Chapter 5 Analytic Trigonometry

Transcription

1 Section 5. Fundamental Identities 0 Cater 5 Analytic Trigonometry Section 5. Fundamental Identities Exloration. cos > sec, sec > cos, and tan sin > cos. sin > csc and tan > cot. csc > sin, cot > tan, and cot cos > sin Quick Review 5. For #, use a calculator rad rad rad rad a - ab + b a - b 6. u + u + u + 7. x - xy - y x + yx - y 8. v - 5v - v + v # y x y - # x y x y - x xy a # y x y + b # x ay + bx y x xy x + y x + y xy x + y # a x + y b xy x # x + y x - y x + y - Section 5. Exercises. sec + tan + > 5>6, so sec ;5>. Ten cos >sec ;>5. But sin, tan 7 0 imlies cos 7 0. So cos >5. Finally, tan sin cos y # x - y x + y x - y x + y x - y sin cos a 5 b 5.. sec + tan + 0, so sec ;0. But cos 7 0 imlies sec 7 0, so sec 0. Finally, tan sec csc csc sec tan sec - - 5, so tan ;5. But sec 7 0, sin 6 0 imlies tan 6 0, so tan -5. And cot >tan -> 5-5>5.. sin - cos , so sin ;0.6. But cos >0, tan <0 imlies sin 6 0, so sin Finally, tan sin >cos -0.6> cos( /- )sin cot tan( /- ) cos( )cos sin( /- ) sin( - /) cot( ) cot tan( /- ) tan( - /) tan x # 0. cot x tan x #. sec y sin a - y b # cos y cos y cos u. cot u sin u # sin ucos u sin u + tan x sec x >cos x sin x. cos x tan x csc x csc x >sin x. - cos u sin u sin u sin u sin u 5. - cos x - cos x sin x sin u + tan u + cos u + tan u sec u 6. sec u sec u sec u sec u 7. csc( x) # sin x 8. sec( x) cos( x) # cos -x cos x cos a cos -x - x b 9. cot( x) cot a # - x b sin -x sin a - x b cos -x # sinx sin -x cosx cos -x 0. cot( x) tan( x) # sin -x sin -x cos -x. sin -x + cos -x. sec -x - tan x sec x - tan x tan a - x b csc x. cot x csc x # csc x

2 0 Cater 5 Analytic Trigonometry + tan x. # + sin x + cot x + cos x + tan x + 5. sec x + csc x - tan x + cot x sec x - tan x + csc x - cot x+ sec u - tan u 6. cos v + sin v 7. ()(tan x+cot x)() a + b a + cos x b sec x 8. sin -tan cos + cos a - u b sin u sin - # cos u+sin sin cos u 9. ()()(tan x)(sec x)(csc x) ()() a ba ba b tan x sec y - tan ysec y + tan y 0. sec y a cos y - sin y cos y ba cos y + sin y cos y b a cos y b + sin y - sin y - sin y - sin y # cos y cos y cos y cos y cos y cos y tan x. csc x + tan x sec x a cos x b + a b # a. b + cos x tan x a # sec cos x b x csc x. sec x + csc x a cos x b + a sin x b # cos x sin x cos x # sin x + cos x sec x. + csc x+ sin x tan x cos x cos x csc x+ csc x sin x sin x cos x sec x 5. cot x - cos x tan x - sec x tan x - sec x sec x - - sec x + - sec x + sec x + sec x - tan x cot x sec x 7. - sec x - - cos x cot x sin x csc x 9. cos x sin x cos x - + sin x -. - cos x - + sin x - (-)(+). - sin x cos x + cos x sin x + csc x tan x - + csc x cot x # tan x - tan x + tan x - tan x + tan x - 6. sec x - sec x + tan x sec x - sec x + sec x - sec x - sec x - sec x + sec x sin x + tan a - tan a - tan a + 8. tan a - + tan a + tan a sin x tan x 50. sec x + sec x - sec x - sec x + sec x + sec x + sec x , so eiter 0 or. Ten x or + n or x 6 + n x 5, n an integer. On te interval: 6 + n x e 6,, 5 6, f

3 Section 5. Fundamental Identities tan x - 0, so eiter tan x 0 or. Ten x n or x ; an + n, n integer. On te interval: x e 0,,, 7 f 5. tan xsin x - 0, so eiter tan x 0 or. Ten x n or x an interger. + n, n However, tan x excludes x, so we ave only + n xn, n an integer. On te interval: x 50, 6 5. tan x - 0, so eiter 0 or tan x. Ten x n or x an integer. Put anoter + n, n way, all multiles of excet for ;, etc., ; On te interval: x e 0,,,, 5, 7 f 55. tan x ;, so x ; an integer. + n, n On te interval: x e,,, 5 f 56. ; an integer., so x + n, n On te interval: x e,, 5, 7 f , so terefore ; x ; an integer. + n, n 58. ( +)(+)0, so - or. Ten x - +n, x - 5 or n x - an integer. + n, n 59. sin usin u - 0, so sin u 0 or sin u. Ten u n, n an integer. 60. sin t- sin t,or sin t+ sin t-0. Tis factors to ( sin t-)(sin t+)0, so sin t or sin t. Ten t or t 5, 6 + n 6 + n n an integer. 6. cos() if n. Only n0 gives a value between and ±, so 0, or xn, n an integer. 6. Tis can be rewritten as ( -)(+)0, so or -. Ten x or 6 + n x 5 an integer. See also # n, n 6. cos L.98, so te solution set is {;.98+n n0, ;, ;,... }. 6. cos L 0.77, so te solution set is {; 0.77+n n0, ;, ;,... }. 65. sin L 0.07 and , so te solution set is {0.07+n or.869+n n0, ;, ;,... }. 66. tan - 5 L.7, so te solution set is {.7+n n0, ;, ;,... } L 0.66, and cos L 0.886, so te solution set is {; n n0, ;, ;,... } L 0.66 and sin L 0.687, so te solution set is {; n n0, ;, ;,... } cos sin u@ 70. tan u sec u@ 7. 9 sec u - tan u@ sin u 6@ cos u@ sec u @ tan u@ 75. True. Since cosine is an even function, so is secant, and tus sec (x- /)sec ( /-x), wic equals csc x by one of te cofunction identities. 76. False. Te domain of validity does not include values of for wic cos 0 and tan sin /cos is undefined, namely all odd integer multiles of /. 77. tan x sec xtan x//cos x Z. Te answer is D. 78. sine, tangent, cosecant, and cotangent are odd, wile cosine and secant are even. Te answer is A. 79. (sec +)(sec -)sec -tan.te answer is C. 80. By te quadratic formula, cos x+-0 imlies - or Tere are tree solutions on te interval (0, ).Te answer is D. 8., ; - sin x, tan x ; - sin x, csc x, sec x ; - sin x 8. 8 tan u + 8 9@ sec u@ - ; - - cot x ; - ; - cos x,, tan x ; - csc x ; - cos x, sec x, cot x ; - cos x 8. Te two functions are arallel to eac oter, searated by unit for every x. At any x, te distance between te two gras is sin x - -cos x sin x + cos x.,

4 06 Cater 5 Analytic Trigonometry [, ] by [, ] 8. Te two functions are arallel to eac oter, searated by unit for every x. At any x, te distance between te two gras is sec x - tan x. 90. Use te int: cos - x cos> - x - > sinx - > Cofunction identity -sin> - x Since sin is odd - Cofunction identity 9. Since A, B, and C are angles of a triangle, A+B -C. So: sin(a+b)sin( -C) sin C 9. Using te identities from Exercises 69 and 70, we ave: sin - x tan( -x) cos - x - -tan x 85. (a) [, ] by [, ] Section 5. Proving Trigonometric Identities Exloration. Te gras lead us to conclude tat tis is not an identity. [ 6, 70] by [0 000, ] (b) Te equation is y, sin(0.997x+.57)+8,855. [, ] by [, ]. For examle, cos( # 0), wereas cos(0).. Yes.. Te gras lead us to conclude tat tis is an identity. [ 6, 70] by [0 000, ] (c) >0.998 L 7. days. Tis is te number of days tat it takes te Moon to make one comlete orbit of te Eart (known as te Moon s sidereal eriod). (d) 5,7 miles (e) y, cos x + 8,855, or y, cos0.997x + 8, Answers will vary. 87. Factor te left-and side: sin u - cos u sin u - cos usin sin u - cos u # u + cos u sin u - cos u 88. Any k satisfying k or k Use te int: sin - x sin> - x - > cosx - > Cofunction identity cos> - x Since cos is even Cofunction identity [, ] by [, ] 5. No. Te gra window can not sow te full gras, so tey could differ outside te viewing window. Also, te function values could be so close tat te gras aear to coincide. Quick Review 5.. csc x+sec x + + sin x + cos x. tan x+cot x + cos x + sin x. # + #

5 Section 5. Proving Trigonometric Identities 07. sin # cos -cos # sin cos -sin sin cos 5. sin x+cos x > + > >cos Å 6. cos Å - sin Å cos Å - sin Å cos Å>sin a cos Å - sin Å cos Å 7. No. (Any negative x.) 8. Yes. 9. No. (Any x for wic <0, e.g. x /.) 0. No. (Any x for wic tan x<0, e.g. x /.). Yes.. Yes. Section 5. Exercises. One ossible roof: x - x - x - x + xx - x - x - x x x - x - x - -x + - x. One ossible roof: x - x a b - a x x b x - x x - x x. One ossible roof: x - x - - x - 9 x + x + x - x + x - - x - x + x + - x - 5. One ossible roof: x - x + - x + x - x + x - - x - x - x + x - - x + x + x sin x + cos x 5..Yes. csc x csc x tan x 6. #.Yes. sec x 7. # cot x # No.. 8. cos a x - Yes. b cos a - x b. sin x 9. (sin x)(+cot x)(sin x)(csc x).yes. sin x 0. No. Confirm graically.. ()(tan x+ cot x) # + # +cos x. ()(cot x+ tan x) # + # +sin x. (-tan x) - tan x+tan x (+tan x)- tan xsec x- tan x. (-) cos x- +sin x (cos x+sin x) One ossible roof: - cos u + cos u - cos u cos u cos u sin u cos u tan u + 6. tan x+sec x cos x - sin x - cos x - -sin x 7. # tan x 8. # sin a sin sec - tan sin sin sin cos b cos sin - sin 9. Multily out te exression on te left side csc x - cos x sin x. (cos t-sin t) +(cos t+sin t) cos t- cos t sin t+sin t+cos t + cos t sin t+sin t cos t+ sin t. sin Å-cos Å(-cos Å)-cos Å- cos Å + tan x sec x. sec x sin x + cos x cos ı cos ı + sin ı. +tan ı sin ı + sin b tan ı cos b cos ı sin ı sec ı csc ı cos ı sin ı cos ı cos ı - sin ı 5. + sin ı cos ı + sin ı cos ı + sin ı - sin ı + sin ı - sin ı cos ı + sin ı cos ı 6. One ossible roof: sec x + sec x + sec x - tan x tan x sec x - sec x - tan xsec x - tan x # tan xsec x - - -

6 08 Cater 5 Analytic Trigonometry tan x sec x - 7. sec x- - sec x + sec x + - cot v - cot v - 8. # tan v cot v tan v - tan v cot v + cot v + tan v cot v tan v + tan v - tan v cos v (Note: cot v tan v # sin v ) + tan v sin v cos v 9. cot x-cos x a -cos x b cos x - sin x cos x # sin x sin x cos x cot x 0. tan -sin a sin -sin cos b sin - cos sin # sin cos cos sin tan. cos x-sin x(cos x+sin x)(cos x-sin x) (cos x-sin x)cos x-sin x. tan t+tan ttan t(tan t+)(sec t-)(sec t) sec t-sec t. (x sin Å+y cos Å) +(x cos Å-y sin Å) (x sin Å+xy sin Å cos Å+y cos Å) +(x cos Å-xy cos Å sin Å+y sin Å) x sin Å+y cos Å+x cos Å+y sin Å (x +y )(sin Å+cos Å)x +y - cos - cos sin. sin sin + cos sin + cos sin + cos tan x tan xsec x + tan xsec x + 5. sec x - sec x - tan x sec x +. See also #6. tan x sin t + cos t sin t + + cos t cos t sin t sin t + cos t sin t + + cos t + cos t + cos t sin t + cos t sin t + cos t csc t sin t sin x - cos x sin x - - sin x sin x + + cos x + sin x sec x + sec x 8. # sec x - - sec x sec x - sec x sec x + (Note: sec x #.) sec x - sin t + cos t sin t + + cos t - cos t cos t sin t sin t - cos t sin t + - cos t - cos t + - cos t sin t - cos t sin t - cos t - cos t + cos t sin t - cos t sin t sin A cos B + cos A sin B 0. cos A cos B - sin A sin B cos A cos B sin A cos B + cos A sin B ± # cos A cos B - sin A sin B cos A cos B sin A cos A + sin B cos B tan A + tan B - tan A tan B - sin A sin B cos A cos B. sin x cos xsin x cos x sin x(-sin x)(sin x-sin x). sin 5 x cos xsin x cos x (sin x) cos x (-cos x) cos x (- cos x+cos x)cos x (cos x- cos x+cos 6 x). cos 5 xcos x (cos x) (-sin x) (- sin x+sin x). sin x cos xsin x cos x sin x (-sin x)(sin x-sin 5 x) tan x cot x cot x - tan x tan x # cot x + # - cot x - tan x a > - + > - b sin x - cos x - sin x + + cos x + +csc x sec x +. Tis involves rewriting a -b as (a-b)(a +ab+b ), were a and b sin x sec x cos x

7 Section 5. Proving Trigonometric Identities 09 tan x tan x cos x - tan x # + - tan x cos x cos x - sin x cos x + sin x + cos x - sin x cos x - sin x + cos x + sin x cos x sin x - cos x cos x(cos x)()(-sin x)() 50. sec x(sec x)(sec x)(+tan x)(sec x) 5. sin 5 x(sin x)()(sin x) () (-cos x) () (- cos x+cos x)() + 5. (b) divide troug by : sec x+tan x (d) multily out: (+sec x)(-) -+sec x-sec x # - cos x sin x -+ - # tan x. 5. (a) ut over a common denominator: sec x+csc x + a a b b a # sin x + cos x cos cos b x sin x x sin x sec x csc x. 55. (c) ut over a common denominator: sin x sec x. cos x 56. (e) multily and divide by : tan x + cot x sin x + cos x a + b. 57. (b) multily and divide by sec x+tan x: # sec x + tan x sec x + tan x sec x - tan x sec x + tan x sec x - tan x sec x + tan x. 58. False. Tere are numbers in te domain of bot sides of te equation for wic equality does not old, namely all negative real numbers. For examle, -, not. 59. True. If x is in te domain of bot sides of te equation, ten x 0. Te equation x x olds for all x 0, so it is an identity. 60. By te definition of identity, all tree must be true. Te answer is E. 6. A roof is - # cos x + sin x + Te answer is E. 6. One ossible roof: tan + sec sin cos + cos sin + cos sin + # sin - cos sin - sin - cos sin - -cos cos sin - -cos sin - cos - sin Te answer is C. 6. k must equal, so f(x) Z 0. Te answer is B. 6. ; cot x # 65. ; tan x # 66. ; + csc x + sec x > > sin x+cos x csc x cot x csc x > >sin x 67. ; - - sec x > cos x - cos x sin x - sin x sin x sin x sin x 68. ;. tan x > 69. ; (sec x)(-sin x) a (cos x) b 70. Since te sum of te logaritms is te logaritm of te roduct, and since te roduct of te absolute values of all six basic trig functions is, te logaritms sum to ln, wic is 0.

8 0 Cater 5 Analytic Trigonometry 7. If A and B are comlementary angles, ten sin A+sin Bsin A+sin ( /-A) sin A+cos A 7. Ceck Exercises 5 for correct identities. 7. Multily and divide by -sin t under te radical: - sin t # - sin t - sin t C + sin t - sin t C - sin t - sin t ƒ - sin t ƒ since a ƒ a ƒ. C cos t ƒ cos t ƒ Now, since -sin t 0, we can disense wit te absolute value in te numerator, but it must stay in te denominator. 7. Multily and divide by +cos t under te radical: + cos t # + cos t + cos t C - cos t + cos t C - cos t + cos t ƒ + cos t ƒ since a ƒ a ƒ. C sin t ƒ sin t ƒ Now, since +cos t 0, we can disense wit te absolute value in te numerator, but it must stay in te denominator. 75. sin 6 x+cos 6 x(sin x) +cos 6 x (-cos x) +cos 6 x (- cos x+ cos x-cos 6 x)+cos 6 x - cos x(-cos x)- cos x sin x. 76. Note tat a -b (a-b)(a +ab+b ). Also note tat a +ab+b a +ab+b -ab (a+b) -ab. Taking acos x and bsin x, we ave cos 6 x-sin 6 x (cos x-sin x)(cos x+cos x sin x+sin x) (cos x-sin x)[(cos x+sin x) -cos x sin x] (cos x-sin x)(-cos x sin x). 77. One ossible roof: ln tan x ln ƒ ƒ ƒ ƒ ln -ln. 78. One ossible roof: ln sec +tan +ln sec -tan ln sec -tan ln (a) Tey are not equal. Sown is te window [,,] by [, ]; graing on nearly any viewing window does not sow any aarent difference but using TRACE, one finds tat te y coordinates are not identical. Likewise, a table of values will sow sligt differences; for examle, wen x, y wile y [, ] by [, ] (b) One coice for is 0.00 (sown). Te function y is a combination of tree sinusoidal functions (000 sin(x+0.00), 000, and ), all wit eriod. [, ] by [ 0.00, 0.00] 80. (a) cos x-sin x (e x +e x ) - (e x -e x ) [e x ++e x -(e x -+e x )] (). sin x cos x - sin x (b) -tan x- cos x cos x, using te result from (a). Tis equals sec x. cos x cos x cos x - sin x (c) cot x- - sin x sin x, using te result from (a). Tis equals csc x. sin x 8. In te decimal window, te x coordinates used to lot te gra on te calculator are (e.g.) 0, 0., 0., 0., etc. tat is, xn/0, were n is an integer. Ten 0 x n, and te sine of integer multiles of is 0; terefore, +sin 0 x+sin n+0. However, for oter coices of x, suc as x, we ave +sin 0 x+sin 0 Z. Section 5. Sum and Difference Identities Exloration. sin u + v -, sin u + sin v. No.. cos u + v, cos u + cos v. No.. tan > + > -, tan > + tan >. (Many oter answers are ossible.) Quick Review # # No. fx + fy ln x + ln y lnxy fxy Z fx + y

9 Section 5. Sum and Difference Identities 8. No. fx + y e x + y e x e y 9. Yes. fx + y x + y x + y 0. No. fx + y x + y + 0 Section 5. Exercises. sin 5 sin5-0 sin 5 cos 0 - cos 5 sin 0. tan 5 tan5-0 - > + > sin 75 sin5 + 0 sin 5 cos 0 + cos 5 sin 0. cos 75 cos5 + 0 cos 5 cos 0 - sin 5 sin 0 5. cos sin sin cos a - b cos cos + 6. tan> - tan> 7. tan 5 tan a - b + tan> tan> tan> + tan> 8. tan tan a + b - tan> tan> fx fy Z fx + y fx + fy fx + y Z fx + fy # - # 6 - # + # 6 + # - # 6 - # + # + 6 sin 7 sin a + b sin cos + # + # 6 + cos 7 cos a b cos 5 6 cos + 0. sin a- b sin a 6 - b sin 6 cos tan 5 - tan 0 + tan 5 tan 0 cos sin sin 5 6 sin - # + # cos 6 sin # - # In #, matc te given exression wit te sum and difference identities.. sin - 7 sin 5. cos9-8 cos 76. sin a b sin 0. sin a - b sin 7 5. tan tan tan a 5-7. cos a 7 - x b cos a x - 7 b 8. cos a x + 7 b 9. sinx - x 0. cos7y + y cos 0y. tany + x. tana - b. b tan - 5 sin a x - b cos - sin # 0 - # -. Using te difference identity for te tangent function, we encounter tan, wic is undefined. However, we can comute tan a x - sinx - > b. From #, cosx - > sin a x -. Since te cosine function is even, b - cos a x - (see Examle, b cos a - x b - or #5). Terefore tis simlifies to. -cot x cos a x - b cos + sin # 0 + # Te simlest way is to note tat a, so tat - x b - y - x - y - x + y cos ca. Now use - x b - y d cos c - x + yd Examle to conclude tat cos c - x + y d sinx + y. 7. sin a x + 6 b cos 6 + sin 6 # + # 8. cos a x - b cos + sin # + # +

10 Cater 5 Analytic Trigonometry 9. tan a + b tan + tan> - tan tan> + tan - tan 0. cos a + b cos cos - sin sin cos # 0 - sin u # -sin. Equations B and F.. Equations C and E.. Equations D and H.. Equations A and G. 5. Rewrite as - 0; te left side equals sin(x-x), so xn, n an integer. 6. Rewrite as cos x -sin x 0; te left side equals cos(x+x), so x ; ten + n x an integer. 8 + n, n 7. sin a - u b sin cos u - cos sin u # cos u - 0 # sin u cos u. 8. Using te difference identity for te tangent function, we encounter tan, wic is undefined. However, we can comute tan a sin> - u - u b cos u cot u. cos> - u sin u Or, use #, and te fact tat te tangent function is odd. 9. cot a cos> - u - u b sin u tan u using sin> - u cos u te first two cofunction identities. 0. sec a csc u using te - u b cos> - u sin u first cofunction identity.. csc a sec u using te - u b sin> - u cos u second cofunction identity.. cos a x + cos a b - sin a b b # 0 - # tan + - tan # -. To write y + in te form y a sinbx + c, rewrite te formula using te formula for te sine of a sum: y asin bx cos c + cos bx sin c a sin bx cos c + a cos bx sin c a cos csin bx + a sin ccos bx. Ten comare te coefficients: a cos c, b, a sin c. Solve for a as follows: a cos c + a sin c + a cos c + a sin c 5 a cos c + sin c 5 a 5 a ;5 If we coose a to be ositive, ten cos c >5 and sin c >5. c cos - >5 sin - >5. So te sinusoid is y 5 sinx + cos - >5 L 5 sinx Follow te stes sown in Exercise (using te formula for te sine of a difference) to comare te coefficients in y a cos csin bx - a sin ccos bx to te coefficients in y 5 - : a cos c 5,b, a sin c. Solve for a as follows: a cos c + a sin c 5 + a cos c + sin c 69 a ; If we coose a to be ositive, ten cos c 5> and sin c >. So te sinusoid is y sinx - cos - 5> L sinx Follow te stes sown in Exercise to comare te coefficients in y a cos csin bx + a sin ccos bx to te coefficients in y cos x + sin x: a cos c, b, a sin c. Solve for a as follows: a cos c + a sin c + a cos c + sin c 5 a ;5 If we coose a to be ositive, ten cos c > 5 and sin c > 5. So te sinusoid is y 5 sinx - cos - > 5 L.6 sinx Follow te stes sown in Exercise to comare te coefficients in y a cos csin bx + a sin ccos bx to te coefficients in y : a cos c -, b, a sin c. Solve for a as follows: a cos c + a sin c - + a cos c + sin c a ; If we coose a to be negative, ten cos c > and sin c ->. So te sinusoid is y - - cos - > sinx sin(x-y)+sin(x+y) ( cos y- sin y) + ( cos y+ sin y) cos y 8. cos(x-y)+cos(x+y) ( cos y+ sin y) + ( cos y- sin y) cos y 9. cos xcos[(x+x)+x] cos(x+x) -sin(x+x) ( - ) -( + ) cos x-sin x - sin x cos x- sin x 50. sin usin[(u+u)+u]sin(u+u) cos u+ cos(u+u) sin u(sin u cos u+cos u sin u) cos u+ (cos u cos u-sin u sin u) sin u cos u sin u+ cos u sin u-sin u cos u sin u-sin u 5. cos x+cos(x+x)+cos(x-x); use #8 wit x relaced wit x and y relaced wit x. 5. sin x+sin(x+x)+sin(x-x); use #7 wit x relaced wit x and y relaced wit x.

11 Section 5. Sum and Difference Identities 5. tan(x+y) tan(x-y) tan x + tan y tan x - tan y a a - tan x tan y b # + tan x tan y b tan x - tan y since bot te numerator and - tan x tan y denominator are factored forms for differences of squares. 5. tan 5u tan utan(u+u) tan(u-u); use #5 wit xu and yu. + y y cos y + sin y cos y - sin y cos y + sin y > cos y # cos y - sin y > cos y cos y> cos y + sin y> cos y cos y> cos y - sin y> cos y > + sin y>cos y > - sin y>cos y tan x + tan y tan x - tan y 56. True. If B -A, ten cos A+cos B cos A+cos ( -A) cos A+cos cos A+sin sin A cos A+( ) cos A+(0) sin A False. For examle, cos +cos 0, but and are not sulementary. And even toug cos ( /)+cos ( /)0, / is not sulementary wit itself. 58. If cos A cos Bsin A sin B, ten cos (A+B) cos A cos B-sin A sin B0. Te answer is A. 59. y + sin (x+x)sin x. Te answer is A. 60. Sin 5 sin5-0 sin 5 cos 0 - cos 5 sin 0 a b - a b 6 - Te answer is D. tan u + tan v 6. For all u, v, tanu + v Te answer is B. - tan u tan v. sinu + v 6. tan(u+v) cosu + v sin u cos v + cos u sin v cos u cos v - sin u sin v sin u cos v cos u sin v + cos u cos v cos u cos v cos u cos v sin u sin v - cos u cos v cos u cos v sin u cos u + sin v cos v sin u sin v - cos u cos v tan u + tan v - tan u tan v sinu - v 6. tan(u-v) cosu - v sin u cos v - cos u sin v cos u cos v + sin u sin v sin u cos v cos u cos v cos u cos v cos u cos v sin u cos u - sin v cos v sin u sin v + cos u cos v tan u - tan v + tan u tan v 6. Te identity would involve tan a wic does not exist. b, sin a x + b tan a x + b cos + sin cos - sin -cot x 65. Te identity would involve tan a, wic does not exit. b tan a x - b sin a x - cos - sin cos + sin cos a x + b # 0 + # # 0 - # b cos a x - b - cos u sin v cos u cos v + sin u sin v cos u cos v # 0 - # # 0 + # -cot x sinx + - cos + sin cos - + sin a cos - b + sin

12 Cater 5 Analytic Trigonometry cosx + - cos - sin cos - - sin a cos - b - sin 68. Te coordinates of all oints must be a cos a k k b, sin a for k0,,,»,. We only bb need to find te coordinates of tose oints in Quadrant I, because te remaining oints are symmetric. We already know te coordinates for te cases wen k0,,,, 6 since tese corresond to te secial angles. k: cos a b cos a - b cos a b cos a b 6 - k5: cos a 5 b cos a - b cos a b cos a b + sin a b sin a b - # + sin a b cos a b - sin a b cos a b # - # a b Coordinates in te first quadrant are (, 0), a sin a 5 b sin a - b a, 6 - b, a,, sin a b sin a b # + # + 6 sin a b sin a - b sin a b cos a b sin a b cos a b # - # # b, a, b, a, b, + 6 b, 0, 69. sina + B sin - C sin cos C - cos sin C 0 # cos C - - sin C sin C 70. cos Ccos( -(A+B)) cos cos(a+b)+sin sin(a+b) ( )(cos A cos B-sin A sin B) + 0 # sina + B sin A sin B-cos A cos B sin A 7. tan A+tan B+tan C cos A + sin B cos B + sin C cos C sin Acos B cos C + sin Bcos A cos C cos A cos B cos C sin Ccos A cos B + cos A cos B cos C cos Csin A cos B + cos A sin B + sin Ccos A cos B cos A cos B cos C cos C sina + B + sin CcosA + B + sin A sin B cos A cos B cos C cos C sin - C + sin Ccos - C + sin A sin B cos A cos B cos C cos C sin C + sin C -cos C + sin C sin A sin B cos A cos B cos C sin A sin B sin C cos A cos B cos C tan A tan B tan C 7. cos A cos B cos C-sin A sin B cos C -sin A cos B sin C-cos A sin B sin C cos A(cos B cos C-sin B sin C) -sin A(sin B cos C+cos B sin C) cos A cos (B+C)-sin A sin(b+c) cos(a+b+c) cos 7. Tis equation is easier to deal wit after rewriting it as cos 5x +sin 5x sin x0. Te left side of tis equation is te exanded form of cos(5x-x), wic of course equals ; te gra sown is simly y. Te equation 0 is easily solved on te interval [, ]: x ; or x ;. Te original gra is so crowded tat one cannot see were crossings occur. [, ] by [.,.] 7. xa cos t a c cos a T b cos d - t a cos d cos a ( a sin d) sin T b B B in + B ref E 0 x cos a t - c b + E 0 x cos a t + c c b c E 0 x a cos t cos c c + cos t cos x c E 0 c a t T + d b sin a t T b sin d d + sin t sin x c x - sin t sin c b a t T b x a cos t cos c b E 0 x cos t cos c c

13 Section 5. Multile-Angle Identities 5 Section 5. Multile-Angle Identities Exloration. sin 8 - cos> - > # -. sin. 8 ; - - B We take te ositive square root because is a firstquadrant angle. 8. sin cos9> - > # - is a tird- 9. sin. 8 ; C We take te negative square root because 8 quadrant angle. Quick Review 5.. tan x wen x +n, n an integer. tan x wen x - +n, n an integer. Eiter 0 or. Te latter imlies te former, so x, n an integer. + n. Eiter 0 or. Te latter imlies te former, so xn, n an integer. 5. wen x - +n, n an integer 6. wen x +n, n an integer 7. Eiter or -. Ten x or 6 + n 5 x or x ;, n an integer. 6 + n + n 8. Eiter or. Ten x + n or x ;, n an integer. + n 9. Te traezoid can be viewed as a rectangle and two triangles; te area is ten A()()+ ()()+ ()() 0.5 square units. 0. View te triangle as two rigt triangles wit yotenuse, one leg, and te oter leg te eigt equal to - 8 Section 5. Exercises. cos ucos(u+u)cos u cos u-sin u sin u cos u-sin u. Starting wit te result of #: cos ucos u-sin u cos u-(-cos u) cos u-. Starting wit te result of #: cos ucos u-sin u (-sin u)-sin u- sin u tan u + tan u. tan utan(u+u) - tan u tan u tan u - tan u , so (-)0; 0 or wen x0 or x ( -)0, So 0 or wen x0,,., or 5 7. sin x+-0, so ( -)(+) 0; or wen x, 6 5 x or x cos x--0, so ( +)(-) 0; - or wen x0, x or x 9. -, so, or Ten 0 or 0 (but Z 0), so x0, x, x, x, 5 7 x or x. ; 5 0. cos x--0, so. Only - 5 is in [, ], so xcos a - 5 b.70 or x-cos a - 5 b.06 For #, any one of te last several exressions given is an answer to te question. In some cases, oter answers are ossible, as well.. sin +cos sin cos +cos (cos )( sin +). sin +cos sin cos +cos -sin sin cos + cos - sin cos +- sin. sin +cos sin cos +cos cos -sin sin sin cos +(cos -sin ) cos - sin cos sin cos +cos - sin cos sin cos + cos - cos

14 6 Cater 5 Analytic Trigonometry. sin +cos sin cos +cos sin +cos -sin sin cos +(cos -sin ) sin +cos -sin sin cos -sin +cos -sin 5. sin xsin (x) 6. cos 6xcos (x) cos x- 7. csc x # csc x tan x sin x 8. cot x tan x tan x - tan x tan x tan x - cot x-tan x 9. sin x + cos x +( cos x-) ()( cos x-) 0. sin x + cos x+(- sin x) ()( cos x+- sin x) ()(- sin x). cos (x)- sin x -( ) -8 sin x cos x. sin xsin (x) ( )( cos x-) ( )( cos x-). cos x+-0, so or, 5 x, x or x. +- sin x+0, so 7 or -, x, x, or x cos x - (- sin x) -( ) - sin x - sin x - sin x Tus te left side can be written as ()(- sin x). Tis equals 0 in [0, ) wen x x, x 5, x, or x 7, x,. 6. Using #9, tis become cos x0, so x0, x., x, or x 7. +sin x+ ()(+ )0. Ten 0 or - ; te solutions in [0, ) are x0, x,, x x., x, x, x, or x 5 8. Wit ux, tis becomes cos u+cos u0, te same as #. Tis means u, u, u 5, etc. i.e., x. Ten x 6, x, x 5 + n 6, x 7. 6, x, x 6 9. Using results from #5, -cos x ( )-(- sin x ) ()( sin x+ -)0. 0 wen x or x, wile te second - ; 5 factor equals zero wen. It turns out as can be observed by noting, e.g., tat sin a b tat tis means x0., x0.9, x., or x Using #, te left side can be rewritten as cos x-sin x+cos x-sin x. Relacing cos x wit -sin x gives sin x- sin x+ + (+)( sin x+ +). Tis equals 0 wen a x, and b ; 5 wen. Tese values turn out to be x0., x0.7, x., and x.9, as can be observed by noting, e.g., tat sin a - 5 b cos 0. sin 5 B a - B b -. Since sin 5 7 0, take te ositive square root. - cos 90. tan 95 - > -. Note sin 90 > tat tan 95 tan 5. + cos 50. cos 75 C a - C b -. Since cos , take te ositive square root. 5 - cos 5>6. sin C a + C b 5 +. Since sin 7 0, take te ositive square root tan. - cos7>6 + > - - sin7>6 -> + cos> 6. cos C a + 8 C b +. Since cos 7 0, take te ositive 8 square root.

15 Section 5. Multile-Angle Identities 7 7. (a) Starting from te rigt side: - cos u ( sin u)sin u. - - sin u (b) Starting from te rigt side: + cos u cos u. cos u + cos u - 8. (a) tan sin u u cos u - cos u> - cos u + cos u> + cos u (b) Te equation is false wen tan u is a negative number. It would be an identity if it were written as - cos u tan u. B + cos u 9. sin x(sin x) c - d - + c d (- +) 8 0. cos x cos x # + +. sin x sin x # - -. sin 5 x()(sin x) () c - d - + c d Alternatively, take sin 5 x sin x and aly te result of #9. -. cos x, so cos x+-0. Ten or. In te interval [0, ), x, 5 x, or x. General solution: +n or x+n, n an integer.. -cos + x, so cos x+-0. Ten or. In te interval [0, ), 5 x, x, orx. General solution: x +n or x+n, n an integer. 5. Te rigt side equals tan (x/); te only way tat tan(x/) tan (x/) is if eiter tan(x/)0 or tan(x/). In [0, ), tis aens wen x0 or x. Te general solution is xn or x n, n an integer cos x -, so cos x+-0, or ( -)(+)0. Ten or. Let Åcos a In te interval b [0, ), xå, x, or x-å. General solution: x_å+n or x+n, n an integer. 7. False. For examle, f(x) as eriod and g(x) as eriod, but te roduct f(x)g(x) as eriod. 8. True. cos x sin a - a x - bb + Te last exression is in te form for a sinusoid. 9. f(x) f(x)g(x). Te answer is D sin.5 sin a 5 b sin a - x b + C - cos 5 C - > C - - Te answer is E. 5. or 0 or 0 x 6 or 5 x 6 or Te answer is E.

16 8 Cater 5 Analytic Trigonometry 5. sin x-cos x- cos x, wic as te same eriod as te function cos x, namely. Te answer is C. 5. (a) In te figure, te triangle wit side lengts x/ and R is a rigt triangle, since R is given as te erendicular distance. Ten te tangent of te angle / is te ratio oosite over adjacent : tan Solving for x x> R gives te desired equation. Te central angle is /n since one full revolution of radians is divided evenly into n sections. u (b) 5.87 R tan, were /, so R 5.87/( tan ) R0. 5. (a) d A x D x E x Call te center of te rombus E. Consider rigt ABE, wit legs d / and d /, and yotenuse lengt x. jabe as measure /, and using sine o adj equals and cosine equals, we ave y y and sin. d > d cos d > d x x x x (b) Use te double angle formula for te sine function: sin sin a b sin cos d # d x x d d x 55. (a) B x C d ft θ θ ft ft Te volume is 0 ft times te area of te end. Te end is made u of two identical triangles, wit area (sin ) (cos ) eac, and a rectangle wit area () (cos ). Te total volume is ten 0 # (sin cos +cos )0 (cos )(+sin ). Considering only -, te maximum value occurs wen 0.5 (in fact, it aens exactly at ). Te maximum value is about.99 ft (a) x y 00 (x, y) x y Te eigt of te tunnel is y, and te widt is x, so te area is xy. Te x- and y-coordinates of te vertex are 0 cos and 0 sin, so te area is (0 cos )(0 sin )00( cos sin )00 sin. (b) Considering 0, te maximum area occurs wen, or about Tis gives x0 cos, or about., for a widt of 0 about 8.8, and a eigt of y 0 L. 57. csc u sin u sin u cos u # # sin u cos u csc u sec u 58. cot u tan u - tan u tan u a - tan u tan u b a cot u cot u b cot u - cot u 59. sec u cos u - sin u u a - sin u bacsc csc u b csc u csc u sec u cos u cos u - u a cos u - basec sec u b sec u - sec u 6. sec u cos u cos u - sin u u csc u a cos u - sin u basec sec u csc u b sec u csc u csc u - sec u 6. Te second equation cannot work for any values of x for wic 6 0, since te square root cannot be negative. Te first is correct since a double angle identity for te cosine gives - sin x; solving for gives sin x -, so tat -. Te absolute value of bot A sides removes te _. 6. (a) Te following is a scatter lot of te days ast January as x-coordinates (L) and te time (in our mode) as y-coordinates (L) for te time of day tat astronomical twiligt began in norteastern Mali in 005. [ 0, 70] by [ 60, 60]