Section 5. Fundmentl Identities 03 Chter 5 Anlytic Trigonometry Section 5. Fundmentl Identities Exlortion. cos > sec, sec > cos, nd tn sin > cos. sin > csc nd tn > cot 3. csc > sin, cot > tn, nd cot cos > sin Quick Review 5. For #, use clcultor...760 rd67.380. 0.973 rd53.30 3..98 rd3.30. 0.398 rd.60 5. - b + b - b 6. u + u + u + 7. x - 3xy - y x + yx - y 8. v - 5v - 3 v + v - 3 9. 0... # y x y - # x y x y - x xy # y x y + b # x y + bx y x xy x + y x + y xy x + y # x + y b xy x # x + y x - y x + y - Section 5. Exercises. sec + tn + 3> 5>6, so sec ;5>. Then cos >sec ;>5. But sin, tn 7 0 imlies cos 7 0. So cos >5. Finlly, tn 3 sin cos 3 y # x - y x + y x - y x + y x - y sin 3 cos 3 5 b 3 5.. sec + tn + 3 0, so sec ;0. But cos 7 0 imlies sec 7 0, so sec 0. Finlly, tn 3 sec csc 3 csc 3 sec 3 0 0 3. 3. tn sec - - 5, so tn ;5. But sec 7 0, sin 6 0 imlies tn 6 0, so tn -5. And cot >tn -> 5-5>5.. sin - cos - 0.8 0.36, so sin ;0.6. But cos >0, tn <0 imlies sin 6 0, so sin -0.6. Finlly, tn sin >cos -0.6>0.8-0.75. 5. cos( /- )sin 0.5 6. cot tn( /- ) 5.3 7. cos( )cos sin( /- ) sin( - /) 0.73 8. cot( ) cot tn( /- ) tn( - /)7.89 9. tn x # 0. cot x tn x #. sec y sin - y b # cos y cos y cos u. cot u sin u # sin ucos u sin u + tn x sec x >cos x sin x 3. cos x tn x csc x csc x >sin x. - cos u sin u sin u sin u sin u 5. - cos 3 x - cos x sin x sin u + tn u + cos u + tn 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 cos -x - x b 9. cot( x) cot # - x b sin -x sin - x b cos -x # sinx sin -x cosx cos -x 0. cot( x) tn( x) # sin -x sin -x cos -x. sin -x + cos -x. sec -x - tn x sec x - tn x tn - x b csc x 3. cot x csc x # csc x
0 Chter 5 Anlytic Trigonometry + tn x. # + sin x + cot x + cos x + tn x + 5. sec x + csc x - tn x + cot x sec x - tn x + csc x - cot x+ sec u - tn u 6. cos v + sin v 7. ()(tn x+cot x)() + b + cos x b sec x 8. sin -tn cos + cos - u b sin u sin - # cos u+sin sin cos u 9. ()()(tn x)(sec x)(csc x) ()() b b b tn x sec y - tn ysec y + tn y 30. sec y cos y - sin y cos y b cos y + sin y cos y b cos y b + sin y - sin y - sin y - sin y # cos y cos y cos y cos y cos y cos y tn x 3. csc x + tn x sec x cos x b + b #. b + cos x tn x # sec cos x b x csc x 3. sec x + csc x cos x b + sin x b # cos x sin x cos x # sin x + cos x sec x 33. + csc x+ sin x tn x cos x cos x csc x+ csc x sin x 3. - + + + - + - + - + - sin x cos x sec x 35. cot x - cos x tn x - sec x tn x - sec x - - 36. sec x - - sec x + - sec x + sec x + sec x - tn x cot x sec x 37. - sec x - - cos x cot x 38. - + - + - - sin x + + - - - - csc x 39. cos x + + + 0. - + sin x -. - + - cos x - + sin x -. - cos x - + sin x - (-)(+) 3. - sin x + - + cos x + cos x + - - +. sin x + csc x + + + + 5. tn x - + csc x cot x # tn x - tn x + tn x - tn x + tn x - 6. sec x - sec x + tn x sec x - sec x + sec x - sec x - sec x - sec x + sec x - 7. - sin x + tn - tn - tn + 8. tn - + tn + tn sin x - + 9. + - + + - tn x 50. sec x + sec x - sec x - sec x + sec x + sec x + sec x- 5. - 0, so either 0 or. Then x or + n or x 6 + n x 5, n n integer. On the intervl: 6 + n x e 6,, 5 6, 3 f - + + -
Section 5. Fundmentl Identities 05 5. tn x - 0, so either tn x 0 or. Then x n or x ; n + n, n integer. On the intervl: x e 0,,, 7 f 53. tn xsin x - 0, so either tn x 0 or. Then x n or x n interger. + n, n However, tn x excludes x, so we hve only + n xn, n n integer. On the intervl: x 50, 6 5. tn x - 0, so either 0 or tn x. Then x n or x n integer. Put nother + n, n wy, ll multiles of excet for ;, etc., ; 3 On the intervl: x e 0,, 3,, 5, 7 f 55. tn x ;3, so x ; n integer. 3 + n, n On the intervl: x e 3, 3, 3, 5 3 f 56. ; n integer., so x + n, n On the intervl: x e, 3, 5, 7 f 57. - 0, so therefore ; x ; n integer. 3 + n, n 58. ( +)(+)0, so - or. Then x - +n, x - 5 or 6 6 + n x - n integer. + n, n 59. sin usin u - 0, so sin u 0 or sin u. Then u n, n n integer. 60. 3 sin t- sin t,or sin t+3 sin t-0. This fctors to ( sin t-)(sin t+)0, so sin t or sin t. Then t or t 5, 6 + n 6 + n n n integer. 6. cos() if n. Only n0 gives vlue between nd ±, so 0, or xn, n n integer. 6. This cn be rewritten s ( -)(+)0, so or -. Then x or 6 + n x 5 n integer. See lso #60. 6 + n, n 63. cos - 0.37 L.98, so the solution set is {;.98+n n0, ;, ;,... }. 6. cos - 0.75 L 0.77, so the solution set is {; 0.77+n n0, ;, ;,... }. 65. sin - 0.30 L 0.307 nd -0.307.8369, so the solution set is {0.307+n or.8369+n n0, ;, ;,... }. 66. tn - 5 L.373, so the solution set is {.373+n n0, ;, ;,... }. 67. 0. L 0.636, nd cos - 0.636 L 0.886, so the solution set is {; 0.886+n n0, ;, ;,... }. 68. 0. L 0.636 nd sin - 0.636 L 0.687, so the solution set is {; 0.687+n n0, ;, ;,... }. 69. - cos u @ sin u@ 70. tn u + @ sec u@ 7. 9 sec u - 9 3@ tn u@ 7. 36-36 sin u 6@ cos u@ 73. 7. 00 sec u - 00 0@ tn u@ 75. True. Since cosine is n even function, so is secnt, nd thus sec (x- /)sec ( /-x), which equls csc x by one of the cofunction identities. 76. Flse. The domin of vlidity does not include vlues of for which cos 0 nd tn sin /cos is undefined, nmely ll odd integer multiles of /. 77. tn x sec xtn x//cos x Z. The nswer is D. 78. sine, tngent, cosecnt, nd cotngent re odd, while cosine nd secnt re even. The nswer is A. 79. (sec +)(sec -)sec -tn.the nswer is C. 80. By the qudrtic formul, 3 cos x+-0 imlies - or 3 There re three solutions on the intervl (0, ).The nswer is D. 8., ; - sin x, tn x ; - sin x, csc x, sec x ; - sin x 8. 8 tn u + 8 9@ sec u@ - ; - 3-3 cot x ; - ; - cos x,, tn x ; - csc x ; - cos x, sec x, cot x ; - cos x 83. The two functions re rllel to ech other, serted by unit for every x. At ny x, the distnce between the two grhs is sin x - -cos x sin x + cos x.,
06 Chter 5 Anlytic Trigonometry [, ] by [, ] 8. The two functions re rllel to ech other, serted by unit for every x. At ny x, the distnce between the two grhs is sec x - tn x. 90. Use the hint: cos - x cos> - x - > sinx - > Cofunction identity -sin> - x Since sin is odd - Cofunction identity 9. Since A, B, nd C re ngles of tringle, A+B -C. So: sin(a+b)sin( -C) sin C 9. Using the identities from Exercises 69 nd 70, we hve: sin - x tn( -x) cos - x - -tn x 85. () [, ] by [, ] Section 5. Proving Trigonometric Identities Exlortion. The grhs led us to conclude tht this is not n identity. [ 6, 70] by [0 000, 60 000] (b) The eqution is y3, sin(0.997x+.57)+38,855. [, ] by [, ]. For exmle, cos( # 0), wheres cos(0). 3. Yes.. The grhs led us to conclude tht this is n identity. [ 6, 70] by [0 000, 60 000] (c) >0.998 L 7.3 dys. This is the number of dys tht it tkes the Moon to mke one comlete orbit of the Erth (known s the Moon s siderel eriod). (d) 5,7 miles (e) y 3, cos - 0.997x + 38,855, or y 3, cos0.997x + 38,855. 86. Answers will vry. 87. Fctor the left-hnd side: sin u - cos u sin u - cos usin sin u - cos u # u + cos u sin u - cos u 88. Any k stisfying k or k -. 89. Use the hint: sin - x sin> - x - > cosx - > Cofunction identity cos> - x Since cos is even Cofunction identity [, ] by [ 3, 3] 5. No. The grh window cn not show the full grhs, so they could differ outside the viewing window. Also, the function vlues could be so close tht the grhs er to coincide. Quick Review 5.. csc x+sec x + + sin x + cos x. tn x+cot x + cos x + sin x 3. # + #
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 cos Å - sin Å cos Å 7. No. (Any negtive x.) 8. Yes. 9. No. (Any x for which <0, e.g. x /.) 0. No. (Any x for which tn x<0, e.g. x /.). Yes.. Yes. Section 5. Exercises. One ossible roof: x 3 - x - x - x + xx - x - x - x x x - x - x - -x + - x. One ossible roof: x - x b - x x b x - x x - x x 3. One ossible roof: x - x - - x - 9 x + 3 x + x - x + 3x - 3 - x - x + 3 x + - x - 3 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 tn x 6. #.Yes. sec x 7. # cot x # No.. 8. cos x - Yes. b cos - x b. sin 3 x 9. (sin 3 x)(+cot x)(sin 3 x)(csc x).yes. sin x 0. No. Confirm grhiclly.. ()(tn x+ cot x) # + # +cos x. ()(cot x+ tn x) # + # +sin x 3. (-tn x) - tn x+tn x (+tn x)- tn xsec x- tn x. (-) cos x- +sin x (cos x+sin x)- - 5. One ossible roof: - cos u + cos u - cos u cos u cos u sin u cos u tn u + 6. tn x+sec x + + + cos x - sin x - cos x - -sin x 7. # tn x 8. # sin sin sec - tn sin sin sin cos b cos sin - sin 9. Multily out the exression on the left side. + + - 0. - + + - + 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 + tn x sec x 3. sec x sin x + cos x cos ı cos ı + sin ı. +tn ı sin ı + sin b tn 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 - tn x tn x sec x - sec x - tn xsec x - tn x # tn xsec x - - -
08 Chter 5 Anlytic Trigonometry tn x sec x - 7. sec x- - sec x + sec x + - cot v - cot v - 8. # tn v cot v tn v - tn v cot v + cot v + tn v cot v tn v + tn v - tn v cos v (Note: cot v tn v # sin v ) + tn v sin v cos v 9. cot x-cos x -cos x b cos x - sin x cos x # sin x sin x cos x cot x 30. tn -sin sin -sin cos b sin - cos sin # sin cos cos sin tn 3. cos x-sin x(cos x+sin x)(cos x-sin x) (cos x-sin x)cos x-sin x 3. tn t+tn ttn t(tn t+)(sec t-)(sec t) sec t-sec t 33. (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 3. sin sin + cos sin + cos sin + cos tn x tn xsec x + tn xsec x + 35. sec x - sec x - tn x sec x +. See lso #6. tn x sin t + cos t sin t + + cos t 36. + + 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 - - + 37. + + sin x - cos x sin x - - sin x sin x + + cos x + sin x - + + + sec x + sec x 38. # 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 39. + - 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 cos B + cos A sin B 0. cos A cos B - sin B cos A cos B cos B + cos A sin B ± # cos A cos B - sin B cos A cos B cos A + sin B cos B tn A + tn B - tn A tn B - sin B cos A cos B. sin x cos 3 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) 3. cos 5 xcos x (cos x) (-sin x) (- sin x+sin x). sin 3 x cos 3 xsin 3 x cos x sin 3 x (-sin x)(sin 3 x-sin 5 x) tn x cot x 5. + - cot x - tn x tn x # cot x + # - cot x - tn x > - + > - b sin 3 x - cos 3 x - sin x + + cos x + +csc x sec x +. This involves rewriting 3 -b 3 s (-b)( +b+b ), where nd b. 6. + + - 3 - + + + - - sin x sec x cos x
Section 5. Proving Trigonometric Identities 09 tn x 7. + - tn x cos x - tn x # + - tn x cos x cos x - sin x cos x + sin x + cos x - sin x cos x - sin x + cos x + sin x - + + + - + - - 3 - cos x + - 8. sin x - cos x + - - + - - 9. cos 3 x(cos x)()(-sin x)() 50. sec x(sec x)(sec x)(+tn x)(sec x) 5. sin 5 x(sin x)()(sin x) () (-cos x) () (- cos x+cos x)() + 5. (b) divide through by : sec x+tn x. + 53. (d) multily out: (+sec x)(-) -+sec x-sec x - + - # - cos x sin x -+ - # tn x. 5. () ut over common denomintor: sec x+csc x + b b # sin x + cos x cos cos b x sin x x sin x sec x csc x. 55. (c) ut over common denomintor: - + + + + - - sin x sec x. cos x 56. (e) multily nd divide by : tn x + cot x sin x + cos x + b. 57. (b) multily nd divide by sec x+tn x: # sec x + tn x sec x + tn x sec x - tn x sec x + tn x sec x - tn x sec x + tn x. 58. Flse. There re numbers in the domin of both sides of the eqution for which equlity does not hold, nmely ll negtive rel numbers. For exmle, -3 3, not 3. 59. True. If x is in the domin of both sides of the eqution, then x 0. The eqution x x holds for ll x 0, so it is n identity. 60. By the definition of identity, ll three must be true. The nswer is E. 6. A roof is - # + - + + - cos x + sin x + The nswer is E. 6. One ossible roof: tn + sec sin cos + cos sin + cos sin + # sin - cos sin - sin - cos sin - -cos cos sin - -cos sin - cos - sin The nswer is C. 63. k must equl, so f(x) Z 0. The nswer is B. 6. ; cot x # 65. ; tn 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. ;. tn x > 69. ; (sec x)(-sin x) (cos x) b 70. Since the sum of the logrithms is the logrithm of the roduct, nd since the roduct of the bsolute vlues of ll six bsic trig functions is, the logrithms sum to ln, which is 0.
0 Chter 5 Anlytic Trigonometry 7. If A nd B re comlementry ngles, then sin A+sin Bsin A+sin ( /-A) sin A+cos A 7. Check Exercises 5 for correct identities. 73. Multily nd divide by -sin t under the rdicl: - sin t # - sin t - sin t C + sin t - sin t C - sin t - sin t ƒ - sin t ƒ since ƒ ƒ. C cos t ƒ cos t ƒ Now, since -sin t 0, we cn disense with the bsolute vlue in the numertor, but it must sty in the denomintor. 7. Multily nd divide by +cos t under the rdicl: + cos t # + cos t + cos t C - cos t + cos t C - cos t + cos t ƒ + cos t ƒ since ƒ ƒ. C sin t ƒ sin t ƒ Now, since +cos t 0, we cn disense with the bsolute vlue in the numertor, but it must sty in the denomintor. 75. sin 6 x+cos 6 x(sin x) 3 +cos 6 x (-cos x) 3 +cos 6 x (-3 cos x+3 cos x-cos 6 x)+cos 6 x -3 cos x(-cos x)-3 cos x sin x. 76. Note tht 3 -b 3 (-b)( +b+b ). Also note tht +b+b +b+b -b (+b) -b. Tking cos x nd bsin x, we hve 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 tn x ln ƒ ƒ ƒ ƒ ln -ln. 78. One ossible roof: ln sec +tn +ln sec -tn ln sec -tn ln 0 79. () They re not equl. Shown is the window [,,] by [, ]; grhing on nerly ny viewing window does not show ny rent difference but using TRACE, one finds tht the y coordintes re not identicl. Likewise, tble of vlues will show slight differences; for exmle, when x, y 0.53988 while y 0.5030 [, ] by [, ] (b) One choice for h is 0.00 (shown). The function y 3 is combintion of three sinusoidl functions (000 sin(x+0.00), 000, nd ), ll with eriod. [, ] by [ 0.00, 0.00] 80. () cosh x-sinh x (e x +e x ) - (e x -e x ) [e x ++e x -(e x -+e x )] (). sinh x cosh x - sinh x (b) -tnh x- cosh x cosh x, using the result from (). This equls sech x. cosh x cosh x cosh x - sinh x (c) coth x- - sinh x sinh x, using the result from (). This equls csch x. sinh x 8. In the deciml window, the x coordintes used to lot the grh on the clcultor re (e.g.) 0, 0., 0., 0.3, etc. tht is, xn/0, where n is n integer. Then 0 x n, nd the sine of integer multiles of is 0; therefore, +sin 0 x+sin n+0. However, for other choices of x, such s x, we hve +sin 0 x+sin 0 Z. Section 5.3 Sum nd Difference Identities Exlortion. sin u + v -, sin u + sin v. No.. cos u + v, cos u + cos v. No. 3. tn >3 + >3-3, tn >3 + tn >3 3. (Mny other nswers re ossible.) Quick Review 5.3. 5 5-30. 75 5 + 30 3. 65 80-5 80 + 30-5 0-5. # 6-3 - 5 5. # 6-3 - 7 6. + 3 3 + 7. No. fx + fy ln x + ln y lnxy fxy Z fx + y
Section 5.3 Sum nd Difference Identities 8. No. fx + y e x + y e x e y 9. Yes. fx + y 3x + y 3x + 3y 0. No. fx + y x + y + 0 Section 5.3 Exercises. sin 5 sin5-30 sin 5 cos 30 - cos 5 sin 30. tn 5 tn5-30 - 3>3 + 3>3 3-3 3-3 - 3 3 + 3 9-3 3. sin 75 sin5 + 30 sin 5 cos 30 + cos 5 sin 30. cos 75 cos5 + 30 cos 5 cos 30 - sin 5 sin 30 5. cos sin 3 sin cos 3 - b cos 3 cos + 6. tn>3 - tn> 7. tn 5 tn 3 - b + tn>3 tn> - 3-3 + + 3-3 3 + 3-3 - tn>3 + tn> 8. tn tn 3 + b - tn>3 tn> - 3 + - 3 3 - + 3-3 + 3-3 9. fx fy Z fx + y fx + fy fx + y Z fx + fy # 3 - # 6 - # 3 + # 6 + # 3 - # 6 - # + 3 # + 6 sin 7 sin 3 + b sin 3 cos + 3 # + # 6 + cos 7 cos 5 6 - b cos 5 6 cos + 0. sin - b sin 6 - b sin 6 cos - - 6-6 tn 5 - tn 30 + tn 5 tn 30 cos 3 sin sin 5 6 sin -3 # + # cos 6 sin # - 3 # In #, mtch the given exression with the sum nd difference identities.. sin - 7 sin 5. cos9-8 cos 76 3. sin 5 + 7 b sin 0. sin 3 - b sin 7 5. tn 9 + 7 tn 66 6. tn 5-3 7. cos 7 - x b cos x - 7 b 8. cos x + 7 b 9. sin3x - x 0. cos7y + 3y cos 0y. tny + 3x. tn3 - b 3. b tn - 5 sin x - b cos - sin # 0 - # -. Using the difference identity for the tngent function, we encounter tn, which is undefined. However, we cn comute tn x - sinx - > b. From #3, cosx - > sin x -. Since the cosine function is even, b - cos x - (see Exmle, b cos - x b - or #5). Therefore this simlifies to. -cot x cos x - b cos + sin # 0 + # 5. 6. The simlest wy is to note tht, so tht - x b - y - x - y - x + y cos c. Now use - x b - y d cos c - x + yd Exmle to conclude tht cos c - x + y d sinx + y. 7. sin x + 6 b cos 6 + sin 6 # 3 + # 8. cos x - b cos + sin # + # +
Chter 5 Anlytic Trigonometry 9. tn + b tn + tn> - tn tn> + tn - tn 30. cos + b cos cos - sin sin cos # 0 - sin u # -sin 3. Equtions B nd F. 3. Equtions C nd E. 33. Equtions D nd H. 3. Equtions A nd G. 35. Rewrite s - 0; the left side equls sin(x-x), so xn, n n integer. 36. Rewrite s cos 3x -sin 3x 0; the left side equls cos(3x+x), so x ; then + n x n integer. 8 + n, n 37. sin - u b sin cos u - cos sin u # cos u - 0 # sin u cos u. 38. Using the difference identity for the tngent function, we encounter tn, which is undefined. However, we cn comute tn sin> - u - u b cos u cot u. cos> - u sin u Or, use #, nd the fct tht the tngent function is odd. 39. cot cos> - u - u b sin u tn u using sin> - u cos u the first two cofunction identities. 0. sec csc u using the - u b cos> - u sin u first cofunction identity.. csc sec u using the - u b sin> - u cos u second cofunction identity.. cos x + cos b - sin b b # 0 - # tn + - tn # - 3. To write y 3 + in the form y sinbx + c, rewrite the formul using the formul for the sine of sum: y sin bx cos c + cos bx sin c sin bx cos c + cos bx sin c cos csin bx + sin ccos bx. Then comre the coefficients: cos c 3, b, sin c. Solve for s follows: cos c + sin c 3 + cos c + sin c 5 cos c + sin c 5 5 ;5 If we choose to be ositive, then cos c 3>5 nd sin c >5. c cos - 3>5 sin - >5. So the sinusoid is y 5 sinx + cos - 3>5 L 5 sinx + 0.973.. Follow the stes shown in Exercise 3 (using the formul for the sine of difference) to comre the coefficients in y cos csin bx - sin ccos bx to the coefficients in y 5 - : cos c 5,b, sin c. Solve for s follows: cos c + sin c 5 + cos c + sin c 69 ;3 If we choose to be ositive, then cos c 5>3 nd sin c >3. So the sinusoid is y 3 sinx - cos - 5>3 L 3 sinx -.76. 5. Follow the stes shown in Exercise 3 to comre the coefficients in y cos csin bx + sin ccos bx to the coefficients in y cos 3x + sin 3x: cos c, b3, sin c. Solve for s follows: cos c + sin c + cos c + sin c 5 ;5 If we choose to be ositive, then cos c > 5 nd sin c > 5. So the sinusoid is y 5 sin3x - cos - > 5 L.36 sin3x - 0.636. 6. Follow the stes shown in Exercise 3 to comre the coefficients in y cos csin bx + sin ccos bx to the coefficients in y 3 - - + 3 : cos c -, b, sin c 3. Solve for s follows: cos c + sin c - + 3 cos c + sin c 3 ;3 If we choose to be negtive, then cos c > 3 nd sin c -3> 3. So the sinusoid is y -3 - cos - > 3-3.606 sinx - 0.988. 7. 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 3xcos[(x+x)+x] cos(x+x) -sin(x+x) ( - ) -( + ) cos x-sin x - sin x cos x-3 sin x 50. sin 3usin[(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 u3 cos u sin u-sin u 5. cos 3x+cos(x+x)+cos(x-x); use #8 with x relced with x nd y relced with x. 5. +sin(3x+x)+sin(3x-x); use #7 with x relced with 3x nd y relced with x.
Section 5.3 Sum nd Difference Identities 3 53. tn(x+y) tn(x-y) tn x + tn y tn x - tn y - tn x tn y b # + tn x tn y b tn x - tn y since both the numertor nd - tn x tn y denomintor re fctored forms for differences of squres. 5. tn 5u tn 3utn(u+u) tn(u-u); use #53 with xu nd yu. + y 55. - 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 tn x + tn y tn x - tn y 56. True. If B -A, then cos A+cos B cos A+cos ( -A) cos A+cos cos A+sin cos A+( ) cos A+(0) 0. 57. Flse. For exmle, cos 3 +cos 0, but 3 nd re not sulementry. And even though cos (3 /)+cos (3 /)0, 3 / is not sulementry with itself. 58. If cos A cos B sin B, then cos (A+B) cos A cos B- sin B0. The nswer is A. 59. y + sin (x+x)sin 3x. The nswer is A. 60. Sin 5 sin5-30 sin 5 cos 30 - cos 5 sin 30 3 b - b 6 - The nswer is D. tn u + tn v 6. For ll u, v, tnu + v The nswer is B. - tn u tn v. sinu + v 6. tn(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 tn u + tn v - tn u tn v sinu - v 63. tn(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 tn u - tn v + tn u tn v 6. The identity would involve tn which does not exist. b, sin x + b tn x + b cos + sin cos - sin -cot x 65. The identity would involve tn 3, which does not exit. b tn x - 3 b sin x - cos 3 - sin 3 cos 3 + sin 3 cos x + b # 0 + # # 0 - # 3 b cos x - 3 b - cos u sin v cos u cos v + sin u sin v cos u cos v # 0 - # # 0 + # -cot x sinx + h - cos h + sin h - 66. h h cos h - + sin h h cos h - b + sin h h h
Chter 5 Anlytic Trigonometry cosx + h - cos h - sin h - 67. h h cos h - - sin h h cos h - b - sin h h h 68. The coordintes of ll oints must be cos k k b, sin for k0,,,», 3. We only bb need to find the coordintes of those oints in Qudrnt I, becuse the remining oints re symmetric. We lredy know the coordintes for the cses when k0,, 3,, 6 since these corresond to the secil ngles. k: cos b cos 3 - b cos 3 b cos b 6 - k5: cos 5 b cos 3-3 b cos 3 b cos 3 b + sin 3 b sin 3 b - # + sin 3 b cos 3 b - sin 3 b cos 3 b # 3 - # - + 6 b Coordintes in the first qudrnt re (, 0), + 6 + - sin 5 b sin 3-3 b, 3 6 - b,,, sin 3 b sin b # + 3 # + 6 sin b sin 3 - b sin 3 b cos b sin b cos 3 b 3 # - # # 3 6-6 - b, 3, b,, 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 B) + 0 # sina + B sin B-cos A cos B 7. tn A+tn B+tn C cos A + sin B cos B + sin C cos C cos 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 C 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 B cos A cos B cos C cos C sin - C + sin Ccos - C + sin B cos A cos B cos C cos C sin C + sin C -cos C + sin C sin B cos A cos B cos C sin B sin C cos A cos B cos C tn A tn B tn C 7. cos A cos B cos C- sin B cos C - cos B sin C-cos A sin B sin C cos A(cos B cos C-sin B sin C) -(sin B cos C+cos B sin C) cos A cos (B+C)- sin(b+c) cos(a+b+c) cos 73. This eqution is esier to del with fter rewriting it s cos 5x +sin 5x 0. The left side of this eqution is the exnded form of cos(5x-x), which of course equls ; the grh shown is simly y. The eqution 0 is esily solved on the intervl [, ]: x ; or x ; 3. The originl grh is so crowded tht one cnnot see where crossings occur. [, ] by [.,.] 7. x cos t c cos T b cos d - t cos d cos ( sin d) sin T b + 75. B B in + B ref E 0 x cos t - c b + E 0 x cos t + c c b c E 0 x cos t cos c c + cos t cos x c E 0 c t T + d b sin t T b sin d d + sin t sin x c x - sin t sin c b t T b x cos t cos c b E 0 x cos t cos c c
Section 5. Multile-Angle Identities 5 Section 5. Multile-Angle Identities Exlortion. sin 8 - cos> - > # -. sin. 8 ; - - B We tke the ositive squre root becuse is firstqudrnt ngle. 8 3. sin 9 8 - cos9> - > # - is third- 9. sin. 8 ; - - - C We tke the negtive squre root becuse 8 qudrnt ngle. Quick Review 5.. tn x when x +n, n n integer. tn x when x - +n, n n integer 3. Either 0 or. The ltter imlies the former, so x, n n integer. + n. Either 0 or. The ltter imlies the former, so xn, n n integer. 5. when x - +n, n n integer 6. when x +n, n n integer 7. Either or -. Then x or 6 + n 5 x or x ;, n n integer. 6 + n 3 + n 3 8. Either or. Then x + n or x ;, n n integer. + n 9. The trezoid cn be viewed s rectngle nd two tringles; the re is then A()(3)+ ()(3)+ ()(3) 0.5 squre units. 0. View the tringle s two right tringles with hyotenuse 3, one leg, nd the other leg the height equl to 3-8 Section 5. Exercises. cos ucos(u+u)cos u cos u-sin u sin u cos u-sin u. Strting with the result of #: cos ucos u-sin u cos u-(-cos u) cos u- 3. Strting with the result of #: cos ucos u-sin u (-sin u)-sin u- sin u tn u + tn u. tn utn(u+u) - tn u tn u tn u - tn u 5. - 0, so (-)0; 0 or when x0 or x. 6. -0 ( -)0, So 0 or when x0,,. 3, or 5 3 7. sin x+-0, so ( -)(+) 0; or when x, 6 5 3 x or x. 6 8. cos x--0, so ( +)(-) 0; - or when x0, x or x 3 3 9. -, so, or - 0 0 0. Then 0 or 0 3 (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 - 5 b.370 or x-cos - 5 b.06 For #, ny one of the lst severl exressions given is n nswer to the question. In some cses, other nswers re ossible, s well.. sin +cos sin cos +cos (cos )( sin +). sin +cos sin cos +cos -sin sin cos + cos - sin cos +- sin 3. sin +cos 3 sin cos +cos cos -sin sin sin cos +(cos -sin ) cos - sin cos sin cos +cos 3-3 sin cos sin cos + cos 3-3 cos
6 Chter 5 Anlytic Trigonometry. sin 3 +cos sin cos +cos sin +cos -sin sin cos +(cos -sin ) sin +cos -sin 3 sin cos -sin 3 +cos -sin 5. sin (x) 6. cos 6xcos (3x) cos 3x- 7. csc x # csc x tn x sin x 8. cot x tn x tn x - tn x tn x tn x - cot x-tn x 9. sin 3x + cos x +( cos x-) ()( cos x-) 0. sin 3x + cos x+(- sin x) ()( cos x+- sin x) ()(3- sin x). cos (x)- sin x -( ) -8 sin x cos x. sin (x) ( )( cos x-) ( )( cos x-) 3. cos x+-0, so or, 5 x, x or x 3 3. +- sin x+0, so 7 or -, x, x, or x. 6 6 5. cos 3x - (- sin x) -( ) - sin x - sin x - sin x Thus the left side cn be written s ()(- sin x). This equls 0 in [0, ) when x x 3, x 5, x 3, or x 7, x,. 6. Using #9, this become cos x0, so x0, x., x, or x 3 7. ++ ()(+ )0. Then 0 or - ; the solutions in [0, ) re x0, x, 3, x x. 3, x, x 3, x 3, or x 5 3 8. With ux, this becomes cos u+cos u0, the sme s #3. This mens u 3, u, u 5 3, etc. i.e., x. Then x 6, x, x 5 3 + n 3 6, x 7. 6, x 3, x 6 9. Using results from #5, -cos 3x ( )-(- sin x ) ()( sin x+ -)0. 3 0 when x or x, while the second - ; 5 fctor equls zero when. It turns out s cn be observed by noting, e.g., tht - + 5 sin b 0.3596 tht this mens x0., x0.9, x.3, or x.7. 30. Using #, the left side cn be rewritten s 3 cos x-sin 3 x+cos x-sin x. Relcing cos x with -sin x gives sin 3 x- sin x+3 + (+)( sin x+ +). This equls 0 when x 3, nd b ; 5 when. These vlues turn out to be x0.3, x0.7, x., nd x.9, s cn be observed by noting, e.g., tht sin - 5 b 0.3596. - cos 30 3. sin 5 B - 3 B b - 3. Since sin 5 7 0, tke the ositive squre root. - cos 390 3. tn 95-3> - 3. Note sin 390 > tht tn 95 tn 5. + cos 50 33. cos 75 C - 3 C b - 3. Since cos 75 7 0, tke the ositive squre root. 5 - cos 5>6 3. sin C + 3 C b 5 + 3. Since sin 7 0, tke the ositive squre root. 7 35. tn. - cos7>6 + 3> - - 3 sin7>6 -> + cos> 36. cos C + 8 C b 3 +. Since cos 7 0, tke the ositive 8 squre root.
Section 5. Multile-Angle Identities 7 37. () Strting from the right side: - cos u ( sin u)sin u. 3 - - sin u (b) Strting from the right side: + cos u cos u. cos u 3 + cos u - 38. () tn sin u u cos u - cos u> - cos u + cos u> + cos u (b) The eqution is flse when tn u is negtive number. It would be n identity if it were written s - cos u tn u. B + cos u 39. sin x(sin x) c - d - + c - + + d - + + 8 (3- +) 8 0. cos 3 x cos x # + +. sin 3 x sin x # - -. sin 5 x()(sin x) () c - d - + c - + + d - + + 8 3 - +. 8 Alterntively, tke sin 5 x sin x nd ly the result of #39. - 3. cos x, so cos x+-0. Then or. In the intervl [0, ), x, 3 5 x, or x. Generl solution: +n or 3 3 x+n, n n integer.. -cos + x, so cos x+-0. Then or. In the intervl [0, ), 5 x, x, orx. Generl solution: 3 3 x +n or x+n, n n integer. 3 5. The right side equls tn (x/); the only wy tht tn(x/) tn (x/) is if either tn(x/)0 or tn(x/). In [0, ), this hens when x0 or x. The generl solution is xn or x n, n n integer. + - 6. cos x -, so cos x+-30, or ( -3)(+)0. Then or 3. Let Åcos 3 0.77. In the intervl b [0, ), xå, x, or x-å. Generl solution: x_å+n or x+n, n n integer. 7. Flse. For exmle, f(x) hs eriod nd g(x) hs eriod, but the roduct f(x)g(x) hs eriod. 8. True. cos x sin - x - bb + The lst exression is in the form for sinusoid. 9. f(x) f(x)g(x). The nswer is D. 50. + + sin.5 sin 5 b sin - x b + C - cos 5 C - > C - - The nswer is E. 5. or 0 or 0 x 6 or 5 x 6 or 3 The nswer is E.
8 Chter 5 Anlytic Trigonometry 5. sin x-cos x- cos x, which hs the sme eriod s the function cos x, nmely. The nswer is C. 53. () In the figure, the tringle with side lengths x/ nd R is right tringle, since R is given s the erendiculr distnce. Then the tngent of the ngle / is the rtio oosite over djcent : tn Solving for x x> R gives the desired eqution. The centrl ngle is /n since one full revolution of rdins is divided evenly into n sections. u (b) 5.87 R tn, where /, so R 5.87/( tn ) 9.9957. R0. 5. () d A x D x E x Cll the center of the rhombus E. Consider right ABE, with legs d / nd d /, nd hyotenuse length x. jabe hs mesure /, nd using sine o dj equls nd cosine equls, we hve hy hy nd sin. d > d cos d > d x x x x (b) Use the double ngle formul for the sine function: sin sin b sin cos d # d x x d d x 55. () B x C d ft θ θ ft ft The volume is 0 ft times the re of the end. The end is mde u of two identicl tringles, with re (sin ) (cos ) ech, nd rectngle with re () (cos ). The totl volume is then 0 # (sin cos +cos )0 (cos )(+sin ). Considering only -, the mximum vlue occurs when 0.5 (in fct, it hens exctly t ). The mximum vlue is bout.99 ft 3. 6 56. () x y 00 (x, y) x y The height of the tunnel is y, nd the width is x, so the re is xy. The x- nd y-coordintes of the vertex re 0 cos nd 0 sin, so the re is (0 cos )(0 sin )00( cos sin )00 sin. (b) Considering 0, the mximum re occurs when, or bout 0.79. This gives x0 cos, or bout., for width of 0 bout 8.8, nd height of y 0 L. 57. csc u sin u sin u cos u # # sin u cos u csc u sec u 58. cot u tn u - tn u tn u - tn u tn u b cot u cot u b cot u - cot u 59. sec u cos u - sin u u - sin u bcsc csc u b csc u csc u - 60. sec u cos u cos u - u cos u - bsec sec u b sec u - sec u 6. sec u cos u cos u - sin u u csc u cos u - sin u bsec sec u csc u b sec u csc u csc u - sec u 6. The second eqution cnnot work for ny vlues of x for which 6 0, since the squre root cnnot be negtive. The first is correct since double ngle identity for the cosine gives - sin x; solving for gives sin x -, so tht -. The bsolute vlue of both A sides removes the _. 63. () The following is sctter lot of the dys st Jnury s x-coordintes (L) nd the time (in hour mode) s y-coordintes (L) for the time of dy tht stronomicl twilight begn in northestern Mli in 005. [ 30, 370] by [ 60, 60]
Section 5.5 The Lw of Sines 9 (b) The sine regression curve through the oints defined by L nd L is y.656 sin(0.05x-0.85)-.73. This is firly good fit, but not relly s good s one might exect from dt generted by sinusoidl hysicl model. [ 30, 370] by [ 60, 60] (c) Using the formul L-Y(L) (where Y is the sine regression curve), the residul list is: {3.6, 7.56, 3.35, 5.9, 9.35, 3.90, 5., 9.3, 3.90,.57, 9.7, 3.}. (d) The following is sctter lot of the dys st Jnury s x-coordintes (L) nd the residuls (the difference between the ctul number of minutes (L) nd the number of minutes redicted by the regression curve (Y)) s y-coordintes (L3) for the time of dy tht stronomicl twilight begn in northestern Mli in 005. The sine regression curve through the oints defined by L nd L3 is y8.856 sin(0.036x+ 0.576)-0.33. (Note: Round L3 to deciml lces to obtin this nswer.) This is nother firly good fit, which indictes tht the residuls re not due to chnce. There is eriodic vrition tht is most robbly due to hysicl cuses. [ 30, 370] by [ 5, 5] (e) The first regression indictes tht the dt re eriodic nd nerly sinusoidl. The second regression indictes tht the vrition of the dt round the redicted vlues is lso eriodic nd nerly sinusoidl. Periodic vrition round eriodic models is redictble consequence of bodies orbiting bodies, but ncient stronomers hd difficult time reconciling the dt with their simler models of the universe. Section 5.5 The Lw of Sines Exlortion. If BC AB, the segment will not rech from oint B to the dotted line. On the other hnd, if BC 7 AB, then circle of rdius BC will intersect the dotted line in unique oint. (Note tht the line only extends to the left of oint A.). A circle of rdius BC will be tngent to the dotted line t C if BCh, thus determining unique tringle. It will miss the dotted line entirely if BC 6 h, thus determining zero tringles. 3. The second oint (C ) is the reflection of the first oint (C ) on the other side of the ltitude.. sin C sin (-C )sin cos C -cos sin C sin C. 5. If BC AB, then BC cn only extend to the right of the ltitude, thus determining unique tringle. Quick Review 5.5. bc/d. bd/c 3. cd/b. dbc/ 7 sin 8 5. L 3.3 sin 3 9 sin 6. L 3.888 sin 7. xsin 0.3 7.58 8. x80 -sin 0.3 6.5 9. x80 -sin ( 0.7).7 0. x360 +sin ( 0.7) 35.573 Section 5.5 Exercises. Given: b3.7, B5, A60 n AAS cse. C80 -(A+B)75 ; b sin B b sin B c sin C b 3.7 sin 60 L.5; sin B sin 5. Given: c7, B5, C0 n AAS cse. A80 -(B+C)5 ; c 7 sin 5 L 3.9; c sin C sin C sin 0 b b c sin B 7 sin 5 L 5. sin B c sin C sin C sin 0 3. Given: A00, C35, n AAS cse. B80 -(A+C)5 ; L 5.8;. Given: A8, B0, b9 n AAS cse. C80 -(A+B)59 ; b sin B c b sin C sin B L.; 5. Given: A0, B30, b0 n AAS cse. C80 -(A+B)0 ; b sin B c b sin C sin B L.9; 6. Given: A50, B6, n AAS cse. C80 -(A+B)68 ; b sin B c sin C b sin B c sin C sin 5 sin 00 sin 35 sin 00 9 sin 8 sin 0 9 sin 59 sin 0 0 sin 0 sin 30 0 sin 0 sin 30 sin 6 sin 50 sin 68 sin 50 c b sin C sin B L.8 L.7 L 8.8 L.6; L.8 3.7 sin 75 sin 5 L 5.
0 Chter 5 Anlytic Trigonometry 7. Given: A33, B70, b7 n AAS cse. C80 -(A+B)77 ; b sin B c b sin C sin B L.; 8. Given: B6, C03, c n AAS cse. A80 -(B+C)6 ; c sin C b c sin B sin C 7 sin 33 sin 70 7 sin 77 sin 70 sin 6 sin 03 sin 6 sin 03 L 7.3 L 0.8; L 3. 9. Given: A3, 7, b n SSA cse. hb 5.8; h<b<, so there is one tringle. B sin - b b sin - 0.3 L 0. C80 -(A+B) 7.9 ; c sin C 7 sin 7.9 L 5.3 sin 3 0. Given: A9, 3, b8 n SSA cse. hb.; h<b<, so there is one tringle. B sin - b b sin - 0.660 L.3 C80 -(A+B)89.7 ; c sin C 3 sin 89.7 L. sin 9. Given: B70, b, c9 n SSA cse. hc sin B 8.5; h<c<b, so there is one tringle. C sin - c sin B b sin - 0.60 L 37. b A80 -(B+C) 7.8 ; b sin 7.8 L. sin B sin 70. Given: C03, b6, c6 n SSA cse. hb sin C.8; h<b<c, so there is one tringle. B sin - b sin C b sin - 0.73 L 7.3 c A80 -(B+C)9.7 ; c 6 sin 9.7 L 3.0 sin C sin 03 3. Given: A36,, b7. hb.; <h, so no tringle is formed.. Given: B8, b7, c5. hc sin B.9; h<c<b, so there is one tringle. 5. Given: C36, 7, c6. h sin C 0.0; h<c<, so there re two tringles. 6. Given: A73,, b8. hb 6.8; <h, so no tringle is formed. 7. Given: C30, 8, c9. h sin C9; hc, so there is one tringle. 8. Given: B88, b, c6. hc sin B 6.0; b<h, so no tringle is formed. 9. Given: A6, 6, b7. hb 5.3; h<<b, so there re two tringles. B sin - b b sin - 0.95 L 7.7 C 80 -(A+B ) 3.3 ; c sin C 6 sin 3.3 L. sin 6 Or (with B obtuse): B 80 -B 07.3 ; C 80 -(A+B ) 8.7 ; c sin C L.7 0. Given: B38, b, c5. hc sin B 5.; h<b<c, so there re two tringles. C sin - c sin B b sin - 0.73 L 7. b A 80 -(B+C ) 9.9 ; b sin 9.9 L 3.0 sin B sin 38 Or (with C obtuse): C 80 -C 3.9 ; A 80 -(B+C ) 9. ; b sin B L 5.. Given: C68, 9, c8. h sin C 7.6; h<c<, so there re two tringles. A sin - sin C b sin - 0.978 L 78. c B 80 -(A+C) 33.8 ; b c sin B 8 sin 33.8 L 0.8 sin C sin 68 Or (with A obtuse): A 80 -A 0.8 ; B 80 -(A +C) 0. ; b c sin B sin C L 3.. Given: B57,, b0. h sin B 9.; h<b<, so there re two tringles. A sin - sin B b sin - 0.9 L 67.3 b C 80 -(A +B) 55.7 ; c b sin C 0 sin 55.7 L 9.9 sin B sin 57 Or (with A obtuse): A 80 -A.7 ; C 80 -(A +B) 0.3 ; c b sin C sin B L. 3. h0 sin 6.69, so: () 6.69 6 b 6 0. (b) b 6.69 or b 0. (c) b 6 6.69. h sin 53 9.58, so: () 9.58 6 c 6. (b) c 9.58 or c. (c) c 6 9.58 5. () No: this is n SAS cse (b) No: only two ieces of informtion given.
Section 5.5 The Lw of Sines 6. () Yes: this is n AAS cse. B80 -(A+C)3 ; sin B 8 sin 3 b L 88.5; sin 9 sin C 8 sin 9 c L 6. sin 9 (b) No: this is n SAS cse. 7. Given: A6, 8, b n SSA cse. hb 8.; <h, so no tringle is formed. 8. Given: B7, 8, b n SSA cse. h sin B 5.9; h<<b, so there is one tringle. A sin - sin B b sin - 0.78 L 6. b C80 -(A+B)6.8 ; c b sin C sin 6.8 L 5.6 sin B sin 7 9. Given: A36, 5, b8 n SSA cse. hb 9.5; <h, so no tringle is formed. 30. Given: C5, b, c7 n SSA cse. hb sin C 0.9; c<h, so no tringle is formed. 3. Given: B, c8, C39 n AAS cse. A80 -(B+C)99 ; c sin C b c sin B sin C L 8.3; 3. Given: A9, b, B7 n AAS cse. C80 -(A+B) ; b sin B c b sin C sin B 8 sin 99 sin 39 8 sin sin 39 sin 9 sin 7 sin sin 7 L 9.8; 33. Given: C75, b9, c8. n SSA cse. hb sin C 7.3; h<c<b, so there re two tringles. B sin - b sin C b sin - 0.986 L 80. c A 80 -(B+C).6 ; c 8 sin.6 L 0.7 sin C sin 75 Or (with B obtuse): B 80 -B 99.6 ; A 80 -(B +C) 5. ; c sin C L.7 3. Given: A5, 3, b5. n SSA cse. hb.; h<<b, so there re two tringles. B sin - b b sin - 0.933 L 69.0 C 80 -(A+B ) 57.0 ; c sin C 3 sin 57.0 L 3.5 sin 5 Or (with B obtuse): B 80 -B.0 ; C 80 -(A+B ) 5.0 ; c sin C L. L 9. L 7.5 35. Cnnot be solved by lw of sines (n SAS cse). 36. Cnnot be solved by lw of sines (n SAS cse). 37. Given: cab56, A7, B53 n ASA cse, so C80 -(A+B)55 c sin B 56 sin 53 () ACb L 5.6 ft. sin C in 55 (b) hb ( sin B) 5.9 ft. 38. Given: c5, A90-38 5, B90-53 37 n ASA cse, so C80 -(A+B)9 nd c 5 sin 5 L 9.7 mi. sin C sin 9 c sin B 5 sin 37 b L 5.0 mi, sin C sin 9 nd finlly hb sin B.9 mi. 39. Given: c6, C90-68, B90 +5 05 n AAS cse. A80 -(B+C)7, so c 6 sin 7 L.9 ft. sin C sin 8 0. Given: c.3, A8, B37 n ASA cse. C80 -(A+B)5 ; c.3 sin 8 L. mi. sin C sin 5 c sin B.3 sin 37 b L.5 mi. sin C sin 5 Therefore, the ltitude is hb (.5) sin 8 0.7 mi or sin B (.) sin 37 mi 0.7 mi... ft 8 0 The length of the brce is the leg of the lrger tringle. sin 8 x, so x.9 ft. 78.75 B.5 5.5 ft x C The center of the wheel (A) nd two djcent chirs 360 (B nd C) form tringle with 5.5, A 6.5, nd BC78.75. This is n ASA cse, so sin B 5.5 sin 78.75 the rdius is bc L 39.7 ft. sin.5 Alterntively, let D be the midoint of BC, nd consider right ^ABD, with mjbad.5 nd BD7.75 ft; then r is the hyotenuse of this tringle, so 7.75 r L 39.7 ft. sin.5
Chter 5 Anlytic Trigonometry 3. Consider the tringle with vertices t the to of the flgole (A) nd the two observers (B nd C). Then 600, B9, nd C (n ASA cse), so A80 -(B+C)0 ; sin B 600 sin 9 b L 303.9; sin 0 sin C 600 sin c L 33.5 sin 0 nd finlly hb sin Cc sin B 08.9 ft.. Consider the tringle with vertices t the to of the tree (A) nd the two observers (B nd C). Then 00, B5, nd C0 (n ASA cse), so A80 -(B+C)5 ; sin B 00 sin 5 b L 80.5; sin 5 sin C 00 sin 0 c L 38.5; sin 5 nd finlly hb sin Cc sin B 6.7 ft. 5. Given: c0, B5, C33 n AAS cse. A80 -(B+C)95, so c 0 sin 95 L 36.6 mi, nd sin C sin 33 c sin B 0 sin 5 b L 8.9 mi. sin C sin 33 6. We use the men (verge) mesurements for A, B, nd AB, which re 79.7, 83.9, nd 5.9 feet, resectively. This gives 6. for ngle C. By the Lw of Sines, 5.9 sin 83.9 AC L 9. feet. sin 6. 7. True. By the lw of sines, sin B, b which is equivlent to (since, sin B Z 0). sin B b 8. Flse. By the lw of sines, the third side of the tringle 0 sin 00 mesures, which is bout 5.3 inches. Tht sin 0 mkes the erimeter bout 0+0+5.335.3, which is less thn 36 inches. 9. The third ngle is 3. By the Lw of Sines, sin 3 sin 53, which cn be solved for x..0 x The nswer is C. 50. With SSA, the known side oosite the known ngle sometimes hs two different ossible ositions. The nswer is D. 5. The longest side is oosite the lrgest ngle, while the shortest side is oosite the smllest ngle. By the Lw of sin 50 sin 70 Sines,, which cn be solved for x. 9.0 x The nswer is A. 5. Becuse BC>AB, only one tringle is ossible. The nswer is B. 53. () Given ny tringle with side lengths, b, nd c, the lw of sines sys tht sin B sin C. b c But we cn lso find nother tringle(using ASA) with two ngles the sme s the first (in which cse the third ngle is lso the sme) nd different side length sy,. Suose tht k for some constnt k. Then for this new tringle, we hve sin B sin C. Since b c k # sin B, we cn see tht # sin B, k b k b so tht b kb nd similrly, c kc. So for ny choice of ositive constnt k, we cn crete tringle with ngles A, B, nd C. (b) Possible nswers:, b 3, c (or ny set of three numbers roortionl to these). (c) Any set of three identicl numbers. 5. In ech roof, ssume tht sides, b, nd c re oosite ngles A, B, nd C, nd tht c is the hyotenuse. sin 90 () c c c o hy sin B sin 90 (b) b c cos> - B b c b cos A c dj hy (c) sin B b sin B b cos A b tn A b o dj 55. () hab (b) BC 6 AB (c) BC AB or BCAB (d) AB 6 BC 6 AB 56. Drwing the line suggested in the hint, nd extending to meet tht line t, sy, D, gives right ^ADC nd right ^ADB. A 8 C 5 B D BC Then AD8 sin 3.0 nd DC8 cos 7., so DBDC-5 nd cab AD + DB L 3.9. Finlly, A(90 - )-sin DB nd AB b L 9. B80 -A-C 8.9.
Section 5.6 The Lw of Cosines 3 57. Given: c., B5, C36.5-5.5. An AAS cse: A80 -(B+C)3.5, so c sin B. sin 5 ACb L 8.7 mi, nd sin C sin.5 c. sin 3.5 BC L. mi. sin C sin.5 The height is h sin 5 b sin 36.5 5. mi. Section 5.6 The Lw of Cosines Exlortion. The semierimeters re 5 nd 50. A 55-55 - 85 - + 5050-50 - 050-86 875.788 ces.,0.595 squre feet 3. 0.00783 squre miles. 0.975 cres 5. The estimte of little over n cre seems questionble, but the roughness of their mesurement system does not rovide firm evidence tht it is incorrect. If Jim nd Brbr wish to mke n issue of it with the owner, they would be well-dvised to get some more relible dt. 6. Yes. In fct, ny olygonl region cn be subdivided into tringles. Quick Review 5.6. Acos 3 53.30 5 b. Ccos ( 0.3) 03.97 3. Acos ( 0.68) 3.8. Ccos.9 50.08 3 b 5. () (b) Acos 6. () cos A 8 - x - y -xy (b) Acos x + y - 8 b xy cos A y - x - 5-0 x - y + 5 b 0 x + y - 8. xy x - y + 5 0 7. One nswer: (x-)(x-)x -3x+. Generlly: (x-)(x-b)x -(+b)x+b for ny two ositive numbers nd b. 8. One nswer: (x-)(x+)x -. Generlly, (x-)(x+b)x -(-b)x-b for ny two ositive numbers nd b. 9. One nswer: (x-i)(x+i)x + 0. One nswer: (x-) x -x+. Generlly: (x-) x -x+ for ny ositive number. Section 5.6 Exercises. Given: B3, c8, 3 n SAS cse. b + c - c cos B L 369.60 L 9.; C cos - + b - c b L cos - 0.99 L 8.3 ; b A 80 - B + C L 30.7.. Given: C, b, n SAS cse. c + b - b cos C L 90.303 L 9.5; A cos - b + c - b L cos - 0.67 L 80.3 ; bc B 80 - A + C L 57.7. 3. Given: 7, b9, c n SSS cse. A cos - b + c - b L cos - 0.8 L 76.8 ; bc B cos - + c - b b L cos - 0.78 L 3. ; c C 80 - A + B L 60.. Given: 8, b35, c7 n SSS cse. A cos - b + c - b L cos - 0.63 L 5. ; bc B cos - + c - b b L cos - 0.59 L 99. ; c C 80 - A + B L 8.6. 5. Given: A55, b, c7 n SAS cse. b + c - bc cos A L 96.639 L 9.8; B cos - + c - b b L cos - 0.0 L 89.3 ; c C 80 - A + B L 35.7. 6. Given: B35, 3, c9 n SAS cse. b + c - c cos B L 87.505 L 9.5; C cos - + b - c b L cos - 0.99 L.7 ; b A 80 - B + C L 3.3. 7. Given:, b, C95 n SAS cse. c + b - b cos C L 68.96 L 5.; A cos - b + c - b L cos - 0.879 L 8.5 ; bc B 80 - A + C L 56.5. 8. Given: b, c3, A8 n SAS cse. b + c - bc cos A L 55.67 L 35.; B cos - + c - b b L cos - 0.788 L 37.9 ; c C 80 - A + B L 60.. 9. No tringles ossible (+cb) 0. No tringles ossible (+b<c). Given: 3., b7.6, c6. n SSS cse. A cos - b + c - b L cos - 0.909 L.6 ; bc B cos - + c - b b L cos - 0.60 L 99. ; c C 80 - A + B L 56..
Chter 5 Anlytic Trigonometry. No tringles ossible (+b<c) Exercises 3 6 re SSA cses, nd cn be solved with either the Lw of Sines or the Lw of Cosines. The lw of cosines solution is shown. 3. Given: A, 7, b0 n SSA cse. Solve the qudrtic eqution 7 0 +c -(0)c cos, or c -(.86 )c+50; there re two ositive + c - b solutions: L9.87 or 5.376. Since cos B : c c 9.87, B cos (0.9) 7.9, nd C 80 -(A+B ) 65., or c 5.376, B cos ( 0.9) 07., nd C 80 -(A+B ) 30.9.. Given: A57,, b0 n SSA cse. Solve the qudrtic eqution 0 +c -(0)C cos 57, or c -(0.893)c-0; there is one ositive + c - b solution c.56. Since cos B, c B cos (0.67) 9.7 nd C80 -(A+B) 73.3. 5. Given: A63, 8.6, b. n SSA cse. Solve the qudrtic eqution 8.6. +c -(.)c cos 63, or c -(0.079)c+9.50; there re no rel solutions, so there is no tringle. 6. Given: A7, 9.3, b8.5 n SSA cse. Solve the qudrtic eqution 9.3 8.5 +c -(8.5)c cos 7, or c -(5.535)c-.0; there is one ositive + c - b solution: c 7.7. Since cos B, c B cos (0.503) 59.8 nd C80 -(A+B) 9.. 7. Given: A7, b3, c9 n SAS cse. b + c - bc cos A L 555.689 L 3.573, so Are L 93.307 L.33 ft (using Heron s formul). Or, use A bc. 8. Given: A5, b, c n SAS cse. b + c - bc cos A L 7.99 L 6.583, so Are L 38.35 L 5.8 m (using Heron s formul). Or, use A bc. 9. Given: B0, 0, c n SAS cse. b + c - c cos B L 667.955 L 5.85, so Are L 659.6 L 07.98 cm (using Heron s formul). Or, use A c sin B. 0. Given: C,.8, b5. n SAS cse. c + b - b cos C L 36.8 L 6.0, so Are L 8. L.6 in. (using Heron s formul). Or, use A b sin C. For # 8, tringle cn be formed if +b<c, +c<b, nd b+c<. 7. s ; Are 66.9375 L 8.8. s ; Are 303.875 L 7. 3. No tringle is formed (+bc).. s7; Are,960 360 L 3.8 5. 36.; Are 6,70.36 L 6.5 6. No tringle is formed (+b<c) 7. s.; Are 98,69.856 L 3.05 8. s3.8; Are0,69. L 0.3 9. Let, b5, nd c6. The lrgest ngle is oosite the lrgest side, so we cll it C. Since cos C + b - c, C cos - b 8 b L 8.89.5 rdins. 30. The shorter digonl slits the rllelogrm into two (congruent) tringles with 6, B39, nd c8. The digonl hs length b + c - c cos B 7.59 L 6.5 ft. 3. Following the method of Exmle 3, divide the hexgon into 6 tringles. Ech hs two -inch sides tht form 60 ngle. 3. Following the method of Exmle 3, divide the nongon into 9 tringles. Ech hs two 0-inch sides tht form 0 ngle. 33. 6 * sin 60 63 L 37. squre inches 9 * 00sin 0 L 89.3 squre inches 30 s s In the figure, nd so s sec 30 83. The re of the hexgon is 6 * 8383sin 60 883 L 98.8 squre inches. 3. 0 s s In the figure, 0 nd so s0 sec 0. The re of the nongon is 9 * 0 sec 0 0 sec 0 sin 0 L 37.6 squre inches.
Section 5.6 The Lw of Cosines 5 35. Given: C5, BC60, ACb0 n SAS cse. AB c + b - b cos C 7,009.959 L 30. ft. 36. () The home-to-second segment is the hyotenuse of right tringle, so the distnce from the itcher s rubber to second bse is 90-60.5 L 66.8 ft. This is bit more thn c 60.5 + 90-60.590 cos 5 059.857 L 63.7 ft. (b) B cos - + c - b b L cos - -0.09 c 9.8. 37. () c 0 + 60-060 cos 5 805.887 L.5 ft. (b) The home-to-second segment is the hyotenuse of right tringle, so the distnce from the itcher s rubber to second bse is 60-0 L.9 ft. (c) B cos - + c - b b L cos - -0.057 c 93.3. 38. Given: 75, b860, nd C78. An SAS cse, so ABc + b - b cos C L 707,63.58 8. ft. 39. () Using right ACE, mjcae tn - 6 8 b tn - 3 b L 8.35. (b) Using A L 8.35, we hve n SAS cse, so DF 9 + - 9 cos A L 0.08.5 ft. (c) EF 8 + - 8 cos A L 58.68 7.6 ft. 0. After two hours, the lnes hve trveled 700 nd 760 miles, nd the ngle between them is.5, so the distnce is 700 + 760-700760 cos.5 8,59.77 L 90.8 mi.. AB 73 + 65-7365 cos 8 56.356 L.5 yd.. mjhab 35,so HB 0 + 0-00 cos 35 365.685 L 37.0 ft. Note tht AB is the hyotenuse of n equilterl right 0 tringle with leg length, nd HC is the 0 hyotenuse of n equilterl right tringle with leg length 0 + 0, so HC 0 + 0 L 8.3 ft. Finlly, using right HAD with leg lengths HA 0 ft nd AD HC L 8.3 ft, we hve HD HA + AD L 5.3 ft. 3. AB c + 3 3, AC b + 3 0, nd BC + 5, so A cos - b + c - mjcab b bc 9 cos - 37.9. 30 b. ABC is right tringle (C90 ), with BC + nd ACb, so ABc nd Bm ABCsin + b 3 j 3 b 9.5. 5. True. By the Lw of Cosines, b +c -bc cos A, which is ositive number. Since b +c -bc cos A>0, it follows tht b +c >bc cos A. 6. True. The digonl oosite ngle slits the rllelogrm into two congruent tringles, ech with re b sin. θ b 7. Following the method of Exmle 3, divide the dodecgon into tringles. Ech hs two -inch sides tht form 30 ngle. * sin 30 3 The nswer is B. 8. The semierimeter is s(7+8+9)/. Then by Heron s Formul, A - 7-8 - 9 5. The nswer is B. 9. After 30 minutes, the first bot hs trveled miles nd the second hs trveled 6 miles. By the Lw of Cosines, the two bots re + 6-6 cos 0 L 3.05 miles rt. The nswer is C. 50. By the Lw of Cosines, 7 +5 -(7)(5) cos, so cos 7 + 5-5.06. 75 The nswer is E. 5. Consider tht n-sided regulr olygon inscribed within circle cn divide into n equilterl tringles, ech with r 360 equl re of sin. (The two equl sides of the n equilterl tringle re of length r, the rdius of the circle.) Then, the re of the olygon is exctly nr 360 sin. n b + c - 5. () b + c - b + c - bc cos A bc bc Lw of Cosines bc cos A bc cos A (b) The identity in () hs two other equivlent forms: cos B + c - b b bc cos C + b - c c bc
6 Chter 5 Anlytic Trigonometry 30. - 5. 53. () Shi A: 5. knots; hr 37. -. Shi B:. knots hrs (b) We use them ll in the roof: cos A + cos B + cos C b c b + c - bc cos A b + c - bc 5. +. - 8.7 5.. A 35.8 (c) c +b -b cos C (9.6) +(60.) -(9.6)(60.) cos (35.8 ).0, so the bots re 3.8 nuticl miles rt t noon. 5. Use the re formul nd the Lw of Sines: A b sin C sin B b sin C sin B sin C sin sin B qlw of Sines b r 55. Let P be the center of the circle. Then, cos P 5 + 5-7 0.0, so P 88.9. The re 55 of the segment is The re of the tringle, however, is 55 sin 88.9.50 in, so the re of the shded region is rox. 6.9 in. Chter 5 Review. sin 00 cos 00 sin 00 + + c - b bc + + b - c bc b + c - + + c - b + + b - c bc + b + c bc r # 88.9 360 L 5 # 0.7 L 9.39 in. tn 0. tn 80 - tn 0 3. ; the exression simlifies to (cos ) +( sin cos ) (cos ) +(sin ).. cos x; the exression cn be rewritten -( ) -() cos x. 5. cos 3xcos(x+x) - (cos x-sin x) -( ) cos 3 x-3 sin x cos 3 x-3(-cos x) cos 3 x-3 +3 cos 3 x cos 3 x-3 6. cos x-cos x(-sin x)-(-sin x) sin x-sin x 7. tn x-sin xsin x - b cos x sin x # sin x tn x cos x 8. sin cos 3 + sin 3 cos ( sin cos )(cos +sin ) ( sin cos )()sin. 9. csc x- cot x - # - cos x tn u + sin u 0. + cos u + cos u ; b tn u A cos u b + tn u. Recll tht tn cot. - tn u + + cot u - cot u + tn u - cot u + + cot u - tn u - tn u - cot u + tn u - cot u - + + cot u - tn u - - tn u - cot u 0 0 - tn u - cot u. sin 3 sin( + )sin cos +cos sin sin cos +(cos -sin ) sin 3 sin cos -sin 3 3. cos t c; B + cos t d + cos t + cos t b sec t sec t b + sec t sec t tn 3 g - cot 3 g. tn g + csc g 5. tn g - cot gtn g + tn g cot g + cot g tn g + csc g tn g - cot gtn g + + cot g tn g + csc g tn g - cot gtn g + csc g tn g - cot g tn g + csc g cos f - tn f + sin f - cot f cos f - tn f bcos f cos f b + sin f - cot f bsin f sin f b cos f sin f sin f - cos f cos f - sin f cos f - sin f + cos f - sin f cos f + sin f cos -z cos -z 6. sec -z + tn -z 3 + sin -z>cos -z cos -z cos z + sin -z - sin z - sin z - sin z + sin z