Trigonometric Identities and Equations examples

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9 Trigonometric Identities and Equations Figure 1 A sine wave models disturbance. (credit: modification of work by Mikael Altemark, Flickr). ChAPTeR OUTl Ine 9.1 Solving Trigonometric equations with Identities 9.2 Sum and difference Identities 9.3 double-Angle, half-Angle, and Reduction Formulas 9.4 Sum-to-Product and Product-to-Sum Formulas 9.5 Solving Trigonometric equations Introduction Math is everywhere, even in places we might not immediately recognize. For example, mathematical relationships describe the transmission of images, light, and sound. The sinusoidal graph in Figure 1 models music playing on a phone, radio, or computer. Such graphs are described using trigonometric equations and functions. In this chapter, we discuss how to manipulate trigonometric equations algebraically by applying various formulas and trigonometric identities. We will also investigate some of the ways that trigonometric equations are used to model real-life phenomena. 695696 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns l eARnIng Obje CTIveS In this section, you will: • Verify the fundamental trigonometric identities. • Simplify trigonometric expressions using algebra and the identities. 9.1 SOlv Ing T RIgOn OmeTRIC eQUATIOn S WITh Iden TITIeS Figure 1 International passports and travel documents In espionage movies, we see international spies with multiple passports, each claiming a die ff rent identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation. In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions. verifying the Fundamental Trigonometric Identities Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the die ff rence of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways. To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. We will begin with the Pythagorean identities (see Table 1), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identie fi s, but now we will also use additional identities. Pythagorean Identities 2 2 2 2 2 2 sin θ + cos θ = 1 1 + cot θ = csc θ 1 + tan θ = sec θ Table 1SECTION 9.1 s ol vi N g t rig o Nome tri c e aqu tio Ns wi t h i de Ntities 697 2 2 e s Th econd and third identities can be obtained by manipulating the first. The identity 1 + cot θ = csc θ is found by rewriting the left side of the equation in terms of sine and cosine. 2 2 Prove: 1 + cot θ = csc θ 2 cos θ 2 _____ Rewrite the left side. 1 + cot θ = 1 +        2 sin θ 2 2 Write both terms with the common sin θ cos θ _____ _____ =      +          2 2 denominator. sin θ sin θ 2 2 sin θ + cos θ ___________ =     2 sin θ 1 ____ =   2 sin θ 2 = csc θ 2 2 Similarly, 1 + tan θ = sec θ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives 2 sin θ 2 ____ 1 + tan θ = 1 +       Rewrite left side.   cos θ 2 2 cos θ sin θ Write both terms with the ____ ____ =      +            cos θ cos θ common denominator. 2 2 cos θ + sin θ ___________ =     2 cos θ 1 _____ =   2 cos θ 2 = sec θ Recall that we determined which trigonometric functions are odd and which are even. The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. (See Table 2). Even-Odd Identities tan(−θ) = −tan θ sin(−θ) = −sin θ cos(−θ) = cos θ cot(−θ) = −cot θ csc(−θ) = −csc θ sec(−θ) = sec θ Table 2 Recall that an odd function is one in which f (−x) = −f (x) for all x in the domain of f. The sine function is an odd function because sin(−θ) = −sin θ. The gr aph of an odd function is symmetric about the origin. For example, consider π π π π __ __ __ __ corresponding inputs of and − . The output of sin is opposite the output of sin −  . Thus,     2 2 2 2 π __ sin = 1   2 and π π __ __ sin −   = −sin     2 2 = −1 This i s shown in Figure 2. y π , 1 2 2 f(x) = sin x x –π π 2π –2π π , – –2 –1 2 Figure 2 graph of y = sin θ Recall that an even function is one in which f (−x) = f (x) for all x in the domain of f698 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns The graph of an even function is symmetric about the y-axis. The cosine function is an even function because π π π __ __ __ cos(−θ) = cos θ. For example, consider corresponding inputs and − . The output of cos is the same as the   4 4 4 π __ output of cos −  . Thus,   π π 4 __ __ cos −  = cos     4 4 ≈ 0.707 See Figure 3. y π π 2 – , , 0.707 0.707 4 4 x –π π 2π –2π f(x) = cos x –2 Figure 3 graph of y = cos θ For all θ in the domain of the sine and cosine functions, respectively, we can state the following: • Since sin(−θ) = −sin θ, sine is an odd function. • Since, cos(−θ) = cos θ, cosine is an even function. The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, tan(−θ) = −tan θ. We can interpret the tangent of a negative angle as sin(−θ) −sin θ ______ ______ tan(−θ) = = = −tan θ. Tangent is therefore an odd function, which means that tan(−θ) = −tan(θ) cos(−θ) cos θ for all θ in the domain of the tangent function. The cotangent identity, cot( −θ) = −cot θ, also follows from the sine and cosine identities. We can interpret the cos(−θ) cos θ _______ _____ cotangent of a negative angle as cot(−θ) = = = −cot θ. Cotangent is therefore an odd function, sin(−θ) −sin θ which means that cot(−θ) = −cot(θ) for all θ in the domain of the cotangent function. The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be 1 1 ______ _____ interpreted as csc(−θ) = = = −csc θ. The c osecant function is therefore odd. sin(−θ) −sin θ Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted 1 1 ______ ____ as sec(−θ) = = = sec θ. The s ecant function is therefore even. cos(−θ) cos θ To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities. The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. See Table 3. Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry. Reciprocal Identities 1 1 ____ ____ sin θ csc θ = = csc θ sin θ 1 1 ____ ____ cos θ sec θ = = sec θ cos θ 1 1 ____ ____ tan θ cot θ = = cot θ tan θ Table 3 The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See Table 4. Quotient Identities sin θ cos θ ____ ____ tan θ cot θ = = cos θ sin θ Table 4SECTION 9.1 s ol vi Ng t rig o Nome tri c e aqu tio N ws i t h i de Ntities 699 e r Th eciprocal and quotient identities are derived from the definitions of the basic trigonometric functions. summarizing trigonometric identities The Pythagorean identities are based on the properties of a right triangle. 2 2 cos θ + sin θ = 1 2 2 1 + cot θ = csc θ 2 2 1 + tan θ = sec θ The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan(−θ) = −tan θ cot(−θ) = −cot θ sin(−θ) = −sin θ csc(−θ) = −csc θ cos(−θ) = cos θ sec(−θ) = sec θ The reciprocal identities define reciprocals of the trigonometric functions. 1 ____ sin θ = csc θ 1 ____ cos θ = sec θ 1 ____ tan θ = cot θ 1 ____ csc θ = sin θ 1 ____ sec θ = cos θ 1 ____ cot θ = tan θ The quotient identities define the relationship among the trigonometric functions. sin θ ____ tan θ = cos θ cos θ ____ cot θ = sin θ Example 1 Graphing the Equations of an Identity 1 1 ____ ____ Graph both sides of the identity cot θ . In other words, on the graphing calculator, graph y = cot θ and y . = = tan θ tan θ Solution See Figure 4. y (4, 3) 5 α x (4, 0) Figure 4700 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns Analysis We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities. How To… Given a trigonometric identity, verify that it is true. 1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build. 2. Look for opportunities to factor expressions, square a binomial, or add fractions. 3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions. 4. If these steps do not yield the desired result, try converting all terms to sines and cosines. Example 2 Verifying a Trigonometric Identity Verify tan θ cos θ = sin θ. Solution We will start on the left side, as it is the more complicated side: sin θ ____ tan θcos θ cos θ =   cos θ sin θ ____ cos θ =   cos θ = sin θ Analysis This identity was fairly simple to verify, as it only required writing tan θ in terms of sin θ and cos θ. Try It 1 Verify the identity csc θ cos θ tan θ = 1. Example 3 Verifying the Equivalency Using the Even-Odd Identities Verify the following equivalency using the even-odd identities: 2 (1 + sin x)1 + sin(−x) = cos x Solution Working on the left side of the equation, we have (1 + sin x)1 + sin(−x) = (1 + sin x)(1 − sin x) Since sin(−x) = −sin x 2 = 1 − sin x Differen ce of squares 2 2 2 = cos x cos x = 1 − sin x 2 Example 4 Verifying a Trigonometric Identity Involving sec θ 2 sec θ − 1 ________ 2 Verify the identity = sin θ 2 sec θ Solution As the left side is more complicated, let’s begin there. 2 2 sec θ − 1 (tan θ + 1) − 1 ________ _____________ 2 2 sec θ = tan θ + 1 = 2 2 sec θ sec θ 2 tan θ _____ = 2 sec θ 1 2 ____ tan θ =  2  sec θ 1 2 2 2 ____ = tan θ(cos θ) cos θ = 2 sec θ 2 2 sin θ sin θ _____ 2 2 _____ (cos θ) tan θ = =   2 2 cos θ cos θ 2 sin θ _____ 2 = (cos θ)   2 cos θ 2 = sin θ SECTION 9.1 s ol vi N g t rig o Nome tri c e aqu tio N ws i t h i de Ntities 701 e Th re is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side. 2 2 sec θ − 1 sec θ 1 ________ _____ ____ = − 2 2 2 sec θ sec θ sec θ 2 = 1 − cos θ 2 = sin θ 2 2 Analysis In the first method, we used the identity sec θ = tan θ + 1 and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same. Try It 2 cot θ ____ Show that = cos θ. csc θ Example 5 Creating and Verifying an Identity Create an identity for the expression 2tan θ sec θ by rewriting strictly in terms of sine. Solution e Th re are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression: sin θ 1 ____ ____ 2 tan θ sec θ 2 =    cos θ cos θ 2 sin θ _____ = 2 cos θ 2 sin θ ________ 2 2 Substitute 1 − sin θ for cos θ = 2 1 − sin θ Thus, 2 sin θ ________ 2tan θ sec θ = 2 1 − sin θ Example 6 Verifying an Identity Using Algebra and Even/Odd Identities Verify the identity: 2 2 sin (−θ) − cos (−θ) _________________ = cos θ − sin θ sin(−θ) − cos(−θ) Solution Let’s start with the left side and simplify: 2 2 2 2 sin (−θ) − cos (−θ) sin(−θ) − cos(−θ) _________________ __________________ = sin(−θ) − cos(−θ) sin(−θ) − cos(−θ) 2 2 sin(−x) = −sin x and cos(−x) = cos x (−sin θ) − (cos θ) ________________ = −sin θ − cos θ 2 2 (sin θ) − (cos θ) ______________ Differen ce of squares = −sin θ − cos θ (sin θ − cos θ)(sin θ + cos θ) ______________________ = −(sin θ + cos θ) (sin θ − cos θ)(sin θ + cos θ) ______________________ = −(sin θ + cos θ) = cos θ − sin θ Try It 3 2 sin θ − 1 sin θ + 1 _____________ _______ Verify the identity = . tan θ sin θ − tan θ tan θ 702 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns Example 7 Verifying an Identity Involving Cosines and Cotangents 2 2 Verify the identity: (1 − cos x)(1 + cot x) = 1. Solution We will work on the left side of the equation 2 cos x 2 2 2 _____ (1 − cos x)(1 + cot x) (1 − cos x) 1 + =   2 sin x 2 2 sin x cos x Fin d the common denominator. 2 _____ _____ (1 − cos x) + =   2 2 sin x sin x 2 2 sin x + cos x 2 ___________ (1 − cos x) =   2 sin x 1 2 ____ = (sin x)  2  sin x 1 = Using Algebra to Simplify Trigonometric expressions We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations. For example, the equation (sin x + 1)(sin x − 1) = 0 resembles the equation (x + 1)(x − 1) = 0, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations. 2 2 Another example is the die ff rence of squares formula, a − b = (a − b)(a + b), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve. Example 8 Writing the Trigonometric Expression as an Algebraic Expression 2 Write the following trigonometric expression as an algebraic expression: 2cos θ + cos θ − 1. 2 Solution Notice that the pattern displayed has the same form as a standard quadratic expression, ax + bx + c. Letting cos θ = x, we can rewrite the expression as follows: 2 2x + x − 1 This expression can be factored as (2 x + 1)(x − 1). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for x. At this point, we would replace x with cos θ and solve for θ. Example 9 Rewriting a T rigonometric Expression Using the Difference of Squares 2 Rewrite the trigonometric expression: 4 cos θ − 1. Solution Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the die ff rence of squares. 2 2 4 cos θ − 1 = (2 cos θ) − 1 = (2 cos θ − 1)(2 cos θ + 1) Analysis If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let cos θ = x, rewrite the 2 expression as 4x − 1, and factor (2x − 1)(2x + 1). The n replace x with cos θ and solve for the angle. Try It 4 2 Rewrite the trigonometric expression using the difference of squares: 25 − 9 sin θ. SECTION 9.1 s ol vi Ng t rig o Nome tri c e aqu tio Ns wi th i de Ntities 703 Example 10 Simplify by Rewriting and Using Substitution Simplify the expression by rewriting and using identities: 2 2 csc θ − cot θ Solution We can start with the Pythagorean Identity. 2 2 1 + cot θ = csc θ 2 2 Now we can simplify by substituting 1 + cot θ for csc θ. We have 2 2 2 2 csc θ − cot θ = 1 + cot θ − cot θ = 1 Try It 5 cos θ 1 − sin θ _______ _______ Use algebraic techniques to verify the identity: . = 1 + sin θ cos θ (Hint: Multiply the numerator and denominator on the left side by 1 − sin θ.) Access these online resources for additional instruction and practice with the fundamental trigonometric identities. • Fundamental Trigonometric Identities (http://openstaxcollege.org/l/funtrigiden) • verifying Trigonometric Identities (http://openstaxcollege.org/l/verifytrigiden)704 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns 9 .1 S e CTIO n exe RCIS e S veRbAl 2. Examine the graph of f (x) = sec x on the interval 1. We know g(x) = cos x is an even function, and −π, π. How can we tell whether the function f (x) = sin x and h(x) = tan x are odd functions. What 2 2 2 is even or odd by only observing the graph of about G(x) = cos x, F (x) = sin x, and H(x) = tan x? f (x) = sec x? Are they even, odd, or neither? Why? 3. Aer exa ft mining the reciprocal identity for sec t, 4. All of the Pythagorean identities are related. explain why the function is undefined at certain Describe how to manipulate the equations to get 2 2 points. from sin t + cos t = 1 to the other forms. Algeb RAIC For the following exercises, use the fundamental identities to fully simplify the expression. 5. sin x cos x sec x 6. sin(−x)cos(−x)csc(−x) 2 7. tan x sin x + sec x cos x 8. csc x + cos x cot(−x) cot t + tan t __________ 3 2 9. 10. 3 sin t csc t + cos t + 2 cos(−t)cos t sec(−t) −sin(−x)cos x sec x csc x tan x ________________________ 11. −tan(−x)cot(−x) 12.   cot x 2 1 + tan θ 1 ________ 2 ____ tan x tan x 1 + tan x 1 ____ ____ ________ _____ 13. sin θ + + 14. + − 2 2    2 2 2 csc θ sec θ csc x sec x 1 + cot x cos x 2 1 − cos x 2 ________ 15. + 2 sin x 2 tan x For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression. tan x + cot x sec x + csc x __________ __________ 16. ; cos x 17. ; sin x csc x 1 + tan x cos x 1 _______ ________ 18. tan x; cos x 19. cot x; cot x + − 1 + sin x sin x cos x 1 cos x _______ _______ 20. ; csc x 21. (sec x + csc x)(sin x + cos x) − 2 − cot x; tan x − 1 − cos x 1 + cos x 1 1 − sin x 1 + sin x __________ _______ _______ 22. ; sec x and tan x 23. ; sec x and tan x − csc x − sin x 1 + sin x 1 − sin x 24. tan x; sec x 25. sec x; cot x 26. sec x; sin x 27. cot x; sin x 28. cot x; csc x For the following exercises, verify the identity. 3 2 29. cos x − cos x = cos x sin x 30. cos x(tan x − sec(−x)) = sin x − 1 2 2 1 sin 1 + sin x x ________ _____ _____ 2 2 31. 1 2 tan x 32. (sin x + cos x) = 1 + 2 sin x cos x = + = + 2 2 2 cos x cos x cos x 2 2 2 2 33. cos x − tan x = 2 − sin x − sec x SECTION 9.1 s ectio N e xercises 705 exTen SIOnS For the following exercises, prove or disprove the identity. 1 1 _______ ___________ 2 2 2 35. csc x(1 + sin x) = cot x 34. − = −2 cot x csc x 1 + cos x 1 − cos( − x) 2 2 sec (−x) − tan x 2 + 2 tan x ______________ _________ 2 36. − 2 sin x cos 2x =     tan x 2 + 2 cot x tan x sec(−x) ____ 2 __________ 37. sin(−x) cos x 38. sin(−x) = = − sec x tan x + cot x 1 + sin x cos x _______ _________ 39. = cos x 1 + sin(−x) For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 2 2 cos θ − sin θ ___________ 2 2 2 2 41. 3 sin θ + 4 cos θ = 3 + cos θ 40. = sin θ 2 1 − tan θ sec θ + tan θ 2 __________ 42. sec θ = cot θ + cos θ 706 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns l eARnIng Obje CTIveS In this section, you will: • Use sum and difference formulas for cosine. • Use sum and difference formulas for sine. • Use sum and difference formulas for tangent. • Use sum and difference formulas for cofunctions. • Use sum and difference formulas to verify identities. 9.2 SUm And dIFFeRen Ce Iden TITIeS Figure 1 Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr) How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances. e t Th rigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. e Th se are special equations or postulates, true for all values input to the equations, and with innumerable applications. In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity. Using the Sum and difference Formulas for Cosine Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure 2. 1 1 , , 120° 90° (0, 1) 2 2 60° 2π π , , 3 π 135° 3 3π π 45° 2 4 4 5π π 1 , 1 , 150° 6 6 30° 2 2 π 0°, (1, 0) (–1, 0) 180° 2π 1 1 , 330° , 2 210° 2 11π 7π 6 6 7π , 5π , 315° 3π 225° 4 4 4π 5π 2 3 3 1 1 , , 240° 300° 2 2 270° (0, –1) Figure 2 The Unit CircleSECTION 9.2 s um a Nd d eiff re Nc e i de Ntities 707 We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if we can break it up into the sum or die ff rence of two of the special angles. See Table 1. Sum formula for cosine cos(α + β) = cos α cos β − sin α sin β Die ff rence formula for cosine cos(α − β) = cos α cos β + sin α sin β Table 1 First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 3. Point P is at an angle α from the positive x-axis with coordinates (cos α, sin α) and point Q is at an angle of β from the positive x-axis with coordinates (cos β, sin β). Note the measure of angle POQ is α − β. Label two more points: A at an angle of (α − β) from the positive x-axis with coordinates (cos(α − β), sin(α − β)); and point B with coordinates (1, 0). Triangle POQ is a rotation of triangle AOB and thus the distance from P to Q is the same as the distance from A to B. y A P α – β Q B α β x O 1 Figure 3 We can find the distance from P to Q using the distance formula. —— 2 2 d = √(cos α − cos β) + (sin α − sin β) PQ ———— 2 2 2 2 = √cos α − 2 cos α cos β + cos β + sin α − 2 sin α sin β + sin β  e Th n w e apply the Pythagorean Identity and simplify. ———— 2 2 2 2 = √(cos α + sin α) + (cos β + sin β) − 2 cos α cos β − 2 sin α sin β —— = √1 + 1 − 2 cos α cos β − 2 sin α sin β —— = √ 2 − 2 cos α cos β − 2 sin α sin β Similarly, using the distance formula we can find the distance from A to B. —— 2 2 d = (cos(α − β ) − 1) + (sin(α − β) − 0) √ AB ——— 2 2 = √cos (α − β) − 2 cos(α − β) + 1 + sin (α − β) Applying the Pythagorean Identity and simplifying we get: ——— 2 2 = √(cos (α − β) + sin (α − β)) − 2 cos( α − β) + 1 —— = √ 1 − 2 cos(α − β) + 1 — = √2 − 2 cos( α − β) Because the two distances are the same, we set them equal to each other and simplify. —— — √ 2 − 2 cos α cos β − 2 sin α sin β = √2 − 2 cos( α − β) 2 − 2 cos α cos β − 2 sin α sin β = 2 − 2 cos(α − β) Finally we subtract 2 from both sides and divide both sides by −2. cos α cos β + sin α sin β = cos(α − β) u Th s, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.708 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns sum and difference formulas for cosine e Th s e formulas can be used to calculate the cosine of sums and die ff rences of angles. cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β How To… Given two angles, find the cosine of the die ff rence between the angles. 1. Write the difference formula for cosine. 2. Substitute the values of the given angles into the formula. 3. Simplify. Example 1 Finding the Exact V alue Using the Formula for the Cosine of the Difference of Two Angles 5π π __ __ Using the formula for the cosine of the die ff rence of two angles, find the exact value of cos − .   4 6 Solution Begin by writing the formula for the cosine of the die ff rence of two angles. Then substitute the given values. cos(α − β) = cos α cos β + sin α sin β 5π π 5π π 5π π ___ __ ___ __ ___ __ cos − = cos cos + sin sin           4 6 4 6 4 6 — — — √2 √3 √ 2 1 ____ ____ ____ __ = −   −          2 2 2 2 — — √ 6 √ 2 ____ ____ = − − 4 4 — — −√ 6 − √2 __________ = 4 Keep in mind that we can always check the answer using a graphing calculator in radian mode. Try It 1 π π __ __ Find the exact value of cos   −   .   3 4 Example 2 Finding the Exact V alue Using the Formula for the Sum of Two Angles for Cosine Find the exact value of cos(75°). Solution As 75° = 45° + 30°, we can evaluate cos(75°) as cos(45° + 30°). cos(α + β) = cos α cos β − sin α sin β cos(45° + 30°) = cos(45°)cos(30°) − sin(45°)sin(30°) — — — √ 2 √3 √ 2 1 ____ ____ ____ __ = −     2 2 2 2 — — √ 6 √ 2 ____ ____ = − 4 4 — — √ 6 − √2 __________ = 4 Keep in mind that we can always check the answer using a graphing calculator in degree mode. Analysis Note that we could have also solved this problem using the fact that 75° = 135° − 60°. cos(α − β) = cos α cos β + sin α sin β cos(135° − 60°) = cos(135°)cos(60°) − sin(135°)sin(60°) — — — √ 2 1 √2 √ 3 ____ ____ ____ ____ = − +        2 2 2 2 — — √2 √6 ____ ____ = − +       4 4 — — √ 6 − √2 _ = 4SECTION 9.2 s u am N d d eiff re Nce i de Ntities 709 Try It 2 Find the exact value of cos(105°). Using the Sum and difference Formulas for Sine e Th sum and die ff rence formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas. sum and difference formulas for sine e Th s e formulas can be used to calculate the sines of sums and die ff rences of angles. sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β How To… Given two angles, find the sine of the die ff rence between the angles. 1. Write the difference formula for sine. 2. Substitute the given angles into the formula. 3. Simplify. Example 3 Using Sum and Dif ference Identities to Evaluate the Difference of Angles Use the sum and die ff rence identities to evaluate the die ff rence of the angles and show that part a equals part b. a. sin(45° − 30°) b. sin(135° − 120°) Solution a. Let’s begin by writing the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(45° − 30°) = sin(45°)cos(30°) − cos(45°)sin(30°) N ext, we need to find the values of the trigonometric expressions. — — — √ 2 √ 3 √2 1 ____ ____ ____ __ sin(45°) = , cos(30°) = , cos(45°) = , sin(30°) =    2 2 2 2 N ow we can substitute these values into the equation and simplify. — — — √2 √3 √2 1 ____ ____ ____ __ sin(45° − 30°) = −       2 2 2 2 — — √ 6 −  √2 _________ = 4 b. Again, we write the formula and substitute the given angles. sin(α − β) = sin α cos β − cos α sin β sin(135° − 120°) = sin(135°)cos(120°) − cos(135°)sin(120°) N ext, we fin d the values of the trigonometric expressions. — — — √ 2 1 √ 2 √ 3 ____ __ ____ ____ sin(135°) = , cos(120°) = −   , cos(135°) = − , sin(120°) = 2 2 2 2 N ow we can substitute these values into the equation and simplify. — — — √2 1 √ 2 √3 ____ __ ____ ____ sin(135° − 120°) = −   −  −        2 2 2 2 — — −√ 2 + √6 __________ = 4 — — √ 6 − √2 _________ = 4710 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns — — — √ 2 1 √2 √3 ____ __ ____ ____ sin(135° − 120°) = −   −  −        2 2 2 2 — — − √2 + √6 __________ = 4 — — √ 6 − √2 _________ = 4 Example 4 Finding the Exact V alue of an Expression Involving an Inverse Trigonometric Function 1 3 −1 __ −1 __ Find the exact value of sin cos + sin . Then check the answer with a graphing calculator.   2 5 1 3 −1 __ −1 __ Solution e pa Th ttern displayed in this problem is sin( α + β). Let α = cos and β = sin . Then we can write 2 5 1 __ cos α =    , 0 ≤ α ≤ π 2 3 π π __ __ __ sin β =    , − ≤ β ≤ 5 2 2 We will use the Pythagorean identities to find sin α and cos β. — — 2 2 sin α = √1 − cos α cos β = √1 − sin β ______ ______ 1 9 __ __ = 1 −    = 1 − √ √ 4 25 __ ___ 3 __ 16 __ = = √ 4 √ 25 — √ 3 _ 4 __ = =   2 5 Using the sum formula for sine, 1 3 −1 __ −1 __ sin cos + sin = sin(α + β)   2 5 = sin α cos β + cos α sin β — √3 4 1 3 ____ __ __ __ = · +   · 2 5 2 5 — 4√ 3 + 3 ________ = 10 Using the Sum and difference Formulas for Tangent Finding exact values for the tangent of the sum or die ff rence of two angles is a little more complicated, but again, it is a matter of recognizing the pattern. Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine sin x ____ and simplifying. Recall, tan x = , cos x ≠ 0. cos x Let’s derive the sum formula for tangent. sin(α + β) _ tan(α + β) = cos(α + β) sin α cos β + cos α sin β __________________   = cos α cos β − sin α sin β sin α cos β + cos α sin β __________________ Divide the numerator and cos α cos β = denominator by cos α cos β cos α cos β − sin α sin β __________________ cos α cos β sin α cos β cos α sin β ________ ________ + cos α cos β cos α cos β = cos α cos β sin α sin β _________ ________ − cos α cos β cos α cos β SECTION 9.2 s u am N d d eiff re Nce i de Ntities 711 sin α sin β ____ ____ + cos α cos β = sin α sin β ________ 1 − cos α cos β tan α + tan β ____________ = 1 − tan α tan β We can derive the die ff rence formula for tangent in a similar way. sum and difference formulas for tangent The sum and die ff rence formulas for tangent are: tan α + tan β tan α − tan β ___________ ___________ tan(α + β) = tan(α − β) = 1 − tan α tan β 1 + tan α tan β How To… Given two angles, find the tangent of the sum of the angles. 1. Write the sum formula for tangent. 2. Substitute the given angles into the formula. 3. Simplify. Example 5 Finding the Exact Value of an Expression Involving Tangent π π __ __ Find the exact value of tan + .   6 4 Solution Let’s first write the sum formula for tangent and substitute the given angles into the formula. tan α + tan β __ tan(α + β) = 1 − tan α tan β π π __ __ tan + tan     π π 6 4 __ __ ___ tan + =     π π 6 4 __ __ 1 − tan tan      6 4 Next, we determine the individual tangents within the formulas: π 1 π __ _ __ tan = tan = 1 —     6 4 √ 3 So we have 1 _ + 1 — π π √ 3 __ __ __ tan + =   6 4 1 _ 1 − (1) —   √3 — 1 + √ 3 _ — √ 3 _ = — √ 3 − 1 _ — √3 — — 1 + √ 3 √3 __ = — —   √ 3 √ 3 − 1 — √3 + 1 _ = — √3 − 1 712 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns Try It 3 2π π __ __ Find the exact value of tan + .   3 4 Example 6 Finding Multiple Sums and Differences of Angles 3 π 5 3π __ __ __ __ Given sin α =    , 0 α , cos β = − , π β , find 5 2 13 2 a. sin(α + β) b. cos(α + β) c. tan(α + β) d. tan(α − β) Solution We can use the sum and die ff rence formulas to identify the sum or die ff rence of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and die ff rence formulas. 3 π __ __ a. To find sin(α + β), we begin with sin α =   and 0 α . The side opposite α has length 3, the hypotenuse 5 2 has length 5, and α is in the first quadrant. See Figure 4. Using the Pythagorean Theorem,we can find the length of side a: 2 2 2 a + 3 = 5 2 a = 16 a = 4 y (4, 3) 5 α x (4, 0) Figure 4 5 3π __ __ Since cos β = − and π β , the side adjacent to β is −5, the hypotenuse is 13, and β is in the third 13 2 quadrant. See Figure 5. Again, using the Pythagorean Theorem, we have 2 2 2 (−5) + a = 13 2 25 + a = 169 2 a = 144 a = ±12 Since β is in the third quadrant, a = –12. y (–5, 0) x β 13 (–5, –12) Figure 5 e Th next step is finding the cosine of α and the sine of β. The cosine of α is the adjacent side over the hypotenuse. 4 __ We can find it from the triangle in Figure 5: cos α =    . W e can also find the sine of β from the triangle in Figure 5, 5 12 __ as opposite side over the hypotenuse: sin β = − . N ow we are ready to evaluate sin(α + β). 13SECTION 9.2 s u am N d d eiff re Nce i de Ntities 713 sin(α + β) = sin αcos β + cos αsin β 3 5 4 12 __ __ __ __ = − + −        5 13 5 13 15 48 __ __ = − − 65 65 63 __ = − 65 b. We can find cos(α + β) in a similar manner. We substitute the values according to the formula. cos(α + β) = cos α cos β − sin α sin β 4 5 3 12 __ __ __ __ = − −  −        5 13 5 13 20 36 __ __ = − + 65 65 16 __ = 65 3 4 __ __ c. For tan(α + β), if sin α =   a nd cos α =    , t hen 5 5 3 __ 5 3 _ __ tan α = =   4 4 __ 5 12 5 __ __ If sin β = − a nd cos β = − , then −12 _ 13 13 13 12 _ __ tan β = = −5 5 _ 13 en, Th tan α + tan β __ tan(α + β) = 1 − tan α tan β 3 12 __ __ + 4 5 __   = 3 12 __ __ 1 −     4 5 63 __ 20 _ = 16 __ −   20 63 __ = − 16 d. To find tan(α − β), we have the values we need. We can substitute them in and evaluate. tan α − tan β __ tan(α − β) = 1 + tan α tan β 3 12 __ __ − 4 5 __   = 3 12 __ __ 1 +     4 5 33 __ −  20 _ = 56 __ 20 33 _ = − 56 714 CHAPTER 9 trigo Nome tric id e Ntities a Nd e tioaqu Ns Analysis A common mistake when addressing problems such as this one is that we may be tempted to think that α and β are angles in the same triangle, which of course, they are not. Also note that sin(α + β) ________ tan(α + β) = cos(α + β) Using Sum and difference Formulas for Cofunctions Now that we can find the sine, cosine, and tangent functions for the sums and die ff rences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two π π __ __ positive angles is , those two angles are complements, and the sum of the two acute angles in a right triangle is , 2 2 so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as θ, then the other acute π __ angle must be labeled − θ .   2 π __ Notice also that sin θ = cos − θ : opposite over hypotenuse. Thus, when two angles are complimentary, we can   2 say that the sine of θ equals the cofunction of the complement of θ. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions. π – θ 2 θ Figure 6 From these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions . cofunction identities The cofunction identities are summarized in Table 2. π π π __ __ __ sin θ = cos − θ cos θ = sin − θ tan θ = cot − θ       2 2 2 π π π __ __ __ sec θ = csc − θ csc θ = sec − θ cot θ = tan − θ       2 2 2 Table 2 Notice that the formulas in the table may also be justified algebraically using the sum and difference formulas. For example, using cos(α − β) = cos αcos β + sin αsin β, we can write π π π __ __ __ cos − θ = cos cos θ + sin sin θ   2 2 2 = (0)cos θ + (1)sin θ = sin θ