so that, which is impossible. In the figure below, we see geometrically why this is the case., no triangle Simply put, if the side a is too short to connect to form a triangle. This means if, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case.  If, then and the Law of Sines gives so that. Here, as required. ,a,csinhcahah90hacacahsinacsinsincasinsin1caaaahahahahsinacsinsinacsinsin1caaa90 285,  Moving along, now suppose. As before, the Law of Sines gives. Since, or which means there are two solutions to : an acute angle which we’ll call, and its supplement. We need
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 manipulate the inequality a bit, we have, which gives and. This proves a triangle can contain both of the angles α, giving us the second triangle predicted in the theorem.  To prove the last case in the theorem, we assume. Then, which forces γ to be an acute angle. Hence, we get only one triangle in this case, completing the proof., two triangles ah90hacsinsincahasincasin1casinsinca00180,a0,cca0000018018001801800180hacac 286, one triangle One last comment before we end this discussion. In the Angle-Side-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. We will next move on to Section 7.2 where we use the Law of Sines to solve application problems. ac 287 7.1 Exercises In Exercises 1 – 12, find the length of side x. Round to the nearest tenth. 1. 3. 5. 2. 4.
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45a42b 19. 21. 23. 25. 27. 29. 31.,,,,,,,,,,,,,, 289 20. 22. 24. 26. 28. 30. 32.,,,,,,,,,,,,,, 33. Find the radius of the circle. Round to the nearest tenth. 34. Find the diameter of the circle. Round to the nearest tenth. 35. Prove that the Law of Sines holds when is a right triangle. 36. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 68.788a92b4217a23.5b68.770a90b307a14b4239a23.5b535328.01c657a100b74.63c�
�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 are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 36. Given and, choose four different values for a so that a) the information yields no triangle b) the information yields exactly one right triangle c) the information yields two distinct triangles d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have,, and your choice of a yield only one triangle where that unique triangle has three acute angles. 3010b3010b 291 7.2 Applications of the Law of Sines Learning Objectives In this section you will:  Find the area of an oblique triangle using the sine function.  Solve applied problems using the Law of Sines. Following our practice with solving triangles for missing values in Section 7.1, we begin Section 7.2 by using some of those values to find the area of an oblique triangle. Finding the Area of an Oblique Triangle The following theorem introduces a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 7.3. Suppose, and are the angle-side opposite pairs of a triangle. Then the area enclosed by the triangle is given by Example 7.2.1. Find the area of the triangle in which, units, and. Solution. This is the triangle from Example 7.1.2 in which we found all three
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7sin15sin120c 292 The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 7.3. 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. Example 7.2.2. Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 15 degrees, whereas the angle of elevation measured by the second station is 35 degrees. Find the altitude of the aircraft and round your answer to the nearest tenth of a mile. Solution. To find the altitude, or height, of the aircraft, we first sketch a triangle which reflects the information given to us in the problem. We then use the triangle to determine the distance from one station to the aircraft. Letting a represent the distance from the first station to the aircraft, we look for an angle-side opposite pair from which we can determine the distance a. We know the measure of two angles in the triangle, but the measure of the angle opposite the side of length 20 miles is missing. Noting that the angles in a triangle add up to 180 degrees, we find the unknown angle measure to be. This gives us an angle-side opposite pair with known values and allows us to set up a Law of Sines relationship. 

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