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Solution Guide for Chapter 8: Relationships Among Trigonometric Functions 8.1 TRIGONOMETRIC IDENTITIES E-1. Half-angle formula for the cosine: By the Pythagorean theorem applied to the triangle in Figure 8.14, r 2 x + 1 − cos α 2 !2 = 1. Hence 1 + cos α 1 − cos α = . x2 = 1 − 2 2 r 1 + cos α α Now we take the square root: x = . Because x = cos , this gives the formula 2 2 r α 1 + cos α cos = . 2 2 E-2. Half-angle formula for the sine: We modify the triangle in Figure r8.14, giving the label α α α 1 + cos α x = sin to the side opposite the angle and the label cos = to the side 2 2 2 2 adjacent to that angle. By the Pythagorean theorem applied to the triangle, r 2 x + 1 + cos α 2 !2 = 1. Hence 1 + cos α 1 − cos α x2 = 1 − = . 2 2 r 1 − cos α α Now we take the square root: x = . Because x = sin , this gives the formula 2 2 r α 1 − cos α sin = . 2 2 E-3. Halving, then doubling angles: As suggested, first we use the double-angle formula for the sine, then we use the half-angle formulas: ! r ! r r α α α 1 − cos α 1 + cos α 1 − cos2 α sin 2 = 2 sin cos =2 =2 . 2 2 2 2 2 4 SECTION 8.1 Trigonometric Identities 571 But 1 − cos2 α = sin2 α, so r 2 1 − cos2 α =2 4 s sin2 α = sin α. 4 E-4. Doubling, then halving angles: As suggested, first we use the half-angle formula for the sine: sin 2α 2 r = 1 − cos(2α) . 2 Now we use the double-angle formula for the cosine and the most basic trig identity: 1 − cos(2α) = 1 − (cos2 α − sin2 α) = 1 − cos2 α + sin2 α = sin2 α + sin2 α = 2 sin2 α. Thus sin 2α 2 r = 1 − cos(2α) = 2 s 2 sin2 α = sin α. 2 S-1. The sine of 75 degrees: Now √ 1 1 1 3 + √ × = 0.97. sin 75 = sin(45 + 30 ) = sin 45 cos 30 + cos 45 sin 30 = √ × 2 2 2 2 ◦ ◦ ◦ ◦ ◦ ◦ ◦ S-2. The cosine of 75 degrees: Now √ 1 3 1 1 cos 75 = cos(45 + 30 ) = cos 45 cos 30 − sin 45 sin 30 = √ × − √ × = 0.26. 2 2 2 2 ◦ ◦ ◦ ◦ ◦ ◦ ◦ S-3. The sine of 15 degrees: Now √ 1 3 1 1 sin 15◦ = sin(45◦ − 30◦ ) = sin 45◦ cos 30◦ − cos 45◦ sin 30◦ = √ × − √ × = 0.26. 2 2 2 2 S-4. The cosine of 15 degrees: Now √ 1 3 1 1 + √ × = 0.97. cos 15◦ = cos(45◦ − 30◦ ) = cos 45◦ cos 30◦ + sin 45◦ sin 30◦ = √ × 2 2 2 2 S-5. Half angle: Now ◦ sin 22.5 = sin 1 ◦ 45 2 r = ◦ 1 − cos 45 = 2 s 1− 2 √1 2 = 0.38. 572 Solution Guide for Chapter 8 S-6. Half angle: Now ◦ cos 22.5 = cos 1 ◦ 45 2 r = ◦ s 1 + cos 45 = 2 1+ √1 2 = 0.92. 2 S-7. Half angle twice: Now ◦ sin 11.25 r 1 1 − cos 22.5◦ ◦ = sin 22.5 = = 2 2 v r u √ u u 1 − 1 + 1/ 2 t 2 = = 0.20. 2 v r u ◦ u u 1 − 1 + cos 45 t 2 2 S-8. Double angle: Now √ √ 3 1 3 sin 120 = sin (2 × 60 ) = 2 sin 60 cos 60 = 2 × × = = 0.87. 2 2 2 ◦ ◦ ◦ ◦ S-9. Double-angle formula: Now 2 1 2 ◦ ◦ 2 ◦ ◦ cos 120 = cos(2 × 60 ) = cos 60 − sin 60 = − 2 √ !2 3 1 =− . 2 2 S-10. Sum formula: Now √ 1 1 1 3 sin 105 = sin(45 + 60 ) = sin 45 cos 60 + cos 45 sin 60 = √ × + √ × = 0.97. 2 2 2 2 ◦ ◦ ◦ ◦ ◦ ◦ ◦ S-11. Finding angles: (a) Because t is between 0 and 90 degrees, cos t is positive. Hence cos t = p 1 − sin2 t = p 1 − 0.72 = 0.71. (b) Using the double-angle formula for cosine, along with the most basic trig identity, gives cos(2t) = cos2 t − sin2 t = (1 − sin2 t) − sin2 t = (1 − 0.72 ) − 0.72 = 0.02. Using the rounded value of cos t from Part (a) gives the value 0.01 here. SECTION 8.1 Trigonometric Identities 573 t t is between 0 and 45 degrees, so sin is 2 2 positive. Using the half-angle formula for sine along with Part (a) gives r t 1 − cos t sin = = 0.38. 2 2 (c) Because t is between 0 and 90 degrees, (d) Using the double-angle formula for sine twice, along with Part (a) and Part (b), gives sin(4t) = 2 sin(2t) cos(2t) = 2(2 sin t cos t) cos(2t) = 0.04. The answer will vary depending on rounding in preceding parts. S-12. Finding trigonometric functions of angles: (a) Because t is between 90 and 180 degrees, cos t is negative. Hence p p cos t = − 1 − sin2 t = − 1 − 0.62 = −0.80. (b) Using the double-angle formula for cosine, along with the most basic trig identity, gives cos(2t) = cos2 t − sin2 t = (1 − sin2 t) − sin2 t = (1 − 0.62 ) − 0.62 = 0.28. t t is between 45 and 90 degrees, so sin 2 2 is positive. Using the half-angle formula for sine along with Part (a) gives r t 1 − cos t sin = = 0.95. 2 2 (c) Because t is between 90 and 180 degrees, (d) Using the double-angle formula for sine twice, along with Part (a) and Part (b), gives sin(4t) = 2 sin(2t) cos(2t) = 2(2 sin t cos t) cos(2t) = −0.54. S-13. Calculating sines of angles: 1 (a) Now sin(−30◦ ) = − sin 30◦ = − . 2 1 ◦ ◦ (b) Now cos(−60 ) = cos 60 = . 2 (c) By the sum formula for the tangent in Example 8.2, tan 105◦ = tan(60◦ + 45◦ ) = But tan 60◦ + tan 45◦ . 1 − tan 60◦ tan 45◦ √ sin 60◦ tan 60 = = cos 60◦ ◦ 3 2 1 2 = √ 3 574 Solution Guide for Chapter 8 and sin 45◦ tan 45 = = cos 45◦ ◦ Hence √1 2 √1 2 = 1. √ tan 60◦ + tan 45◦ 3+1 √ = −3.73. tan 105 = = ◦ ◦ 1 − tan 60 tan 45 1− 3 ◦ S-14. Area: The area is 1 1 1 × 5 × 6 × sin 30◦ = × 5 × 6 × = 7.5. 2 2 2 S-15. Area: The area is 1 × 5 × 8 × sin 25◦ = 8.45. 2 S-16. Going backward: We want to find t so that sin t = 0.78. We use the crossing-graphs method with a horizontal span from 0 to 90 using degrees and a vertical span from 0 to 1. We see from the graph below that t = 51.26 degrees. S-17. Going backward: We want to find t so that cos t = 0.45. We use the crossing-graphs method with a horizontal span from 0 to 90 using degrees and a vertical span from 0 to 1. We see from the graph below that t = 63.26 degrees. SECTION 8.1 Trigonometric Identities 575 S-18. Going backward: We want to find t so that cos t = −0.34. We use the crossing-graphs method with a horizontal span from 0 to 180 using degrees and a vertical span from −1 to 1. We see from the graph below that t = 109.88 degrees. 1. Unit circle: According to the unit-circle definition of sine and cosine, in magnitude sin t and cos t are the legs of a right triangle whose hypotenuse is 1 (the radius of the unit circle). By the Pythagorean theorem, sin2 t + cos2 t = 1. 2. Sum formula for the cosine: (a) Because sine is odd, sin(x − 90◦ ) = − sin(90◦ − x) = − cos x. (b) Because cosine is even, cos(x − 90◦ ) = cos(90◦ − x) = sin x. (c) Now cos(α + β) = − sin(α + β − 90◦ ) (by Part (a)) = − sin((α − 90◦ ) + β) = − sin(α − 90◦ ) cos β − cos(α − 90◦ ) sin β (by the sum formula) = cos α cos β − sin α sin β (by Part (a) and Part (b)). 3. Double angle for the cosine: Using the sum formula for the cosine gives cos(2t) = cos(t + t) = cos t cos t − sin t sin t = cos2 t − sin2 t. 4. Difference formula for the sine: Using the sum formula for the sine gives sin(α − β) = sin(α + (−β)) = sin α cos(−β) + cos α sin(−β) = sin α cos β − cos α sin β because sine is odd and cosine is even. 576 Solution Guide for Chapter 8 5. Difference formula for the cosine: Using the sum formula for the cosine gives cos(α − β) = cos(α + (−β)) = cos α cos(−β) − sin α sin(−β) = cos α cos β + sin α sin β because sine is odd and cosine is even. 6. An identity: (a) We use the double-angle formulas: cot(2x) = = cos2 x − sin2 x cos2 x sin2 x cos x sin x cos(2x) = = − = − sin(2x) 2 sin x cos x 2 sin x cos x 2 sin x cos x 2 sin x 2 cos x 1 (cot x − tan x). 2 (b) We start with the right-hand side: 1 cot2 x − 1 = 2 cot x 2 cot2 x 1 − cot x cot x = 1 (cot x − tan x) = cot(2x) 2 by Part (a). 7. Changing how things look: Using the double-angle formula for cosine, along with the most basic trig identity, gives cos(2x) = cos2 x − sin2 x = cos2 x − (1 − cos2 x) = 2 cos2 x − 1. Hence 2 cos2 x = 1 + cos(2x), so cos2 x = 1 (1 + cos(2x)). 2 8. Changing how things look: As suggested, we start with the right-hand side of the equation, and we apply the sum formula and the difference formula for the sine: 1 (sin(x + y) + sin(x − y)) 2 = = 1 1 (sin x cos y + cos x sin y + sin x cos y − cos x sin y) = (2 sin x cos y) 2 2 sin x cos y. 9. An important identity: (a) By the sum formula for the cosine, cos 3t = cos(2t + t) = cos 2t cos t − sin 2t sin t. SECTION 8.1 Trigonometric Identities 577 (b) Using the double-angle formulas in Part (a) gives cos 3t = (cos2 t − sin2 t)(cos t) − (2 sin t cos t)(sin t) = cos3 t − 3 cos t sin2 t. If we replace sin2 t by 1 − cos2 t, this becomes cos3 t − 3 cos t(1 − cos2 t) = 4 cos3 t − 3 cos t. (c) By Part (b) we have 4 cos3 20◦ − 3 cos 20◦ = cos(3(20◦ )) = cos(60◦ ) = 1 , 2 so 1 = 0. 2 1 Hence cos 20◦ is a zero of the polynomial 4x3 − 3x − . 2 4 cos3 20◦ − 3 cos 20◦ − 10. Double-angle formula for the tangent: By the sum formula for the tangent we have tan x + tan x 2 tan x = . 1 − tan x tan x 1 − tan2 x tan(2x) = tan(x + x) = 11. Sum formula for the cotangent: As in Example 8.2 we have sin(α + β) cos(α + β) sin α cos β + cos α sin β . cos α cos β − sin α sin β tan(α + β) = = Now we divide each term of the top and bottom by sin α sin β to get tan(α + β) = sin α cos β sin α sin β cos α cos β sin α sin β + − cos α sin β sin α sin β sin α sin β sin α sin β . Cancelling gives tan(α + β) = cos β sin β + cos α sinα cos β cos α sin α = sin β −1 cot α + cot β . cot α cot β − 1 But then cot(α + β) = 1 cot α cot β − 1 = . tan(α + β) cot α + cot β 12. Double-angle formula for the cotangent again: By Exercise 11 we have cot(2x) = cot(x + x) = cot x cot x − 1 cot2 x − 1 = . cot x + cot x 2 cot x 578 Solution Guide for Chapter 8 13. Area of an isosceles triangle: (a) Now cotangent is adjacent divided by opposite, so cot H= a α cot . 2 2 α H = a . Solving for H gives 2 2 (b) Now Area = 1 1 × Length of base × Length of altitude = × a × H. 2 2 Using the formula for H from Part (a) gives Area = 1 a α 1 α × a × cot = a2 cot . 2 2 2 4 2 14. Area of a regular polygon: As suggested, we divide the polygon into n isosceles trian360 degrees opposite a base of length a. By Exercise 13 the area gles, each with angle of n of each triangle is 1 2 360◦ 1 180◦ a cot = a2 cot . 4 2n 4 n 1 180◦ There are n such triangles, so the area of the polygon is na2 cot . 4 n 8.2 LAWS OF SINES AND COSINES E-1. Thales revisited: For an angle inscribed in a semicircle the degree measure of the central angle is 180◦ . Thus the measure of the inscribed angle is half of 180◦ , so it is 90◦ . Thus the inscribed angle is a right angle. E-2. Converse: Given a right triangle, form the circumscribed circle. The hypotenuse is a chord of that circle, and the angle subtended by the chord is the right angle of the triangle, so the subtended angle has measure 90◦ . Because the degree measure of the subtended angle is half the degree measure of the central angle, the central angle is 2 × 90◦ = 180◦ . Hence the chord is a diameter of the circle. E-3. Special value: (a) Because the degree measure of any subtended angle is half the degree measure of the central angle, the degree measure of β is half the degree measure of α. Now the measure of β is 30◦ , so the measure of α is 60◦ . SECTION 8.2 Laws of Sines and Cosines 579 (b) We consider the angles of the triangle ∆AOB, namely OAB, OBA, and α. Now OB and OA are radii, so they have the same length. Thus the angles opposite these sides, namely OAB and OBA, are equal. By Part (a) the measure of α is 60◦ . Now the angle sum of ∆AOB is 180 degrees, and because OAB and OBA are equal the 1 measure of each must be (180 − 60) = 60 degrees. Hence all angles of ∆AOB 2 must be 60◦ , so the triangle is equilateral. (c) By Part (b), the side AB has the same length as the side OA, and the latter is a radius. Hence the length of AB equals the radius. (d) Now the length of AB divided by sin β equals the diameter, and it follows that sin β is the length of AB divided by the diameter. But by Part (c) the length of AB equals the radius, so it equals half the diameter. Thus sin β is half the diameter divided by 1 the diameter. Because the measure of β is 30◦ , this says that sin 30◦ = . 2 S-1. First we find a: Because the angle sum of the triangle is 180 degrees, a = 180 − 40 − 80 = 60 degrees. Now we use the law of sines to find B: B 12 = , sin 40◦ sin 60◦ so B = sin 40◦ × 12 = 8.91. sin 60◦ Similarly, C 12 = , sin 80◦ sin 60◦ so C = sin 80◦ × 12 = 13.65. sin 60◦ π π S-2. First we find a: Because the angle sum of the triangle is π radians, a = π − − = 6 4 7π = 1.83 radians. Now we use the law of sines to find B: 12 12 B = , sin π6 sin 7π 12 so B = sin π 12 = 6.21. × 6 sin 7π 12 Similarly, 12 C , = sin π4 sin 7π 12 so C = sin π 12 = 8.78. × 4 sin 7π 12 580 Solution Guide for Chapter 8 S-3. First we find b: Because the angle sum of the triangle is 180 degrees, b = 180−55−65 = 60 degrees. Now we use the law of sines to find A: A 10 = , sin 55◦ sin 60◦ so A = sin 55◦ × 10 = 9.46. sin 60◦ Similarly, C 10 = , sin 65◦ sin 60◦ so C = sin 65◦ × 10 = 10.47. sin 60◦ S-4. First we find b: Because the angle sum of the triangle is π radians, b = π −1.5−0.2 = 1.44 radians. Now we use the law of sines to find A: 7 A = , sin 1.5 sin 1.44 so A = sin 1.5 × 7 = 7.04. sin 1.44 Similarly, C 7 = , sin 0.2 sin 1.44 so C = sin 0.2 × 7 = 1.40. sin 1.44 S-5. First we use the law of cosines to find C: C 2 = 82 + 52 − 2 × 8 × 5 cos 30◦ , so C= p 82 + 52 − 2 × 8 × 5 cos 30◦ = 4.44. Now we use the law of sines to find b (the angle opposite B, the smaller of the two remaining sides): 5 4.44 = , sin b sin 30◦ so sin b = 5 × sin 30◦ . 4.44 SECTION 8.2 Laws of Sines and Cosines 581 Using the crossing-graphs method gives b = 34.27 degrees. (There are two solutions between 0 and 180 degrees, but, because B isn’t the longest side of the triangle, the angle b opposite it must be less than 90 degrees.) Because the angle sum of the triangle is 180 degrees, a = 180 − 34.27 − 30 = 115.73 degrees. S-6. First we use the law of cosines to find C: C 2 = 82 + 52 − 2 × 8 × 5 cos π , 7 so r π C = 82 + 52 − 2 × 8 × 5 cos = 4.11. 7 Now we use the law of sines to find b (the angle opposite B, the smaller of the two remaining sides): 5 4.11 = , sin b sin π7 so sin b = 5 × sin π7 . 4.11 Using the crossing-graphs method gives b = 0.56 radian. (There are two solutions between 0 and π radians, but, because B isn’t the longest side of the triangle, the angle π b opposite it must be less than radians.) Because the angle sum of the triangle is π 2 π radians, a = π − 0.56 − = 2.13 radians. 7 S-7. First we use the law of sines to find c: 5 10 = , sin c sin 110◦ so sin c = 5 × sin 110◦ . 10 Using the crossing-graphs method gives c = 28.02 degrees. (There are two solutions between 0 and 180 degrees, but, because b is greater than 90 degrees, c must be less than 90 degrees.) Now we find a: Because the angle sum of the triangle is 180 degrees, a = 180 − 110 − 28.02 = 41.98 degrees. Now we use the law of sines to find A: A 10 = , sin 41.98◦ sin 110◦ so A = sin 41.98◦ × 10 = 7.12. sin 110◦ (We can also find A using the law of cosines.) 582 Solution Guide for Chapter 8 S-8. First we use the law of sines to find c: 5.5 10 = , sin c sin 2.3 so sin c = 5.5 × sin 2.3 . 10 Using the crossing-graphs method gives c = 0.42 radian. (There are two solutions beπ tween 0 and π radians, but, because a is greater than radians, c must be less than 2 π radians.) Now we find b: Because the angle sum of the triangle is π radians, b = 2 π − 2.3 − 0.42 = 0.42 radian. Now we use the law of sines to find B: 10 B = , sin 0.42 sin 2.3 so B = sin 0.42 × 10 = 5.47. sin 2.3 (We can also find B using the law of cosines.) Note that, because b = c, we expect that B = C. The difference is due to rounding. S-9. First we find c: Because the angle sum of the triangle is 180 degrees, c = 180 − 30 − 120 = 30 degrees. Because a and c are equal, A and C are equal, so C = 10. (We can also find C using the law of sines.) Now we use the law of sines to find B: 10 B = , ◦ sin 120 sin 30◦ so B = sin 120◦ × √ 10 . sin 30◦ 3 Now sin 120 = sin 60 = because the reference angle for 120◦ is 180 − 120 = 60 2 1 degrees. Also, sin 30◦ = . Hence 2 √ √ 3 10 B= × 1 = 10 3. 2 2 ◦ ◦ (We can also find B using the law of cosines.) S-10. First we find c: Because the angle sum of the triangle is 180 degrees, c = 180−60−60 = 60 degrees. Because a, b, and c are all equal, this is an equilateral triangle. Hence A, B, and C are all equal, so B = 5 and C = 5. (We can also find B and C using the law of sines.) SECTION 8.2 Laws of Sines and Cosines 583 1. Finding angles: First we use the law of cosines to find c because C is the longest side: 102 = 72 + 92 − 2 × 7 × 9 cos c, so cos c = 72 + 92 − 102 . 2×7×9 Using the crossing-graphs method gives c = 76.23 degrees. (The horizontal span is 0 to 180 using degrees.) Now we use the law of sines to find a: 7 10 = , sin a sin 76.23◦ so sin a = 7 × sin 76.23◦ . 10 Using the crossing-graphs method gives a = 42.83 degrees. (There are two solutions between 0 and 180 degrees, but, because A isn’t the longest side of the triangle, the angle a opposite it must be less than 90 degrees.) Because the angle sum of the triangle is 180 degrees, b = 180 − 42.83 − 76.23 = 60.94 degrees. 2. Finding distance: First we find the angle at city 3: Because the angle sum of the triangle is 180 degrees, the angle at city 3 is 180 − 80 − 70 = 30 degrees. Now we use the law of sines to find the distance in miles from city 1 to city 3, which we denote by A: 200 A = , sin 70◦ sin 30◦ so A = sin 70◦ × 200 = 375.88. sin 30◦ Thus the distance from city 1 to city 3 is about 376 miles. Similarly, if B denotes the distance (in miles) from city 2 to city 3, then B 200 = , sin 80◦ sin 30◦ so B = sin 80◦ × 200 = 393.92. sin 30◦ Thus the distance from city 2 to city 3 is about 394 miles. 3. Mapping a mountain: We use the law of cosines: If C denotes the length (in yards) of the side of the hill, then C 2 = 3002 + 5002 − 2 × 300 × 500 cos 20◦ , so C= p 3002 + 5002 − 2 × 300 × 500 cos 20◦ = 241.02. Thus the length is about 241 yards. 584 Solution Guide for Chapter 8 4. A lost airplane: The two airports and the airplane determine a triangle. The angle at OKC is 90 − 44 = 46 degrees, and the angle at DFW is 90 − 62 = 28 degrees. Because the angle sum of the triangle is 180 degrees, the angle at the airplane is 180 − 46 − 28 = 106 degrees. Now we use the law of sines to find the distance in miles from DFW to the airplane, which we denote by A: A 200 = , ◦ sin 46 sin 106◦ so A = sin 46◦ × 200 = 149.67. sin 106◦ Thus the distance from DFW to the airplane is about 150 miles. Similarly, if B denotes the distance (in miles) from OKC to the airplane, then B 200 = , ◦ sin 28 sin 106◦ so B = sin 28◦ × 200 = 97.68. sin 106◦ Thus the distance from OKC to the airplane is about 98 miles. 5. The base of an isosceles triangle: (a) By the law of cosines we have C 2 = r2 + r2 − 2 × r × r × cos γ = 2r2 (1 − cos γ) = 4r2 (b) Taking the square root in Part (a) gives r C = 2r 1 − cos γ 2 . 1 − cos γ , 2 and by the half-angle formula for sine this says γ C = 2r sin . 2 6. An inscribed polygon: (a) As suggested, we divide the polygon into n isosceles triangles, each of which is 360◦ described by Figure 8.33 with γ = . By Part (b) of Exercise 5 we have n C = 2r sin γ 180◦ = 2r sin . 2 n There are n such triangles, so the perimeter of the polygon is nC = 2nr sin 180◦ . n SECTION 8.2 Laws of Sines and Cosines 585 (b) The limiting value, as the number of sides increases, of the perimeters of inscribed polygons should be the circumference of the circle, namely 2πr. In fact, if we put r = 1 (say) in the formula from Part (a) and examine a table of values for the resulting function of n, namely 2n sin 180◦ , n we see that the limiting value is about 6.28, an approximation of 2π. 7. A baseball field: The center fielder, home plate, and third base determine a triangle. The angle of this triangle at home plate is 45 degrees because the side of the triangle from home plate to the center fielder includes a diagonal of the square shown in the figure. If C denotes the distance (in feet) from the center fielder to third base, then by the law of cosines C 2 = 3802 + 902 − 2 × 380 × 90 cos 45◦ , so C= p 3802 + 902 − 2 × 380 × 90 cos 45◦ = 322.70. Thus the distance is about 323 feet. 8. An identity: (a) Now S −C = 1 1 1 1 1 1 (A+B +C)−C = (A+B)+ C −C = (A+B)− C = (A+B −C). 2 2 2 2 2 2 (b) Now by the definition of S and Part (a) S(S−C) = 1 1 1 1 (A+B+C)× (A+B−C) = ((A+B)+C)((A+B)−C) = ((A+B)2 −C 2 ). 2 2 4 4 Here we used the identity from algebra (x + y)(x − y) = x2 − y 2 . Hence S(S −C) = 1 1 1 ((A+B)2 −C 2 ) = ((A2 +2AB +B 2 )−C 2 ) = (A2 +B 2 +2AB −C 2 ). 4 4 4 Here we used the identity from algebra (x + y)2 = x2 + 2xy + y 2 . 586 Solution Guide for Chapter 8 (c) Now the law of cosines gives C 2 = A2 + B 2 − 2AB cos γ. Substituting this into the result from Part (b) gives S(S − C) = = = = 1 2 (A + B 2 + 2AB − C 2 ) 4 1 2 (A + B 2 + 2AB − (A2 + B 2 − 2AB cos γ)) 4 1 (2AB + 2AB cos γ) 4 1 AB(1 + cos γ). 2 (d) By Part (c) and the half-angle formula for the cosine, S(S − C) = 1 γ AB(1 + cos γ) = AB cos2 , 2 2 so γ S(S − C) = cos2 . AB 2 Taking the square root gives r S(S − C) γ = cos . AB 2 γ (Note that γ is between 0 and 180 degrees, so is between 0 and 90 degrees, and 2 γ thus cos is nonnegative.) 2 9. Another identity: (a) Now S −A = 1 1 1 1 1 1 (A+B +C)−A = A+ (B +C)−A = − A+ (B +C) = (C +B −A). 2 2 2 2 2 2 This can also be written as 1 (C + (B − A)). 2 A similar computation gives S−B = 1 1 (C − B + A) = (C − (B − A)). 2 2 Hence (S − A)(S − B) = 1 1 (C + (B − A)) (C − (B − A)). 2 2 SECTION 8.2 Laws of Sines and Cosines 587 (b) We use Part (a) along with the following identity from algebra: (x + y)(x − y) = x2 − y 2 . We get (S − A)(S − B) = 1 1 1 (C + (B − A)) (C − (B − A)) = (C 2 − (B − A)2 ). 2 2 4 Expanding using the identity (x − y)2 = x2 − 2xy + y 2 gives (S−A)(S−B) = 1 2 1 1 (C −(B−A)2 ) = (C 2 −(B 2 −2AB+A2 )) = (C 2 −B 2 −A2 +2AB). 4 4 4 (c) Now the law of cosines gives C 2 = A2 + B 2 − 2AB cos γ. Substituting this into the result from Part (b) gives (S − A)(S − B) 1 2 (C − B 2 − A2 + 2AB) 4 1 2 (A + B 2 − 2AB cos γ − B 2 − A2 + 2AB) 4 1 (2AB − 2AB cos γ) 4 1 AB(1 − cos γ). 2 = = = = (d) By Part (c) and the half-angle formula for the sine, (S − A)(S − B) = 1 γ AB(1 − cos γ) = AB sin2 , 2 2 so (S − A)(S − B) γ = sin2 . AB 2 Taking the square root gives r (S − A)(S − B) γ = sin . AB 2 γ (Note that γ is between 0 and 180 degrees, so is between 0 and 90 degrees, and 2 γ thus sin is nonnegative.) 2 588 Solution Guide for Chapter 8 10. Heron’s area formula: (a) Part (c) of Exercise 8 says S(S − C) = 1 AB(1 + cos γ), 2 and Part (c) of Exercise 9 says (S − A)(S − B) = 1 AB(1 − cos γ). 2 Thus, multiplying the earlier results gives S(S−A)(S−B)(S−C) = 1 1 1 AB(1+cos γ) AB(1−cos γ) = A2 B 2 (1+cos γ)(1−cos γ). 2 2 4 Now (1 + cos γ)(1 − cos γ) = 1 − cos2 γ = sin2 γ. Hence S(S − A)(S − B)(S − C) = 1 2 2 1 A B (1 + cos γ)(1 − cos γ) = A2 B 2 sin2 γ. 4 4 p 1 (b) The area of the triangle is AB sin γ, and by Part (a) this equals S(S − A)(S − B)(S − C). 2 11. Getting area: In the notation of Exercise 10, S = 1 (7 + 8 + 9) = 12, and by Heron’s 2 formula the area of the triangle is p √ 12(12 − 7)(12 − 8)(12 − 9) = 720 = 26.83. 12. Getting area: In the notation of Exercise 10, S = 1 (6 + 9 + 13) = 14, and by Heron’s 2 formula the area of the triangle is p √ 14(14 − 6)(14 − 9)(14 − 13) = 560 = 23.66. 13. No such triangle: If there were such a triangle then by the law of sines we would have 12 4 = , sin α sin 20◦ so sin α = 12 × sin 20◦ = 1.03. 4 But this is impossible because the sine function is never larger than 1. SECTION 8.2 589 Laws of Sines and Cosines 14. No such triangle: If there were such a triangle then by the law of sines we would have 1 11 = , sin α sin 70◦ so sin α = 11 × sin 70◦ = 10.34. But this is impossible because the sine function is never larger than 1. 15. Two such triangles: By the law of sines we have 22 25 = , sin α sin 55◦ so sin α = 25 × sin 55◦ . 22 Using the crossing-graphs method with a horizontal span from 0 to 180 degrees gives two solutions, α = 68.57 degrees and α = 111.43 degrees. When α = 68.57 degrees we get γ = 180 − 68.57 − 55 = 56.43 degrees (because the angle sum of the triangle is 180 degrees). By the law of sines, C 22 = , sin 56.43◦ sin 55◦ so C = sin 56.43◦ × 22 = 22.38. sin 55◦ When α = 111.43 degrees we get γ = 180 − 111.43 − 55 = 13.57 degrees. By the law of sines, C 22 = , ◦ sin 13.57 sin 55◦ so C = sin 13.57◦ × 22 = 6.30. sin 55◦ The two triangles are shown below. o γ=13.57 o A=25 γ=56.43 A=25 B=22 o o β=55 α=68.57 C=22.38 B=22 β=55 o o α=111.43 C=6.30 590 Solution Guide for Chapter 8 8.3 INVERSE TRIGONOMETRIC FUNCTIONS E-1. Adding up lots of terms: The sum of the first 10 terms of the series is 4 1 1 4 1 1 4 1 4 − − − + · · · − − . 5 239 3 53 2393 19 519 23919 The calculator’s displayed value for this is 3.141592654, which is the same as the calculator’s displayed value for π. Adding up the first 10 terms of the alternating sum of the reciprocals of the odd integers (times 4) gives 1 1 4 1 − + ··· − . 3 19 The calculator’s displayed value for this is 3.041839619, which differs from the correct value for π (and thus with the above value from Machin’s formula) already in the first decimal place. E-2. The error: If we add up the first 15 terms in Machin’s formula we will miss the true value of π by no more than 4 31 4 1 − 31 5 23931 . This is about 1.11 × 10−22 . E-3. Getting Machin’s formula: (a) As suggested, we apply the double-angle formula for the tangent twice: 2 tan x 2 2 1−tan x 2 tan(2x) = tan(4x) = tan(2(2x)) = 2 . 1 − tan2 (2x) 2 tan x 1 − 1−tan2 x Now to simplify this we multiply both the numerator and the denominator of the big fraction by (1 − tan2 x)2 . The result is tan(4x) = 2(2 tan x)(1 − tan2 x) 4 tan x(1 − tan2 x) = . 2 2 2 (1 − tan x) − (2 tan x) (1 − tan2 x)2 − 4 tan2 x 1 1 in Part (a), so tan x = : 5 5 1 1 2 4 1 − 5 5 1 120 tan 4 arctan = 2 = 119 . 5 1 2 1 2 1− 5 −4 5 (b) We use x = arctan SECTION 8.3 Inverse Trigonometric Functions 591 (c) As suggested, we apply the sum formula for the tangent: 1 tan 4 arctan 15 + tan − arctan 239 1 1 . tan 4 arctan − arctan = 1 5 239 1 − tan 4 arctan 15 tan − arctan 239 Now in general tan(−x) = − tan x because sine is odd and cosine is even. Thus 1 1 1 tan − arctan = − tan arctan =− . 239 239 239 Using this and the result of Part (b) gives 120 1 1 1 119 −239 tan 4 arctan − arctan = = 1. 1 5 239 1 + 120 119 239 √ √ 3 3 ◦ S-1. We want an angle between −90 and 90 degrees whose sine is . Now sin 60 = , 2 2 √ ! 3 π so arcsin equals 60 degrees or radians. 2 3 S-2. By Part 6 of Key Idea 8.5, √ ! √ ! 3 3 . = − arcsin arcsin − 2 2 √ ! 3 π In Exercise S-1 we found that arcsin equals 60 degrees or radians. Hence 2 3 ! √ 3 π arcsin − equals −60 degrees or − radians. 2 3 √ √ 3 3 S-3. We want an angle between 0 and 180 degrees whose cosine is . Now cos 30◦ = , 2 2 ! √ 3 π so arccos equals 30 degrees or radian. 2 6 S-4. By Part 6 of Key Idea 8.5, in terms of degrees √ ! √ ! 3 3 ◦ arccos − = 180 − arccos . 2 2 √ ! √ ! 3 3 ◦ = 30 . Because 180−30 = 150, arccos − In Exercise S-3 we found that arccos 2 2 5π equals 150 degrees or radians. 6 S-5. We want an angle between 0 and 180 degrees whose cosine is 0. Now cos 90◦ = 0, so π arccos 0 equals 90 degrees or radians. 2 S-6. By Part 6 of Key Idea 8.5, in terms of degrees 1 1 ◦ arccos − = 180 − arccos . 2 2 1 1 1 ◦ ◦ Now cos 60 = , so arccos = 60 . Because 180 − 60 = 120, arccos − equals 2 2 2 2π 120 degrees or radians. 3 592 Solution Guide for Chapter 8 S-7. By Part 6 of Key Idea 8.5, 1 1 arcsin − = − arcsin . 2 2 1 1 π Now sin 30◦ = , so arcsin − equals −30 degrees or − radian. 2 2 6 sin α , cos α ◦ so we look for α so that sin α = 0. Now sin 0 = 0, so arctan 0 equals 0 degrees or 0 S-8. We want an angle α between −90 and 90 degrees so that tan α = 0. Now tan α = radians. √ sin α S-9. We want an angle α between −90 and 90 degrees so that tan α = 3. Now tan α = , cos α √ √ 3 √ 3 1 and cos α = . The angle and as noted 3 = 21 , so we look for α so that sin α = 2 2 2 √ π ◦ α = 60 satisfies both of these conditions, so arctan 3 equals 60 degrees or radians. 3 S-10. By Part 3 of Key Idea 8.6, √ √ arctan − 3 = − arctan 3. √ √ π In Exercise S-9 we found that arctan 3 equals 60 degrees or radians. Hence arctan − 3 3 π equals −60 degrees or − radians. 3 1. Using inverse trig functions: Let α = arctan x, so tan α = x. We want to represent this situation by a right triangle with an acute angle of α so that tan α = x. Because tan α = Opposite , Adjacent we can accomplish this by assigning the length x to the side opposite α and the length 1 √ to the side adjacent to α. Then by the Pythagorean theorem the hypotenuse is 1 + x2 , so cos α = Thus cos(arctan x) = √ Adjacent 1 =√ . Hypotenuse 1 + x2 1 . 1 + x2 2. Using inverse trig functions: Let α = arcsin x, so sin α = x. We want to represent this situation by a right triangle with an acute angle of α so that sin α = x. Because sin α = Opposite , Hypotenuse we can accomplish this by assigning the length x to the side opposite α and the length 1 √ to the hypotenuse. Then by the Pythagorean theorem the side adjacent to α is 1 − x2 , so √ Thus cot(arcsin x) = Adjacent = cot α = Opposite 1 − x2 . x √ 1 − x2 . x SECTION 8.3 Inverse Trigonometric Functions 593 3. Using inverse trig functions: Let α = arccos x, so cos α = x. We want to represent this situation by a right triangle with an acute angle of α so that cos α = x. Because cos α = Adjacent , Hypotenuse we can accomplish this by assigning the length x to the side adjacent to α and the length √ 1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is 1 − x2 , so Opposite tan α = = Adjacent √ Thus tan(arccos x) = √ 1 − x2 . x 1 − x2 . x 4. Using inverse trig functions: Let α = arccos(2x), so cos α = 2x. We want to represent this situation by a right triangle with an acute angle of α so that cos α = 2x. Because cos α = Adjacent , Hypotenuse we can accomplish this by assigning the length 2x to the side adjacent to α and the length 1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is p √ 1 − (2x)2 = 1 − 4x2 , so √ p Opposite 1 − 4x2 sin α = = = 1 − 4x2 . Hypotenuse 1 p Thus sin(arccos(2x)) = 1 − 4x2 . 5. A basketball player: In this situation the angle α he must incline his eyes is an acute angle of a right triangle with opposite side oflength 10 − 6 = 4 and adjacent side of 4 4 length 25. Thus tan α = . Then α = arctan , and the calculator gives the value 25 25 4 arctan = 9.09 degrees. Thus the angle is 9.09 degrees. An alternative approach 25 4 is to solve tan α = using the crossing-graphs method with a horizontal span of 0 to 25 90 in terms of degrees. 594 Solution Guide for Chapter 8 6. A soccer player: In this situation the angle α is an angle of a triangle with opposite side of length 24 and adjacent sides of length 20 and 40. By the law of cosines, 242 = 402 + 202 − 2 × 40 × 20 × cos α, so 402 + 202 − 242 . 2 × 40 × 20 cos α = Then α = arccos 402 + 202 − 242 2 × 40 × 20 , and for this the calculator gives the value α = 27.13 degrees. Thus the angle is 27.13 degrees. An alternative approach is to solve cos α = 402 + 202 − 242 2 × 40 × 20 using the crossing-graphs method with a horizontal span of 0 to 180 in terms of degrees. 7. Solving trigonometric equations: By inspection we can see that t = 1 is one solution of sin 2t = sin(t + 1). Another way to solve this is to take the arcsine of both sides: arcsin(sin 2t) = arcsin(sin(t + 1)). (This is legitimate because the sine function takes values between −1 and 1.) Because both 2t and t + 1 lie between 0 and 90 degrees, arcsin(sin 2t) = 2t and arcsin(sin(t+1)) = t+1. Thus the equation simplifies to 2t = t+1. Solving this gives t = 1. 8. Solving trigonometric equations: By the double angle formula for the sine, sin(2t) = 2 sin t cos t. Thus the equation can be written as sin(2t) = 0.5. Taking the arcsine of both sides gives arcsin(sin(2t)) = arcsin(0.5). Because t lies between 0 and 45 degrees, 2t lies between 0 and 90 degrees, so arcsin(sin(2t)) = 2t. Hence the equation simplifies to 2t = arcsin(0.5). Now arcsin(0.5) = 30◦ because sin 30◦ = 0.5. Hence 2t = 30◦ , and so t = 15◦ . sin t 1 and tan t = , so the given 9. Solving trigonometric equations: Now sec t = cos t cos t 1 sin t equation says = 2× . Multiplying both sides by cos t gives 1 = 2 sin t or cos t cos t 1 1 sin t = . Taking the arcsine of both sides gives arcsin(sin t) = arcsin . Because 2 2 t lies between 0 and 90 degrees, arcsin(sin t) = t. Thus the equation simplifies to t = 1 1 1 arcsin . Now arcsin = 30◦ because sin 30◦ = . Hence t = 30◦ . 2 2 2 10. Solving inverse trig equations: We take the tangent of both sides: tan(arctan x) = tan 9◦ . Now the equation tan(arctan x) = x is always true, so x = tan 9◦ = 0.16. 11. Solving inverse trig equations: We take the sine of both sides: sin(arcsin x) = sin x. Now the equation sin(arcsin x) = x is always true, so x = sin x. SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 595 8.4 COMPLEX NUMBERS AND DeMOIVRE’S THEOREM E-1. Which graph is it? Because θ is the counterclockwise angle from the positive horizontal axis, the equation θ = c says that the angle always equals the constant c. This describes a ray from the origin that makes an angle c with the positive x-axis. E-2. A missing part: Think of θ as starting at 0 and decreasing, so we are moving in a clockwise direction. This corresponds to a second hand on a clock that runs forward as usual, but with the second hand changing in length as it turns. Its tip will trace out a spiral, but to describe the spiral we note that the negative values of θ lead to negative values of r, and we locate these by reflecting through the origin. The result is the spiral obtained by reflecting through the origin the spiral shown in Figure 8.55. Below we show the graph of both spirals (using radians). The part for negative values of θ is the darker curve. E-3. An interesting graph: Experimenting by graphing r = sin(nθ) for different positive integral values of n leads to the following conclusions: If n is an odd integer, then the graph looks like a flower with n petals. If n is an even integer, the flower has 2n petals. √ Below we show the graph of r = sin( 2θ) for θ between 0 and 50 using radians. The graph has overlapping petals that eventually fill the circle of radius 1. 596 Solution Guide for Chapter 8 cos(2θ) . Below we cos θ show the graph of the strophoid. (Note that we have made the graph in “dot” mode. In E-4. The strophoid: The equation of the strophoid can be written as r = “connected” mode there is an extraneous vertical line x = −1.) We now describe, using radians to measure θ, how the graph is made. Think of θ as starting at 0 and increasing, so we are moving in a counterclockwise direction. As the π angle θ increases toward , the distance r decreases to 0, because the numerator ap4 π π proaches cos 2 × = cos = 0. Hence the graph passes through the origin at 4 π π2 π θ = . As we move past θ = toward θ = , the numerator cos(2θ) becomes negative 4 4 2 and the denominator cos θ stays positive but approaches 0. Over this range r is negative and growing in size without limit. Reflecting across the origin to account for the negative value of r gives the lower part of the graph on the left that goes down. As we move π past θ = , the numerator cos(2θ) stays negative, and the denominator cos θ decreases 2 from 0 to become negative. Hence r is positive, and no reflection is necessary. This gives 3π the upper part of the graph on the left. Continuing on through θ = to θ = π gives 4 the rest of the graph, which passes through the origin again and returns to the starting point. E-5. Coming up with a polar graph: We apply the law of cosines to the triangle in the figure. 1 The side opposite the angle θ has length x, and the adjacent sides have length and 2 cos θ. Hence 2 1 1 1 1 2 x = + cos2 θ − 2 × × cos θ × cos θ = + cos2 θ − cos2 θ = . 2 2 4 4 1 1 1 1 Thus x2 = , so x = . Hence the curve is a circle with center , 0 and radius . 4 2 2 2 SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 597 S-1. Calculating with complex numbers: (a) Now (3 + 7i) − (2 + 4i) = (3 − 2) + (7 − 4)i = 1 + 3i. (b) Now (3 + 2i)(5 − 6i) = 3 × 5 − 3 × 6i + 2i × 5 − 2i × 6i = 15 − 8i − 12i2 = 27 − 8i because i2 = −1. (c) Now (6 + i)(6 − i) = 6 × 6 − 6 × i + i × 6 − i × i = 36 − i2 = 37 because i2 = −1. This can also be written as 37 + 0i. S-2. A complex quotient: Following the suggestion, we multiply top and bottom of by 1 − i: 4 + 2i 4 + 2i 1 − i (4 + 2i)(1 − i) 4 − 2i − 2i2 6 − 2i = × = = = = 3 − i. 2 1+i 1+i 1−i (1 + i)(i − i) 1−i 2 S-3. Locating complex numbers: The numbers are located in the following figure. (d) (c) _ 7i 2+4i (a) 6+2i (e) 8 (b) 3-i 4 + 2i 1+i 598 Solution Guide for Chapter 8 S-4. Locating complex numbers: The numbers are located in the following figure. (b) 3 (a) 3π/4 1 π/4 _ π/4 1 (d) 2 (c) S-5. Standard representation: By the Euler formula √ π π + i4 sin = 2 + 2 3i = 2 + 3.46i 3 3 √ 1 π 3 π (using radians, of course). because cos = and sin = 2 3 2 3 4eiπ/3 = 4 cos S-6. Standard representation: By the Euler formula π π 6 6 6e−iπ/4 = 6 cos − + i6 sin − = √ − √ i = 4.24 − 4.24i, 4 4 2 2 using radians, of course. S-7. Standard representation: By the Euler formula 2e4i = 2 cos 4 + i2 sin 4 = −1.31 − 1.51i, using radians, of course. S-8. Standard representation: As suggested, we use De Moivre’s theorem with n = −1. This says (cos θ + i sin θ)−1 = cos(−θ) + i sin(−θ) = cos θ − i sin θ. Thus 1 1 1 1 1 = (eiθ )−1 = (cos θ + i sin θ)−1 = cos θ − i sin θ. reiθ r r r r SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 599 S-9. Verification: Now (cos 3 + i sin 3)2 = (cos 3 + i sin 3)(cos 3 + i sin 3) = cos2 3 − sin2 3 + 2i cos 3 sin 3 because i2 = −1. The calculator gives the values cos2 3 − sin2 3 = 0.96 and 2 cos 3 sin 3 = −0.28. Thus (cos 3 + i sin 3)2 = 0.96 − 0.28i. Also, the calculator gives cos 6 = 0.96 and sin 6 = −0.28. Thus 2 (cos 3 + i sin 3) = cos 6 + i sin 6. S-10. Verification: Now (cos 1.5+i sin 1.5)2 = (cos 1.5+i sin 1.5)(cos 1.5+i sin 1.5) = cos2 1.5−sin2 1.5+2i cos 1.5 sin 1.5 because i2 = −1. The calculator gives the values cos2 1.5 − sin2 1.5 = −0.99 and 2 cos 1.5 sin 1.5 = 0.14. Thus (cos 1.5 + i sin 1.5)2 = −0.99 + 0.14i. Also, the calculator gives cos 3 = −0.99 and sin 3 = 0.14. Thus 2 (cos 1.5 + i sin 1.5) = cos 3 + i sin 3. 600 Solution Guide for Chapter 8 1. Standard representation: By the Euler formula 5ei5π/6 = 5 cos √ 5π 5π 5 3 5 + 5i sin =− + i = −4.33 + 2.5i 6 6 2 2 √ 5π 3 5π 1 because cos =− and sin = (using radians, of course). 6 2 6 2 4 4 2. Exponential form: The modulus of √ + i √ is 2 2 s 2 2 4 4 √ = 4. + √ r= 2 2 We want an angle θ so that cos θ = and sin θ = √4 2 4 1 =√ 2 √4 2 1 =√ . 4 2 π satisfies this, so an exponential form is 4eπi/4 . 4 √ 3. Exponential form: The modulus of −1 + i 3 is r √ 2 2 r = (−1) + 3 = 2. Now θ = We want an angle θ so that cos θ = 1 −1 =− 2 2 and √ sin θ = Now θ = 3 . 2 2π satisfies this, so an exponential form is 2e2πi/3 . 3 SECTION 8.4 4. Powers: Now cos Complex Numbers and DeMoivre’s Theorem 601 √ π 3 π 1 = and sin = using radians. Thus we have 6 2 6 2 !35 √ 3 1 π π 35 +i = cos + i sin . 2 2 6 6 By De Moivre’s theorem, Now cos π π 35 35π 35π + i sin = cos + i sin . 6 6 6 6 √ 35π 11π 11π 3 1 35π cos + i sin = cos + i sin = −i . 6 6 6 6 2 2 Hence √ 3 1 +i 2 2 √ !35 = 3 1 −i . 2 2 √ π 3 π 1 5. Roots: Now cos = and sin = using radians. Thus we want to find the fourth 6 2 π 6 π2 roots of cos + i sin = eiπ/6 . Here they are: 6 6 1 π π π First 4th root = ei 4 6 = cos + i sin 24 24 1 π 13π 13π Second 4th root = ei 4 ( 6 +2π) = cos + i sin 24 24 1 π 25π 25π Third 4th root = ei 4 ( 6 +4π) = cos + i sin 24 24 1 π 37π 37π Fourth 4th root = ei 4 ( 6 +6π) = cos + i sin . 24 24 6. Powers: The calculator gives the values cos 70◦ = 0.34 and sin 70◦ = 0.94. Thus we have 103 (0.34 + 0.94i)103 = (cos 70◦ + i sin 70◦ ) . By De Moivre’s theorem, 103 (cos 70◦ + i sin 70◦ ) = cos(103 × 70◦ ) + i sin(103 × 70◦ ) = 0.98 + 0.17i. Hence (0.34 + 0.94i)103 = 0.98 + 0.17i. 602 Solution Guide for Chapter 8 7. Powers: The calculator gives the values cos 0.7 = 0.76 and sin 0.7 = 0.64 using radians. Thus we have 8 (0.76 + 0.64i)8 = (cos 0.7 + i sin 0.7) . By De Moivre’s theorem, 8 (cos 0.7 + i sin 0.7) = cos(8 × 0.7) + i sin(8 × 0.7) = 0.78 − 0.63i. Hence (0.76 + 0.64i)8 = 0.78 − 0.63i. √ π 1 π 3 using radians. Thus we have 8. Powers: Now cos − = and sin − =− 2 3 2 3 √ !30 π π 30 1 3 = cos − + i sin − . −i 2 3 3 2 By De Moivre’s theorem, π π 30 π π cos − + i sin − = cos 30 × − + i sin 30 × − . 3 3 3 3 Now π 30 × − = −10π, 3 and cos (−10π) + i sin (−10π) = 1 because −10π is an integral multiple of 2π. Hence √ !30 1 3 −i = 1. 2 2 √ π 1 π 3 using radians. Thus we want to find 9. Roots: Now cos − = and sin − =− 2 3 π2 3π the fourth roots of cos − + i sin − = e−iπ/3 . Here they are: 3 3 π π 1 π π π First 4th root = ei 4 (− 3 ) = cos − + i sin − = cos − i sin 12 12 12 12 5π 5π i 14 (− π +2π) 3 Second 4th root = e = cos + i sin 12 12 11π 11π i 41 (− π +4π) 3 Third 4th root = e = cos + i sin 12 12 17π 17π i 14 (− π +6π) 3 Fourth 4th root = e = cos + i sin . 24 24 SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 603 10. Roots: The calculator gives the values cos 0.6 = 0.83 and sin 0.6 = 0.56 using radians. Thus we want to find the sixth roots of cos 0.6 + i sin 0.6 = e0.6i . Here they are: First 6th root Second 6th root Third 6th root Fourth 6th root Fifth 6th root Sixth 6th root 1 = ei 6 (0.6) = cos 0.1 + i sin 0.1 1 π π = ei 6 (0.6+2π) = cos 0.1 + + i sin 0.1 + 3 3 2π 2π i 61 (0.6+4π) = cos 0.1 + + i sin 0.1 + = e 3 3 1 = ei 6 (0.6+6π) = cos (0.1 + π) + i sin (0.1 + π) 4π 4π i 16 (0.6+8π) = cos 0.1 + + i sin 0.1 + = e 3 3 1 5π 5π + i sin 0.1 + . = ei 6 (0.6+10π) = cos 0.1 + 3 3 11. Roots: The calculator gives the values cos 2 = −0.42 and sin 2 = 0.91 using radians. Thus we want to find the cube roots of cos 2 + i sin 2 = e2i . Here they are: First cube root 1 2 2 + i sin 3 3 2 2π 2 2π = cos + + i sin + 3 3 3 3 2 4π 2 4π + + i sin + . = cos 3 3 3 3 = ei 3 (2) = cos 1 Second cube root = ei 3 (2+2π) Third cube root = ei 3 (2+4π) 1 12. Roots: The values cos π = −1 and sin π = 0 (in terms of radians) can be found using the unit-circle definitions or the calculator. Thus we want to find the cube roots of cos π + i sin π = eiπ . Here they are: First cube root Second cube root Third cube root π π 1 = e = cos + i sin = + i 3 3 2 i 13 (π+2π) = e = cos π + i sin π = −1 1 5π 5π = ei 3 (π+4π) = cos + i sin = 3 3 i 31 (π) √ 3 2 √ 3 1 . −i 2 2 604 Solution Guide for Chapter 8 3π 3π = 0 and sin = −1 (in terms of radians) can be found 2 2 using the unit-circle definitions or the calculator. Thus we want to find the fifth roots 3π 3π of cos + i sin = ei3π/2 . Here they are: 2 2 13. Roots: The values cos 1 3π 2 ) 1 3π 2 +2π) 1 3π 2 +4π) 1 3π 2 +6π) 1 3π 2 +8π) First fifth root = ei 5 ( Second fifth root = ei 5 ( Third fifth root = ei 5 ( Fourth fifth root = ei 5 ( Fifth fifth root = ei 5 ( 3π 3π + i sin 10 10 7π 7π = cos + i sin 10 10 11π 11π = cos + i sin 10 10 15π 15π = cos + i sin = −i 10 10 19π 19π = cos + i sin . 10 10 = cos π π 1 = sin = √ (in terms of radians) can be found using special 4 4 2 values of sine and cosine or the calculator. Thus we want to find the fourth roots of π π cos + i sin = eiπ/4 . Here they are: 4 4 1 π π π First fourth root = ei 4 ( 4 ) = cos + i sin 16 16 1 π 9π 9π Second fourth root = ei 4 ( 4 +2π) = cos + i sin 16 16 17π 17π i 14 ( π +4π) Third fourth root = e 4 + i sin = cos 16 16 25π 25π i 14 ( π +6π) Fourth fourth root = e 4 + i sin . = cos 16 16 14. Roots: The values cos 15. Exponential form: The modulus of 1 + i tan θ is r= p p 12 + (tan θ)2 = 1 + tan2 θ. But tan2 θ = sec2 θ − 1 by Example 8.1 in Section 8.1, so r= p 1 + sec2 θ − 1 = sec θ. Factoring this out gives 1 + i tan θ = sec θ × Thus the polar form is (sec θ)eθi . 1 + i tan θ = sec θ(cos θ + i sin θ). sec θ SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 605 16. The modulus of a complex number: As suggested, by the Euler formula we have a + bi = reiθ = r cos θ + ir sin θ. Because two complex numbers are equal exactly when their real and imaginary parts are equal, we have a = r cos θ and b = r sin θ. Then a2 + b2 = (r cos θ)2 + (r sin θ)2 = r2 (cos2 θ + sin2 θ) = r2 . Hence √ a2 + b2 = r. 17. The complex conjugate: (a) In the notation of the exercise, a = 1 and b = 2. Thus the complex conjugate of 1 + 2i is 1 − 2i. (b) In the notation of the exercise, a = 1 and b = −3. Thus the complex conjugate of 1 − 3i is 1 + 3i. (c) In the notation of the exercise, a = 7 and b = 0. Thus the complex conjugate of 7 is 7 − 0i = 7. (d) In the notation of the exercise, a = 0 and b = −3. Thus the complex conjugate of −3i is 0 − (−3)i = 3i. (e) The standard representation of 3eiπ/7 is 3 cos π π + i3 sin . Hence the complex 7 7 conjugate of 3eiπ/7 is 3 cos π π π π − i3 sin = 3 cos − + i3 sin − = 3e−iπ/7 . 7 7 7 7 18. More on the modulus of a complex number: If a and b are real, then multiplying a + ib by its conjugate a − ib gives (a + bi)(a − ib) = a2 − iab + iab − i2 b2 = a2 + b2 , which is the square of the modulus of a + ib. 606 Solution Guide for Chapter 8 19. Double-angle formulas: By De Moivre’s theorem, (cos θ + i sin θ)2 = cos 2θ + i sin 2θ. Direct multiplication gives (cos θ + i sin θ)2 = (cos θ + i sin θ)(cos θ + i sin θ) = (cos2 θ − sin2 θ) + i(2 sin θ cos θ). If we equate the two expressions for (cos θ + i sin θ)2 and recall that two complex numbers are equal exactly when their real and imaginary parts are equal, we get the doubleangle formulas: cos 2θ = cos2 θ − sin2 θ and sin 2θ = 2 sin θ cos θ. 20. Triple-angle formulas: (a) By De Moivre’s theorem, (cos θ + i sin θ)3 = cos 3θ + i sin 3θ. Direct multiplication gives (cos θ + i sin θ)3 = (cos θ + i sin θ)2 (cos θ + i sin θ) = ((cos2 θ − sin2 θ) + i(2 sin θ cos θ))(cos θ + i sin θ) = (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ). If we equate the two expressions for (cos θ + i sin θ)3 and recall that two complex numbers are equal exactly when their real and imaginary parts are equal, we get cos 3θ = cos3 θ − 3 cos θ sin2 θ sin 3θ = 3 cos2 θ sin θ − sin3 θ. (b) Because sin2 θ + cos2 θ = 1, we have sin2 θ = 1 − cos2 θ. Thus, by Part (a), cos 3θ = cos3 θ − 3 cos θ sin2 θ = cos3 θ − 3 cos θ(1 − cos2 θ) = 4 cos3 θ − 3 cos θ. In a similar way, from sin2 θ + cos2 θ = 1 we get cos2 θ = 1 − sin2 θ. Hence, by Part (a), sin 3θ = 3 cos2 θ sin θ − sin3 θ = 3(1 − sin2 θ) sin θ − sin3 θ = 3 sin θ − 4 sin3 θ. SECTION 8.4 Complex Numbers and DeMoivre’s Theorem 607 21. Sum formulas: As suggested, we start with the equation ei(α+β) = eiα eiβ . Expanding the expression on the left using the Euler formula gives ei(α+β) = cos(α + β) + i sin(α + β). In a similar way we find eiα eiβ = (cos α + i sin α)(cos β + i sin β). Direct multiplication in the latter expression gives (cos α + i sin α)(cos β + i sin β) = (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β). Now we equate the expressions we found for both sides of the equation ei(α+β) = eiα eiβ and recall that two complex numbers are equal exactly when their real and imaginary parts are equal. The result is cos(α + β) = cos α cos β − sin α sin β and sin(α + β) = sin α cos β + cos α sin β. These are the sum formulas. 22. Lots of values: As suggested, we start with the observation that −1 = ei(π+2kπ) for every integer k. This implies that, for every integer k, ei(π+2kπ)×i = e−(π+2kπ) is a value of (−1)i . This gives infinitely many values of (−1)i and represents a sequence of positive real numbers with limiting value 0. 608 Solution Guide for Chapter 8 Chapter 8 Review Exercises 1. Sum formula: Now √ 3 1 1 1 = 0.97. sin 105 = sin(45 + 60 ) = sin 45 cos 60 + cos 45 sin 60 = √ × + √ × 2 2 2 2 ◦ ◦ 2. Area: The area is ◦ ◦ ◦ ◦ ◦ 1 × 3 × 2 × sin 35◦ = 1.72. 2 3. Identity: We use the half-angle formula 1 + cos t t 2 = cos 2 2 with t = 2x to obtain cos2 x = 1 + cos(2x) . 2 Then cos4 x = (cos2 x)2 = 1 + cos(2x) 2 2 = 1 1 + 2 cos(2x) + cos2 (2x) . 4 Now we use the half-angle formula cos2 t 1 + cos t = 2 2 with t = 4x to obtain cos2 (2x) = 1 + cos(4x) . 2 Thus 1 cos x = 4 4 1 + cos(4x) 1 + 2 cos(2x) + 2 = 3 1 1 + cos(2x) + cos(4x). 8 2 8 This can also be done by starting with the right-hand side and using the double-angle formula for the cosine. 4. Identity: Now the sum formula says cos(x + y) = cos x cos y − sin x sin y, and the difference formula says cos(x − y) = cos x cos y + sin x sin y. Adding these gives cos(x + y) + cos(x − y) = 2 cos x cos y. Dividing by 2 gives cos x cos y = 1 (cos(x + y) + cos(x − y)). 2 Chapter 8 Review Exercises 609 5. Sides and angles: We use the law of cosines to find C: C 2 = 32 + 42 − 2 × 3 × 4 cos 80◦ , so C= p 32 + 42 − 2 × 3 × 4 cos 80◦ = 4.56. Now we use the law of sines to find a (the angle opposite A): 3 4.56 = , sin a sin 80◦ so sin 80◦ . 4.56 sin a = 3 × To solve this equation we first note that the angle a must be less than 90 degrees because A isn’t the longest side of the triangle. Then using the crossing-graphs method or the arcsine function gives a = 40.38 degrees. Because the angle sum of the triangle is 180 degrees, the angle opposite B is 180 − 40.38 − 80 = 59.62 degrees. 6. Sides and angles: First we use the law of sines to find b (the angle opposite B): 20 10 = , sin b sin 130◦ so sin b = 10 × sin 130◦ . 20 To solve this equation we first note that the angle b must be less than 90 degrees because the angle opposite A is greater than 90 degrees. Then using the crossing-graphs method or the arcsine function gives b = 22.52 degrees. Because the angle sum of the triangle is 180 degrees, the angle opposite C is 180 − 130 − 22.52 = 27.48 degrees. Now we use the law of sines to find C: C 20 = , ◦ sin 27.48 sin 130◦ so C = sin 27.48◦ × 20 = 12.05. sin 130◦ (We can also find C using the law of cosines.) 7. Slide: The slide forms one side of a triangle for which the opposite angle is 20 degrees, with the other sides having lengths 50 and 90 (both in feet). By the law of cosines the length of the slide is p 502 + 902 − 2 × 50 × 90 cos 20◦ = 46.29 feet. 610 Solution Guide for Chapter 8 8. Pole: The top of the pole and the two anchoring points form a triangle. Because the angle sum of the triangle is 180 degrees, the angle at the top is 180 − 60 − 70 = 50 degrees. Now we use the law of sines to find the lengths of the wires. If A is the length (in feet) of the wire opposite the 60-degree angle, then A 7.5 = , ◦ sin 60 sin 50◦ so A = sin 60◦ × 7.5 = 8.48 feet. sin 50◦ If B is the length (in feet) of the wire opposite the 70-degree angle, then B 7.5 = , sin 70◦ sin 50◦ so B = sin 70◦ × 7.5 = 9.20 feet. sin 50◦ 9. Inverse sine: We want an angle between −90 and 90 degrees whose sine is 0. Now sin 0◦ = 0, so arcsin 0 equals 0 degrees or 0 radians. 10. Inverse tangent: We want an angle α between −90 and 90 degrees so that tan α = 1. 1 sin α , so we look for α so that sin α = cos α. Now sin 45◦ = cos 45◦ = √ , Now tan α = cos α 2 π so arctan 1 equals 45 degrees or radian. 4 11. Evaluate: Let α = arccos x, so cos α = x. We want to represent this situation by a right triangle with an acute angle of α so that cos α = x. Because cos α = Adjacent , Hypotenuse we can accomplish this by assigning the length x to the side adjacent to α and the length √ 1 to the hypotenuse. Then by the Pythagorean theorem the side opposite α is 1 − x2 , so cot α = Thus cot(arccos x) = √ x . 1 − x2 Adjacent x . =√ Opposite 1 − x2 Chapter 8 Review Exercises 611 12. Solve: Let α = arccos x, so x = cos α. Now the arccosine takes values between 0 and 180 degrees, and we are told that x is positive and less than 1, so α is positive and less than 90 degrees. In terms of α the original equation says sin(2α) = sin α. By the double-angle formula for the sine, this says 2 sin α cos α = sin α. By the restrictions on α we know that sin α is not zero. Hence we can divide both sides 1 1 by 2 sin α to get cos α = . Thus the solution is x = . 2 2 13. Calculating with complex numbers: (a) Now (−2 + 3i) − (5 + i) = (−2 − 5) + (3 − 1)i = −7 + 2i. (b) Now i(5 + 7i) = i × 5 + i × 7i = 5i + 7i2 = −7 + 5i because i2 = −1. (c) Now (1 − i)2 = (1 − i)(1 − i) = 1 × 1 − 1 × i − i × 1 + i × i = 1 − 2i + i2 = −2i because i2 = −1. 14. Standard representation: By the Euler formula 3π 3π −3iπ/2 + 7i sin − = 7i 7e = 7 cos − 2 2 3π 3π because (by the unit-circle definition or the calculator) cos − = 0 and sin − = 2 2 1 (using radians, of course). 612 Solution Guide for Chapter 8 π π 1 1 15. Powers: Because cos − = √ and sin − = √ (using radians), we have 4 4 2 2 4 π π 4 1 1 √ − i√ = cos − + i sin − . 4 4 2 2 By De Moivre’s theorem, π π 4 cos − + i sin − 4 4 π π cos 4 × − + i sin 4 × − 4 4 = cos (−π) + i sin (−π) = = −1. Hence 1 1 √ − i√ 2 2 4 = −1. π π 16. Roots: Now cos = 0 and sin = 1 using radians. Thus we want to find the square 2 π 2 π roots of cos + i sin = eiπ/2 . Here they are: 2 2 First square root Second square root 1 π 1 π π 1 1 π + i sin = √ + i √ 4 4 2 2 5π 5π 1 1 = cos + i sin = −√ − i√ . 4 4 2 2 = ei 2 ( 2 ) = cos = ei 2 ( 2 +2π)