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Section 6.1 Verifying Trigonometric Identities Do you Enjoy Solving Puzzles? We have already established some basic relationships among the trigonometric functions. The reciprocal, quotient, and Pythagorean identities follow easily from the definitions of the trigonometric functions. The even-odd identities are established using properties of the Cartesian coordinate system. In this section, we see how new identities can be verified using identities that we already know. The process will involve some trial and error and may remind you of solving a puzzle. Objective #1: Use fundamental trigonometric identities to verify identities. Solved Problem #1 1a. Verify the identity: csc x tan x sec x. In many cases, it is helpful to rewrite all the trigonometric functions on one side in terms of sines and cosines using reciprocal and quotient identities. In general, we start by working with the more complicated side, so we 1 begin with the left side and rewrite it in terms of sines and cosines using the reciprocal identity csc x sin x sin x and the quotient identity tan x . cos x csc x tan x 1 sin x 1 sin x 1 sec x sin x cos x sin x cos x cos x 1 in the last step. Also note how we started with the left side cos x of the given identity, csc x tan x sec x, and ended with the right side. Note the use of the reciprocal identity sec x Pencil Problem #1 1a. Verify the identity: sin x sec x tan x. Copyright © 2014 Pearson Education Inc. 217 Algebra and Trigonometry 5e 1b. Verify the identity: cos x cot x sin x csc x. Once again we will rewrite all the trigonometric functions on one side in terms of sines and cosines. We start by working with the more complicated side, the left side, and rewrite it using the quotient identity cos x cot x . sin x cos x cot x sin x 1b. Verify the identity: csc θ sin θ cot θ cos θ . [Hints: The Pythagorean identity cos2 θ sin 2 θ 1 can also be used in the form 1 sin 2 θ cos2 θ and a fraction of the form can be rewritten as a2 b a a .] b 1 cos x cos x sin x 1 sin x cos2 x sin x sin x sin x 1 sin x cos2 x sin 2 x sin x sin x 2 cos x sin 2 x sin x 1 sin x csc x Note how we expressed both terms on the left side as fractions with a common denominator and then added; this technique is often helpful when one side of the given identity has two terms and the other side has only one term. Also note the use of the Pythagorean identity cos2 x sin 2 x 1. 1c. Verify the identity: sin x sin x cos 2 x sin 3 x. Sometimes factoring out a common factor is helpful. We start by working with the more complicated side, the left side, and factor out the common factor sin x. 1c. Verify the identity by using factoring first: sec x sec x sin 2 x cos x. [Hint: After factoring out the common factor on the left side and using a Pythagorean identity as we did in Solved Problem #1c, rewrite secant in terms of cosine.] Then we will use the Pythagorean identity cos2 x sin 2 x 1 in the form 1 cos2 x sin 2 x. sin x sin x cos2 x sin x (1 cos2 x ) sin x sin 2 x sin 3 x 218 Copyright © 2014 Pearson Education Inc. Section 6.1 1d. Verify the identity: 1 cos θ csc θ cot θ . sin θ There is usually more than one way to verify an identity. Here we will verify the identity in two ways. In the first method, we will start with the left side and work toward the right. In the second method, we will start with the right side and work toward the left. 1d. Verify the identity in two different ways as in 1 sin θ Solved Problem #1d: sec θ tan θ . cos θ Method 1: Start with the left side and write the single fraction as the sum of two fractions. Then use appropriate reciprocal and quotient identities. 1 cos θ 1 cos θ csc θ cot θ sin θ sin θ sin θ Method 2: Start with the right side and rewrite the given functions in terms of sines and cosines. Then add the resulting fractions. csc θ cot θ 1 cos θ 1 cos θ sin θ sin θ sin θ Although the equation in Method 2 could be obtained by reversing the equation in Method 1, the thought processes are different. Perhaps one direction is more obvious to you than the other. 1e. Verify the identity: sin x 1 cos x 2 csc x. 1 cos x sin x 1e. Verify the identity: cos x 1 sin x 2 sec x. 1 sin x cos x We will work with the left side and begin by getting a common denominator so that we can combine the two fractions into one fraction. The LCD is (1 cos x )sin x. Watch for opportunities to factor and simplify within the fraction and to apply basic identities. The Pythagorean identity cos2 x sin 2 x 1 will be used to replace cos2 x sin 2 x with 1 in a key step along the way. (continued on next page) Copyright © 2014 Pearson Education Inc. 219 Algebra and Trigonometry 5e sin x 1 cos x sin x sin x 1 cos x 1 cos x 1 cos x sin x 1 cos x sin x sin x 1 cos x sin 2 x (1 cos x ) 2 (1 cos x ) sin x (1 cos x ) sin x sin 2 x (1 cos x )2 (1 cos x )sin x sin 2 x 1 2 cos x cos 2 x (1 cos x )sin x (sin 2 x cos2 x ) 1 2 cos x (1 cos x )sin x 1 1 2 cos x (1 cos x ) sin x 2 2 cos x (1 cos x ) sin x 2(1 cos x ) (1 cos x ) sin x 2( 1 cos x ) ( 1 cos x ) sin x 2 sin x 2 csc x Answers for Pencil Problems (Textbook Exercise references in parentheses): sin x 1 sin x tan x (6.1 #1) 1 cos x cos x 1 sin θ sin θ 1 sin 2 θ 1 sin 2 θ cos2 θ cos θ cos θ csc θ sin θ cot θ cos θ . (6.1 #11) sin θ 1 sin θ sin θ sin θ sin θ sin θ sin θ 1 1 sec x sec x sin2 x sec x(1 sin 2 x) sec x cos2 x cos2 x cos x (6.1 #7) cos x 1 sin θ 1 sin θ 1 sin θ 1 sin θ (6.1 #24) sec θ tan θ ; sec θ tan θ cos θ cos θ cos θ cos θ cos θ cos θ cos x 1 sin x cos x cos x (1 sin x )(1 sin x ) cos 2 x (1 sin x ) 2 cos 2 x 1 2 sin x sin 2 x 1 sin x cos x (1 sin x ) cos x (1 sin x ) cos x (1 sin x ) cos x (1 sin x ) cos x 1a. sin x sec x 1b. 1c. 1d. 1e. 220 (cos 2 x sin 2 x ) 1 2 sin x 1 1 2 sin x 2 2 sin x 2(1 sin x ) 2 2 sec x (1 sin x ) cos x (1 sin x ) cos x (1 sin x ) cos x (1 sin x ) cos x cos x Copyright © 2014 Pearson Education Inc. 6.1 #31 Section 6.2 Sum and Difference Formulas More Identities? In this section, you will learn to use another set of identities, known as the sum and difference formulas. The first of these is established using the definitions of the trigonometric functions and the properties of the Cartesian plane. Once the first formula is established, the other three can be verified using it and other basic identities. Once verified, these formulas can be used to find function values as well as verify other identities. In the Exercise Set, you will see how these formulas can be used to simplify functions that model sound vibrations. Objective #1: Use the formula for the cosine of the difference of two angles. Solved Problem #1 3 . Use the formula for 2 the cosine of the difference of two angles to find the exact value of cos 30 cos(90 60). 1a. We know that cos 30 Pencil Problem #1 1a. Use the formula for the cosine of the difference of two angles to find the exact value of cos15 cos(45 30). The formula is cos(α β ) cos α cos β sin α sin β. In cos(90 60), α 90 and β 60. Apply the formula above, substitute exact values for the four resulting trigonometric expressions, and simplify. cos(90 60) cos 90 cos 60 sin 90 sin 60 0 1 3 3 3 1 0 2 2 2 2 Note that this result is the same as the known value of cos 30 3 , as it should be. 2 Copyright © 2014 Pearson Education Inc. 221 Algebra and Trigonometry 5e 1b. Use the formula for the cosine of the difference of two angles to find the exact value of cos 70 cos 40 sin 70 sin 40. 1b. Use the formula for the cosine of the difference of two angles to find the exact value of cos50 cos 20 sin 50 sin 20. The formula is cos(α β ) cos α cos β sin α sin β. Notice that cos 70 cos 40 sin 70 sin 40 looks like the right side of the formula with α 70 and β 40. cos 70 cos 40 sin 70 sin 40 sin(70 40) sin 30 1 2 1c. Verify the identity: cos(α β ) 1 tan α tan β. cos α cos β 1c. Verify the identity: We begin with the left side and apply the formula for the cosine of the difference of two angles in the numerator. We then rewrite the fraction as the sum of two fractions and simplify. cos(α β ) cos α cos β sin 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 sin α sin β 1 cos α cos β 1 tan α tan β 222 Copyright © 2014 Pearson Education Inc. cos(α β ) tan α cot β. cos α sin β Section 6.2 Objective #2: Use sum and difference formulas for cosines and sines. Solved Problem #2 2. We are given that sin α 4 for a quadrant II angle 5 1 for a quadrant I angle β. Use this 2 information in Solved Problems #2a−d. α and sin β 2a. Find the exact value of cos α . Pencil Problem #2 2. We are given that sin α 3 for a quadrant I angle 5 5 for a quadrant II angle β. Use this 13 information in Pencil Problems #2a−d. α and sin β 2a. Find the exact value of cos α . The value of cos α is negative, since α is a quadrant II angle. Use the Pythagorean identity sin 2 α cos2 α 1 to find cos α , choosing the negative value. 2 4 2 cos α 1 5 2 25 16 9 4 cos2 α 1 5 25 25 25 cos α 9 3 25 5 2b. Find the exact value of cos β. 2b. Find the exact value of cos β. The value of cos β is positive, since β is a quadrant I angle. Use the Pythagorean identity sin 2 β cos2 β 1 to find cos β , choosing the positive value. 2 1 2 cos β 1 2 2 4 1 3 1 cos2 β 1 2 4 4 4 cos β 3 3 4 2 Copyright © 2014 Pearson Education Inc. 223 Algebra and Trigonometry 5e 2c. Find the exact value of cos(α β ). 2c. Find the exact value of cos(α β ). We use the sum formula for cosine and substitute 4 1 the given values, sin α and sin β , as well 5 2 3 as the values we just found, cos α and 5 3 cos β . 2 cos(α β ) cos α cos β sin α sin β 3 3 4 1 5 2 5 2 3 3 4 3 3 4 10 10 10 2d. Find the exact value of sin(α β ). 2d. Find the exact value of sin(α β ). We use the sum formula for sine and substitute the 4 1 3 values sin α , sin β , cos α , and 5 2 5 3 cos β . 2 sin(α β ) sin α cos β cos α sin β 224 4 3 3 1 5 2 5 2 4 3 3 4 33 10 10 10 Copyright © 2014 Pearson Education Inc. Section 6.2 Objective #3: Use sum and difference formulas for tangents. Solved Problem #3 3. Verify the identity: tan( x π ) tan x. Pencil Problem #3 3. Verify the identity: tan(2π x ) tan x. Start with the left side and apply the sum formula for tangent. Use the fact that tan π 0. tan x tan π 1 tan x tan π tan x 0 1 tan x 0 tan x 1 tan x tan( x π ) Copyright © 2014 Pearson Education Inc. 225 Algebra and Trigonometry 5e Answers for Pencil Problems (Textbook Exercise references in parentheses): 1a. 1b. 1c. 6 2 (6.2 #1) 4 3 (6.2 #5) 2 cos(α β ) cos α cos β sin α sin β cos α cos β sin α sin β cot β tan α tan α cot β cos α sin β cos α sin β cos α sin β cos α sin β 12 63 16 2c. (6.2 #57a) 2d. (6.2 #57b) 13 65 65 tan 2π tan x 0 tan x tan x 3. tan(2π x ) tan x (6.2 #37) 1 tan 2π tan x 1 0 tan x 1 2a. 226 4 5 2b. Copyright © 2014 Pearson Education Inc. (6.2 #9) Section 6.3 Double Angle, Power-Reducing, and Half-Angle Formulas How Far Can You Throw That? When you throw an object, the distance that the object will travel before hitting the ground depends on the initial speed of the object as well as the angle at which the object leaves your hand. This distance can be modeled by a formula that involves both sines and cosines. In the Exercise Set, you will see how a trigonometric identity introduced in this section can be used to rewrite a formula involving two trigonometric functions as a formula involving only one such function. The simpler form can be used to determine an angle that maximizes throwing distance. Objective #1: Use the double-angle formulas. Solved Problem #1 4 and θ lies in quadrant II, find the exact 5 value of sin 2θ , cos 2θ , and tan 2θ . 1a. If sin θ Pencil Problem #1 15 and θ lies in quadrant II, find the 17 exact value of sin 2θ , cos 2θ , and tan 2θ . 1a. If sin θ The formulas sin 2θ 2 sin θ cos θ and cos 2θ cos2 θ sin 2 θ require that we know both 4 sin θ and cos θ . We are given that sin θ . We 5 need to find cos θ . Since θ lies in quadrant II, the value of cos θ is negative. Use the Pythagorean identity sin 2 θ cos 2 θ 1 to find cos θ , choosing the negative value. 2 4 2 cos θ 1 5 2 25 16 9 4 cos2 θ 1 5 25 25 25 cos θ 9 3 25 5 Copyright © 2014 Pearson Education Inc. 227 Algebra and Trigonometry 5e Now substitute the values into the formulas. 24 4 3 sin 2θ 2 sin θ cos θ 2 5 5 25 cos 2θ cos2 θ sin 2 θ 2 2 9 16 7 3 4 5 5 25 25 25 The formula tan 2θ 2 tan θ 1 tan 2 θ requires that we know tan θ . Since we know sin θ 4 and 5 3 cos θ , we can use a quotient identity to find 5 tan θ . 4 sin θ 4 5 4 5 tan θ 3 cos θ 5 3 3 5 Now we can use this value in the formula. tan 2θ 2 tan θ 1 tan 2 θ 4 2 3 4 1 3 2 8 8 8 9 24 3 3 16 7 3 7 7 1 9 9 Note that we can also find tan 2θ using the values of sin 2θ and cos 2θ and a quotient identity. sin 2θ tan 2θ cos 2θ 228 24 25 24 7 7 25 Copyright © 2014 Pearson Education Inc. Section 6.3 1b. Find the exact value of 2 sin15 cos15. 1b. Find the exact value of cos2 15 sin 2 15. The given expression looks like the right side of the double-angle formula cos 2θ cos2 θ sin 2 θ with θ = 15°. We use the formula to rewrite the expression and then evaluate. cos2 15 sin 2 15 cos(2 15) cos 30 3 2 1c. Verify the identity: sin 3θ 3sin θ 4 sin 3 θ . Notice that the sine on the left side has 3θ as its argument while the ones on the right have just θ, so we need to use one or more identities that allow us to change the argument. The right side may look more complicated, but we will start with the left side because it contains the more complicated argument and we see more opportunities to apply identities. 1c. Verify the identity: sin 4t 4 sin t cos3 t 4 sin 3 t cos t. We first rewrite sin 3θ as sin(2θ θ ) and use the sum formula for sine. Then we use double-angle formulas to simplify the occurrences of sin 2θ and cos 2θ that result from the first step. Since the right side contains only sine, we choose to use the form of the double-angle formula for cosine that expresses cos 2θ in terms of sine only. Still keeping in mind that the right side contains only sine, we replace an occurrence of cos2 θ with 1 sin 2 θ , using a form of the Pythagorean identity sin 2 θ cos 2 θ 1. sin 3θ sin(2θ θ ) sin 2θ cos θ cos 2θ sin θ (2 sin θ cos θ ) cos θ (1 2 sin 2 θ ) sin θ 2 sin θ cos2 θ sin θ 2 sin 3 θ 2 sin θ (1 sin 2 θ ) sin θ 2 sin 3 θ 2 sin θ 2 sin 3 θ sin θ 2 sin 3 θ 3sin θ 4 sin 3 θ Copyright © 2014 Pearson Education Inc. 229 Algebra and Trigonometry 5e Objective #2: Use the power-reducing formulas. Solved Problem #2 2. Write an equivalent expression for sin 4 x that does not contain powers of trigonometric functions greater than 1. Pencil Problem #2 2. Write an equivalent expression for sin 2 x cos2 x that does not contain powers of trigonometric functions greater than 1. We begin by writing sin 4 x as (sin 2 x ) 2 and applying the power-reducing formula for sine. After some simplification, we will use the power-reducing formula for cosine. sin 4 x (sin 2 x )2 1 cos 2 x 2 2 1 2 cos 2 x cos2 2 x 4 1 1 1 cos 2 x cos 2 2 x 4 2 4 1 1 1 1 cos 2(2 x ) cos 2 x 4 2 4 2 1 1 1 1 cos 2 x cos 4 x 4 2 8 8 3 1 1 cos 2 x cos 4 x 8 2 8 230 Copyright © 2014 Pearson Education Inc. Section 6.3 Objective #3: Use the half-angle formulas. Solved Problem #3 3. 3 and a half-angle formula to 2 find the exact value of cos105. Use cos 210 Pencil Problem #3 3. 2 and a half-angle formula to 2 find the exact value of cos157.5. Use cos 315 Since an angle that measures 105° lies in quadrant II and cosine is negative in quadrant II, we choose the negative sign in the half-angle formula for cosine. 210 2 1 cos 210 2 cos105 cos 3 1 2 2 2 3 4 2 3 2 3 1 2 we Note that to simplify the fraction 2 multiplied the numerator and denominator both by 3 3 1 2 1 2 2 2 2 2 3 2: . 22 4 4 Copyright © 2014 Pearson Education Inc. 231 Algebra and Trigonometry 5e Answers for Pencil Problems (Textbook Exercise references in parentheses): 1a. sin 2θ 240 161 240 ; cos 2θ ; tan 2θ 289 289 161 (6.3 #7) 1 (6.3 #15) 2 1c. sin 4t sin(2 2t ) 2sin 2t cos 2t 2(2 sin t cos t )(cos2 t sin 2 t ) 4 sin t cos3 t 4 sin 3 t cos t 1b. 2. 1 1 cos 4 x (6.3 #37) 8 8 3. 232 2 2 2 (6.3 #41) Copyright © 2014 Pearson Education Inc. (6.3 #33) Section 6.4 Product-to-Sum and Sum-to-Product Formulas Music to Your Ears Each time you push a button on a touch-tone phone a sound is produced. This sound can be modeled by the sum of two sine functions. In the Exercise Set, you will learn how to play Mary Had a Little Lamb on your phone and you will write the sum of sines producing a note as a product. Objective #1: Use the product-to-sum formulas. Solved Problem #1 1a. Express sin 5 x sin 2 x as a sum or difference. Pencil Problem #1 1a. Express sin 6 x sin 2 x as a sum or difference. We use the formula for the product of two sines: 1 sin α sin β [cos(α β ) cos(α β )]. 2 1 [cos(5 x 2 x ) cos(5 x 2 x )] 2 1 [cos 3 x cos 7 x ] 2 sin 5 x sin 2 x 1b. Express cos 7 x cos x as a sum or difference. 1b. Express sin x cos 2 x as a sum or difference. We use the formula for the product of two cosines: 1 cos α cos β [cos(α β ) cos(α β )]. 2 1 [cos(7 x x ) cos(7 x x )] 2 1 [cos 6 x cos8 x ] 2 cos 7 x cos x Copyright © 2014 Pearson Education Inc. 233 Algebra and Trigonometry 5e Objective #2: Use the sum-to-product formulas. Solved Problem #2 Pencil Problem #2 2a. Express sin 7 x sin 3x as a product. 2a. Express sin 6 x sin 2 x as a product. We use the formula for the sum of two sines: α β α β sin α sin β 2 sin cos . 2 2 7 x 3x 7 x 3x cos 2 2 10 x 4x 2 sin cos 2 2 2 sin 5 x cos 2 x sin 7 x sin 3x 2 sin 2b. Express cos 3x cos 2 x as a product. 2b. Express cos 4 x cos 2 x as a product. We use the formula for the sum of two cosines: α β αβ cos α cos β 2 cos cos . 2 2 3x 2 x 3x 2 x cos 2 2 x 5x 2 cos cos 2 2 cos 3 x cos 2 x 2 cos Answers for Pencil Problems (Textbook Exercise references in parentheses): 1a. 1 [cos 4 x cos8 x ] (6.4 #1) 2 2a. 2 sin 4 x cos 2 x 234 (6.4 #9) 1b. 1 [sin 3x sin x ] (6.4 #5) 2 2b. 2 cos 3x cos x (6.4 #13) Copyright © 2014 Pearson Education Inc. Section 6.5 Trigonometric Equations How Many Solutions are There? We have seen that trigonometric functions can be used to model cyclic phenomena such as tides, temperatures, and hours of daylight. For example, we have worked with models where we could estimate the number of hours of daylight at a location on a particular day of the year by evaluating the function for a value of the independent variable representing the day. In this section’s Exercise Set, you will solve trigonometric equations to find the dates on which a certain city has 10.5 hours of daylight. In general, the equation has infinitely many solutions, but when we restrict ourselves to one cycle, one year, there are exactly two solutions. In this section, pay attention to whether you are looking for all solutions or just the solutions that are in a given interval. Objective #1: Find all solutions of a trigonometric equation. Solved Problem #1 1. Find all solutions of 5sin x 3sin x 3. Pencil Problem #1 1. Find all solutions of 4 sin θ 1 2 sin θ . First isolate sin x on the left side by first subtracting 3sin x from both sides and then dividing both sides by the resulting coefficient of sin x. 5sin x 3sin x 3 5sin x 3sin x 3sin x 3sin x 3 2 sin x 3 3 2 sin x 2 2 sin x 3 2 The sine function is positive in quadrants I and II π 3 at in quadrant I. In quadrant 3 2 π 2π 3 . In one period of II, sine equals at π 3 3 2 the sine function, there are two solutions of the π 2π and x given equation, x . 3 3 and sine equals (continued on next page) Copyright © 2014 Pearson Education Inc. 235 Algebra and Trigonometry 5e We are asked to find all solutions. Because of the periodic property of the sine function, if we add or subtract any integer multiple of 2π to either of the solutions on the previous page, we will get another solution. Thus, the solutions are x π 3 2π 2nπ , 3 2nπ and x where n is any integer. Objective #2: Solve equations with multiple angles. Solved Problem #2 2a. Solve the equation: tan 2 x 3, 0 x 2π . Pencil Problem #2 2a. Solve the equation: tan 3 x The equation already has the tangent expression isolated on the left. Notice that the argument, 2x, is a double-angle. We solve for 2x first and then for x. We also note that we are only looking for solutions x that satisfy 0 x 2π . However, we first identify all solutions and then select those in the given interval. In Solved Problem #1, we worked with sine which has period 2π. In this problem, we are working with tangent which has period π. In one period, tangent equals 2x π 3 3 only once, at π 3 . Thus, nπ , where n is any integer. Dividing by 2, we find all solutions of the given equation: nπ , where n is any integer. 6 2 (continued on next page) 236 Copyright © 2014 Pearson Education Inc. x π 3 , 0 x 2π . 3 Section 6.5 We want to find the solutions x that satisfy 0 x 2π . For any negative integer n, x is negative and does not satisfy 0 x 2π . For any integer n ≥ 4, x ≥ 2π and does not satisfy 0 x 2π . Here are the values of x for n = 0, 1, 2, and 3: π 0π π x 6 2 6 π 1 π π 3π 4π 2π x 6 2 6 6 6 3 π 2 π π 6π 7π x 6 2 6 6 6 π 3 π π 9π 10π 5π x 6 2 6 6 6 3 The solutions that satisfy 0 x 2π are π 2π 7π 5π , , , and . 6 3 6 3 2b. Solve the equation: sin x 1 , 0 x 2π . 3 2 2b. Solve the equation: sin 2θ 1, 0 θ 2π . 3 The expression involving sine is already isolated on 1 π at the left. In one period, the value of sine is 2 6 π 5π in quadrant I and at π in quadrant II. 6 6 Thus, x π x 5π 2nπ or 2nπ , 3 6 3 6 where n is any integer. Multiplying by 3 in each, we obtain π 5π x 6nπ or x 6nπ , 2 2 where n is any integer. Letting n = 0 in x π 2 6nπ , we obtain x results in a value of x π 2 π 2 . No other integer n 6nπ that satisfies 5π 6nπ , there is no integer 2 n that results in a value of x that satisfies 0 x 2π . 0 x 2π . For x The only solution is π 2 . Copyright © 2014 Pearson Education Inc. 237 Algebra and Trigonometry 5e Objective #3: Solve trigonometric equations quadratic in form. Solved Problem #3 3a. Solve the equation: 2sin 2 x 3sin x 1 0, 0 x 2π . Pencil Problem #3 3a. Solve the equation: 2sin 2 x sin x 1 0, 0 x 2π . Notice that if we replace sin x with u on the left side of the equation we obtain 2u 2 3u 1, which factors as (2u 1)(u 1). Thus, the left side of the equation factors as (2 sin x 1)(sin x 1). Thus, we can solve this equation by factoring and setting each factor equal to 0. 2 sin 2 x 3sin x 1 0 (2 sin x 1)(sin x 1) 0 2 sin x 1 0 or sin x 1 0 sin x 1 2sin x 1 sin x 1 2 The equation sin x 0 ≤ x < 2π: π 6 and 1 has two solutions satisfying 2 5π . The equation sin x 1 has 6 one solution satisfying 0 ≤ x < 2π: are π 6 , π 2 , and π 2 . The solutions 5π . 6 3b. Solve the equation: 4 cos2 x 3 0, 0 x 2π . 3b. Solve the equation: 4 cos2 x 1 0, 0 x 2π . We solve the equation by isolating the squared expression and then using the square root property. 4 cos2 x 3 0 4 cos 2 x 3 3 cos 2 x 4 cos x 238 3 3 3 3 or cos x 4 2 4 2 Copyright © 2014 Pearson Education Inc. Section 6.5 3 has two solutions 2 π 11π . The equation and satisfying 0 ≤ x < 2π: 6 6 3 has two solutions satisfying cos x 2 5π 7π 0 ≤ x < 2π: . The solutions are and 6 6 π 5π 7π 11π , , and . , 6 6 6 6 The equation cos x Objective #4: Use factoring to separate different functions in trigonometric equations. Solved Problem #4 4. Solve the equation: sin x tan x sin x, 0 x 2π . Pencil Problem #4 4. Solve the equation: sin x 2sin x cos x 0, 0 x 2π . We begin by subtracting sin x from both sides, obtaining 0 on the right. Then we factor out the common factor of sin x from the terms on the left and set each factor equal to 0. sin x tan x sin x sin x tan x sin x 0 sin x (tan x 1) 0 sin x 0 or tan x 1 0 tan x 1 The equation sin x 0 has solutions 0 and π satisfying 0 ≤ x < 2π. The equation tan x 1 has π 5π and satisfying 0 ≤ x < 2π. The solutions 4 4 π 5π solutions are 0, , π, and . 4 4 Copyright © 2014 Pearson Education Inc. 239 Algebra and Trigonometry 5e Objective #5: Use identities to solve trigonometric equations. Solved Problem #5 5a. Solve the equation: cos 2 x sin x 0, 0 x 2π . Pencil Problem #5 5a. Solve the equation: sin 2 x cos x, 0 x 2π . We use the double-angle formula cos 2 x 1 2 sin 2 x to rewrite the left side in terms of sine only so the equation becomes quadratic in form. cos 2 x sin x 0 1 2 sin 2 x sin x 0 2 sin 2 x sin x 1 0 2 sin 2 x sin x 1 0 (2 sin x 1)(sin x 1) 0 2 sin x 1 0 or sin x 1 0 sin x 1 2 sin x 1 sin x 1 2 The equation sin x 1 7π has solutions and 2 6 11π satisfying 0 ≤ x < 2π. The equation sin x 1 6 has only the solution solutions are 240 π 2 , π 2 satisfying 0 ≤ x < 2π. The 7π 11π , and . 6 6 Copyright © 2014 Pearson Education Inc. Section 6.5 5b. Solve the equation: cos x sin x 1, 0 x 2π . 5b. Solve the equation: sin x cos x 1, 0 x 2π . We begin by squaring each side of the equation and then applying an identity. Remember that squaring may introduce extraneous solutions, so it is important to check all proposed solutions. cos x sin x 1 (cos x sin x ) 2 ( 1) 2 cos 2 x 2 cos x sin x sin 2 x 1 (cos2 x sin 2 x ) 2 cos x sin x 1 Next we apply the identity sin 2 x cos 2 x 1. 1 2 cos x sin x 1 2 cos x sin x 0 cos x 0 or sin x 0 3π 2 2 satisfying 0 ≤ x < 2π. The second equation has solutions 0 and π satisfying 0 ≤ x < 2π. The first equation has solutions Check π 2 : cos π 2 sin π and π 1? 2 0 1 1? 1 1, true Check 3π 3π 3π : cos sin 1? 2 2 2 0 ( 1) 1? 1 1, false Check 0: cos 0 sin 0 1? 1 0 1? 1 1, false Check π: cos π sin π 1? 1 0 1? 1 1, true The solutions are π 2 and π. Copyright © 2014 Pearson Education Inc. 241 Algebra and Trigonometry 5e Objective #6: Use a calculator to solve trigonometric equations. Solved Problem #6 6a. Solve tan x 3.1044, correct to four decimal places, for 0 ≤ x < 2π. Pencil Problem #6 6a. Solve sin x 0.8246, correct to four decimal places, for 0 ≤ x < 2π. Note that tangent is positive in quadrants I and III and that tan 1 3.1044 will be a value in quadrant I. The solution in quadrant I is x tan 1 3.1044 1.2592. The solution in quadrant III is x π 1.2592 4.4008. The solutions are 1.2592 and 4.4008. 6b. Solve sin x 0.2315, correct to four decimal places, for 0 ≤ x < 2π. 6b. Solve tan x 3, correct to four decimal places, for 0 ≤ x < 2π. Sine is negative in quadrants III and IV and sin 1 0.2315 0.2336 is in quadrant I. The solution in quadrant III is x π 0.2336 3.3752. The solution in quadrant IV is x 2π 0.2336 6.0496. The solutions are 3.3752 and 6.0496. Answers for Pencil Problems (Textbook Exercise references in parentheses): 6a. 5π 2nπ , where n is any integer (6.5 #21) 6 6 π 7π 13π 19π 25π 31π , , , , , (6.5 #29) 2b. no solution (6.5 #33) 18 18 18 18 18 18 π 7π 11π π 2π 4π 5π , , , , , (6.5 #39) 3b. (6.5 #47) 2 6 6 3 3 3 3 2π 4π 0, , π, (6.5 #59) 3 3 π π 5π 3π π , , , (6.5 #69) 5b. 0, (6.5 #77) 6 2 6 2 2 0.9695, 2.1721 (6.5 #85) 6b. 1.8926, 5.0342 (6.5 #89) 242 Copyright © 2014 Pearson Education Inc. 1. θ 2a. 3a. 4. 5a. π 2nπ , θ