Chapter 5 Analytic Trigonometry

Transcription

1 Section 5. Fundamental Identities 5 Chater 5 Analytic Trigonometry Section 5. Fundamental Identities Exloration. cos / sec, sec / cos, and tan sin / cos. sin / csc and tan / cot 3. csc / sin, cot / tan, and cot cos / sin Quick Review 5. For #, use a calculator rad rad a - ab + b (a - b) 7. x - 3xy - y (x + y)(x - y) 9.. # y x y - # x y x y - x xy x + y x + y (x + y) # a xy x + y b xy Section 5. Exercises. sec u + tan u + (3/) 5/6, so sec u ;5/. Then cos u /sec u ;/5. But sin u, tan u 7 0 imlies cos u 7 0. So cos u /5. Finally, tan u 3 sin u cos u 3 sin u 3 cos u 3 a 5 b tan u sec u - - 5, so tan u ;5. But sec u 7 0, sin u 6 0 imlies tan u 6 0, so tan u -5. Finally cot u /tan u -/5-5/5. 5. cos( /- )sin cos( )cos sin( /- ) sin( - /) tan x #. sec y sin a - yb # cos y cos y + tan x sec x /cos x sin x 3. cos x tan x csc x csc x /sin x 5. - cos 3 x ( - cos x) sin x 7. csc( x) # sin (-x) 9. cot( x) cota cos (-x) sin (x) # sin (-x) cos (x). sin (-x) + cos (-x) cos (-x) - xb # cos a - xb sin (-x) sin a - xb tan a - xb csc x cot x 3. csc x # csc x 5. (sec x + csc x) - (tan x + cot x) (sec x - tan x) + (csc x - cot x) + 7. ()(tan x+cot x)() a + b a + cos x sec x ()() b 9. ()()(tan x)(sec x)(csc x) ()() a ba ba b tan x tan x 3. csc x + tan x sec x a cos x b() + a b # a b( + cos x) tan x sec x csc x+ sin x tan x cos xa cos x b csc x+ csc x sin x 35. cot x - cos x ()(tan x) - ()(sec x) ()(tan x - sec x) ()(-) - sec x sec x - - cos x cot x 39. cos x + + ( + ). - + ( - cos x) - + sin x ( - ) 3. - sin x cos x + cos x + - ( - )( + ) 5. tan x - + csc x cot x # tan x - tan x + tan x - tan x + ( tan x - ) Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

2 6 Chater 5 Analytic Trigonometry - sin x ( - )( + ) sin x ( - )( + ) ()( - ) 0, so either 0 or. Then x or + n or x 6 + n cot x ; The two functions are arallel to each other, searated by unit for every x. At any x, the distance between the two grahs is sin x - (-cos x) sin x + cos x. x n, n an integer. On the interval: x e 6,, 5 6, 3 f 53. (tan x)(sin x - ) 0, so either tan x 0 or. Then x n or x an integer. + n, n 85. (a) [, ] by [, ] However, tan x excludes x, so we have only + n xn, n an integer. On the interval: x e 0, f 55. tan x ;3, so x ; an integer. 3 + n, n On the interval: x e 3, 3, 3, 5 3 f [ 6, 70] by [0 000, ] (b) The equation is y3, sin(0.997x+.57)+38, ( - ) 0, so therefore ; x ; an integer. 3 + n, n 59. (sin u)(sin u - ) 0, so sin u 0 or sin u. Then u n, n an integer. 6. cos() if n. Only n0 gives a value between and ±, so 0, or xn, n an integer. 63. cos L.98, so the solution set is {;.98+n n0, ;, ;,... }. 65. sin L and , so the solution set is {0.307+n or.8369+n n0, ;, ;,... } L 0.636, and cos L 0.886, so the solution set is {; n n0, ;, ;,... } cos u ƒsin uƒ 7. 9 sec u - 9 3ƒtan uƒ tan u + 8 9ƒsec uƒ 75. True. Since cosine is an even function, so is secant, and thus sec (x- /)sec ( /-x), which equals csc x by one of the cofunction identities. 77. tan x sec xtan x//cos x Z. The answer is D. 79. (sec +)(sec -)sec -tan. The answer is C. 8., ; - sin x, tan x ; - sin x, csc x, sec x; - sin x [ 6, 70] by [0 000, ] (c) ()/0.998 L 7.3 days. This is the number of days that it takes the Moon to make one comlete orbit of the Earth (known as the Moon s sidereal eriod). (d) 5,7 miles (e) y 3, cos ( x) + 38,855, or y 3, cos (0.997x) + 38, Factor the left-hand side: sin u - cos u (sin u - cos u)(sin u + cos u) (sin u - cos u) # sin u - cos u 89. Use the hint: sin ( - x) sin (/ - (x - /)) cos (x - /) Cofunction identity cos (/ - x) Since cos is even Cofunction identity 9. Since A, B, and C are angles of a triangle, A+B -C. So: sin(a+b)sin( -C) sin C Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

3 Section 5. Proving Trigonometric Identities Exloration. The grahs lead us to conclude that this is not an identity. [, ] by [, ]. For examle, cos( # 0), whereas cos(0). 3. Yes.. The grahs lead us to conclude that this is an identity. [, ] by [ 3, 3] 5. No. The grah window can not show the full grahs, so they could differ outside the viewing window. Also, the function values could be so close that the grahs aear to coincide. Quick Review 5.. csc x+sec x + + cos x + sin x 3. # + # 5. sin x+cos x / + / 7. No. (Any negative x.) 9. No. (Any x for which <0, e.g. x /.). Yes. Section 5. Exercises. One ossible roof: x 3 - x - (x - )(x + ) x(x - x) - (x - ) x x x - x - (x - ) -x + - x 3. One ossible roof: x - x - - x - 9 x + 3 (x + )(x - ) (x + 3)(x - 3) - x - x + 3 x + - (x - 3) 5 Section 5. Proving Trigonometric Identities 7 sin x + cos x 5..Yes. csc x csc x 7. # cot x # No.. 9. (sin 3 x)(+cot x)(sin 3 x)(csc sin 3 x x).yes. sin x. ()(tan x+ cot x) # + # +cos x 3. (-tan x) - tan x+tan x (+tan x)- tan xsec x- tan x 5. One ossible roof: ( - cosu)( + cos u) - cos u cos u cos u sin u cos u tan u cos x - -sin x 7. - # tan x 9. Multily out the exression on the left side.. (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 + tan x sec x 3. sec x sin x + cos x cos b cos b - sin b 5. + sin b cos b( + sin b) cos b( + sin b) ( - sin b)( + sin b) - sin b cos b( + sin b) cos b tan x sec x - 7. sec x- - sec x + sec x cot x-cos x a -cos x b cos x( - sin x) sin cos x # x sin x cos x cot x 3. cos x-sin x(cos x+sin x)(cos x-sin x) (cos x-sin x)cos x-sin x 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 tan x tan x(sec x + ) tan x(sec x + ) 35. sec x - sec x - tan x sec x +. See also Exercise 6. tan x - ( - )( + ) ( + ) sin x - cos x sin x - ( - sin x) sin x + + cos x Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

4 8 Chater 5 Analytic Trigonometry 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 x cos 3 xsin x cos x sin x(-sin x)(sin x-sin x) 3. cos 5 xcos x (cos x) (-sin x) (- sin x+sin x) tan x cot x cot x - tan x tan x # cot x + # - cot x - tan x a / - + / - b sin 3 x - cos 3 x ( - ) sin x + + cos x + +csc x sec x +. This involves rewriting a 3 -b 3 as (a-b)(a +ab+b ), where a and b. 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 ( - )( + ) ( + ) + ( - )( + ) - 9. cos 3 x(cos x)()(-sin x)() 5. sin 5 x(sin x)()(sin x) () (-cos x) () (- cos x+cos x)() 53. (d) multily out: (+sec x)(-) -+sec x-sec x # - cos x sin x -+ - # tan x. 55. (c) ut over a common denominator: sin x sec x. cos x 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. 59. True. If x is in the domain of both sides of the equation, then x Ú 0. The equation (x) x holds for all x Ú 0, so it is an identity. 6. A roof is - # ( + ) - cos x ( + ) sin x + The answer is E. 63. k must equal, so f(x) Z 0. The answer is B. 65. ; tan 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 69. ; (sec x)(-sin x) a (cos x) b 7. If A and B are comlementary angles, then sin A+sin Bsin A+sin ( /-A) sin A+cos A 73. Multily and divide by -sin t under the radical: - sin t # - sin t ( - sin t) B + sin t - sin t B - sin t ( - sin t) ƒ - sin tƒ since a ƒaƒ. B cos t ƒcos tƒ Now, since -sin t Ú 0, we can disense with the absolute value in the numerator, but it must stay in the denominator. 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. ƒƒ 77. One ossible roof: ln tan x ln ƒƒ ln -ln. 79. (a) They are not equal. Shown is the window [,,] by [, ]; grahing on nearly any viewing window does not show any aarent difference but using TRACE, one finds that the y coordinates are not identical. Likewise, a table of values will show slight differences; for examle, when x, y while y [, ] by [, ] Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

5 (b) One choice for h is 0.00 (shown). The function y 3 is a combination of three sinusoidal functions (000 sin(x+0.00), 000, and ), all with eriod. 8. In the decimal window, the x coordinates used to lot the grah on the calculator are (e.g.) 0, 0., 0., 0.3, etc. that is, xn/0, where n is an integer. Then 0 x n, and the sine of integer multiles of is 0; therefore, +sin 0 x+sin n+0. However, for other choices of x, such as x, we have +sin 0 x+sin 0 Z. Section 5.3 Sum and Difference Identities Exloration. sin (u + v) -, sin u + sin v. No.. cos (u + v), cos u + cos v. No. 3. tan (/3 + /3) -3, tan /3 + tan /3 3. (Many other answers are ossible.) Quick Review # No. (f(x) + f(y) ln x + ln y ln (xy) 9. Yes. (f(x + y) 3(x + y) 3x + 3y Section 5.3 Exercises. sin 5 sin(5-30 ) sin 5 cos 30 - cos 5 sin cos cos a 3 - b cos 3 cos + sin 3 sin 7. [, ] by [ 0.00, 0.00] -f(xy) Z f(x + y)) f(x) + f(y)) # 3 - # 6 - sin 75 sin ( ) sin 5 cos 30 + cos 5 sin 30 # 3 + # 6 + # + 3 # tan 5 tan a 3 - tan (/3) - tan (/) b + tan (/3) tan (/) (3 + ) Section 5.3 Sum and Difference Identities 9 cos 7 cos a b cos 5 6 cos + sin 5 6 sin In #, match the given exression with the sum and difference identities.. sin ( - 7 ) sin 5 3. sin a b sin 0 5. tan (9 + 7 ) tan cos a 7 - xb cos ax - 7 b 9. sin (3x - x). tan (y + 3x) 3. sin ax - b cos - sin 5. cos ax - b cos + sin 7. sin ax + 6 b cos 6 + sin 6 # 3 + # 9. tan au + b tan u + tan (/) - tan u tan (/) tan u + - tan u # + tan u - tan u 3. Equations B and F. 33. Equations D and H. 35. Rewrite as - 0; the left side equals sin(x-x), so xn, n an integer. 37. sina - ub sin cos u - cos sin u - 6 # 0 - # - # 0 + # # cos u - 0 # sin u cos u. - 3 # + # 39. cot a cos (/ - u) - ub tan u using the sin (/ - u) sin u cos u first two cofunction identities.. csc a sec u using the - ub sin (/ - u) cos u second cofunction identity. 3. To write y 3 + in the form y a sin (bx + c), rewrite the formula using the formula for the sine of a sum: y a((sin bx cos c) + (cos bx sin c)) a sin bx cos c + a cos bx sin c (a cos c)sin bx + (a sin c)cos bx. Then comare the coefficients: a cos c 3, b, a sin c. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

6 0 Chater 5 Analytic Trigonometry Solve for a as follows: 59. y + sin (x+x)sin 3x. (a cos c) + (a sin c) 3 + The answer is A. a cos c + a sin c 5 tan u + tan v a (cos c + sin c) 5 6. For all u, v, tan (u + v) The answer is B. - tan u tan v. a 5 sin (u - v) a ;5 63. tan(u-v) cos (u - v) If we choose a to be ositive, then cos c 3/5 and sin u cos v - cos u sin v sin c /5. c cos - (3/5) sin - (/5). So the sinusoid is cos u cos v + sin u sin v y 5 sin(x + cos - (3/5)) L 5 sin (x ). sin u cos v cos u sin v 5. Follow the stes shown in Exercise 3 to comare the - cos u cos v cos u cos v coefficients in y (a cos c)sin bx + (a sin c)cos bx to the cos u cos v sin u sin v coefficients in y cos 3x + sin 3x: a cos c, + b3, a sin c. cos u cos v cos u cos v Solve for a as follows: sin u (a cos c) + (a sin c) + cos u - sin v cos v a (cos c + sin sin u sin v c) 5 + cos u cos v a ;5 tan u - tan v If we choose a to be ositive, then cos c /5 and + tan u tan v sin c /5. So the sinusoid is y 5 sin (3x - cos The identity would involve tan a 3, which does not exist. (/5)) b L.36 sin (3x ) sin(x-y)+sin(x+y) tan ax - 3 sin ax - ( cos y- sin y) + ( cos y+ sin y) b b cos y cos ax - 3 b 9. cos 3xcos[(x+x)+x] cos 3 cos(x+x) -sin(x+x) - sin 3 ( - ) -( + ) cos 3 + sin 3 cos x-sin x - sin x cos x-3 sin x # 0 - # ( ) # 0 + # ( ) 5. cos 3x+cos(x+x)+cos(x-x); use Exercise 8 with x relaced with x and y relaced with x. -cot x 53. tan(x+y) tan(x-y) cos (x + h) - cos h - sin h - tan x + tan y tan x - tan y 67. a a - tan x tan y b # + tan x tan y b h h (cos h - ) - sin h tan x - tan y h since both the numerator and - tan x tan y a cos h - b - sin h denominator are factored forms for differences of squares. h h sin (x + y) 69. sin (A + B) sin ( - C) 55. sin cos C - cos sin C sin (x - y) 0 # cos C - (-) sin C cos y + sin y sin C cos y - sin y cos y + sin y /( cos y) 7. tan A+tan B+tan C # cos A + sin B cos B + sin C cos C cos y - sin y /( cos y) (cos B cos C) + sin B(cos A cos C) ( cos y)/( cos y) + ( sin y)/( cos y) cos A cos B cos C ( cos y)/( cos y) - ( sin y)/( cos y) sin C(cos A cos B) + (/) + (sin y/cos y) cos A cos B cos C (/) - (sin y/cos y) cos C( cos B + cos A sin B) + sin C(cos A cos B) tan x + tan y cos A cos B cos C tan x - tan y cos C sin (A + B) + sin C(cos (A + B) + sin B) 57. False. For examle, cos 3 +cos 0, but 3 and cos A cos B cos C are not sulementary. And even though cos C sin ( - C) + sin C(cos ( - C) + sin B) cos (3 /)+cos (3 /)0, 3 / is not sulementary cos A cos B cos C cos C sin C + sin C(-cos C) + sin C sin B with itself. cos A cos B cos C Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

7 Section 5. Multile-Angle Identities sin B sin C cos A cos B cos C tan A tan B tan C 73. This equation is easier to deal with after rewriting it as cos 5x cos x+sin 5x sin x0. The left side of this equation is the exanded form of cos(5x-x), which of course equals ; the grah shown is simly y. The equation 0 is easily solved on the interval [, ]: x ; or x ; 3. The original grah is so crowded that one cannot see where crossings occur. [, ] by [.,.] 75. B B in + B ref E 0 c E 0 v x acos v t cos c c + cos v t cos v x c E 0 c v x cos av t - c b + E 0 v x cos av t + c c b v x a cos v t cos c b E 0 v x cos v t cos c c Section 5. Multile-Angle Identities Exloration. sin - cos (/) 8 - (/) # -. sin. 8 ; - - B We take the ositive square root because is a firstquadrant angle. 3. sin 9 - cos (9/) 8 - (/) # - 9. sin. 8 ; B We take the negative square root because is a thirdquadrant angle. + sin v t sin v x c - sin v t sin v x c b 8 8 Quick Review 5.. tan x when x +n, n an integer. 3. Either 0 or. The latter imlies the former, so x, n an integer. + n 5. when x - +n, n an integer. 7. Either or -. Then x or 6 + n 5 x or x ;, n an integer. 6 + n 3 + n 9. The traezoid can be viewed as a rectangle and two triangles; the area is then A()(3)+ ()(3)+ ()(3) 0.5 square units. Section 5. Exercises. cos ucos(u+u)cos u cos u-sin u sin u cos u-sin u 3. Starting with the result of Exercise : cos ucos u-sin u(-sin u)-sin u - sin u , so (-)0; 0 or when x0 or x. 7. sin x+-0, so ( -)(+) 0; or when x, x or x , so, or ( - ) Then 0 or 0 3 (but Z 0), so x0, x, x, x, 5 7 x or x. For #, any one of the last several exressions given is an answer to the question. In some cases, other answers are ossible as well.. sin +cos sin cos +cos (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 5. sin xsin (x) 7. csc x # csc x tan x sin x Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

8 Chater 5 Analytic Trigonometry 9. sin 3x + cos x +( cos x-) ()( cos x-). cos xcos (x)- sin x -( ) -8 sin x cos x 3. cos x+-0, so or, 5 x, x or x cos 3x - (- sin x) -( ) - sin x - sin x - sin x Thus the left side can be written as ()(- sin x). This equals 0 in [0, ) when x x 3 or x 7, x 5, x 3, x,,. 7. +sin x+ ()(+ )0. Then 0 or 9. - ; the solutions in [0, ) are x0, x, 3, x x or x 5. 3, x, x 3, x 3, 3 [0, ] by [ 60,60] The solutions are {0.3,.57,.83,.08,.7, 5.3} which corresond to {0., 0.5, 0.9,.3,.5,.7}. - cos sin 5 ; ; a - B B b - 3. Since sin 5 7 0, take the ositive square root. + cos cos 75 a - B B b - 3. Since cos , take the ositive square root tan. - cos(7/6) + 3/ sin (7/6) -/ 37. (a) Starting from the right side: ( - cos u) ( sin u)sin u. [ - ( - sin u)] (b) Starting from the right side: ( + cos u) cos u. (cos u) [ + ( cos u - )] 39. sin x(sin x) c ( - ) d ( - + ) c - + ( + cos x) d ( cos x) 8 (3- +cos x) 8. sin 3 x sin x # ( - cos x) ()( - cos x) - 3. cos x, so cos x+-0. Then or. In the interval [0, ), x, 3 5 x, or x. General solution: x +n or 3 3 x+n, n an integer. 5. The right side equals tan (x/); the only way that tan(x/) tan (x/) is if either tan(x/)0 or tan(x/). In [0, ), this haens when x0 or x. The general solution is xn or x n, n an integer False. For examle, f(x) has eriod and g(x) has eriod, but the roduct f(x)g(x) has eriod. 9. f(x) f(x)g(x). The answer is D. 5. or 0 or x or x or 6 6 The answer is E. 53. (a) In the figure, the triangle with side lengths x/ and R is a right triangle, since R is given as the erendicular distance. Then the tangent of the angle / is the ratio u oosite over adjacent : tan Solving for x x/ R gives the desired equation. The central angle is /n since one full revolution of radians is divided evenly into n sections. u (b) 5.87 R tan, where /, so R 5.87/ a tan R0. b Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

9 Section 5.5 The Law of Sines (a) ft θ θ ft ft The volume is 0 ft times the area of the end. The end is made u of two identical triangles, with area (sin ) (cos ) each, and a rectangle with area () (cos ). The total volume is then 0 # (sin cos +cos )0 (cos )(+sin ). Considering only -, the maximum value u occurs when 0.5 (in fact, it haens exactly at ). The maximum value is about.99 ft csc u sin u sin u cos u # # sin u cos u csc u sec u 59. sec u cos u - sin u u a - sin u bacsc csc u b csc u csc 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 63. (a) The following is a scatter lot of the days ast January as x-coordinates (L) and the time (in -hour mode) as y-coordinates (L) for the time of day that the sunset occurred in Honolulu in 009. [ 30,370] by [ 60,60] (b) The sine regression curve through the oints defined by L and L is y.5 sin (0.05x - 0.8) This is a fairly good fit, but not really as good as one might exect from data generated by a sinusoidal hysical model. [ 30,370] by [ 60,60] (c) using the formula L Y(L) (where Y is the sine regression curve), the residual list is {3.7, 7.8, 3.05, -6.50, , , 5.63, 9.79,.5, -3.66, -8.6, -.39}. (d) The following is a scatter lot of the days ast January as x-coordinates (L) and the residuals (the differences between the actual number of minutes (L) and the number of minutes redicted by the regression line (Y)) as y-coordinates (L3) for the time of day that the sunset occurred in Honolulu in 009. The sine regression curve is y 8.73 sin (0.03x + 0.6) (Note: Round L3 to decimal laces to obtain this answer.) This is another fairly good fit, which indicates that the residuals are not due to chance. There is a eriodic variation that is most robably due to hysical causes. [ 30,370] by [ 5,5] (e) The first regression indicates that the data are eriodic and nearly sinusoidal. The second regression indicates that the variation of the data around the redicted values is also eriodic and nearly sinusoidal. Periodic variation around eriodic models is a redictable consequence of bodies orbiting bodies, but ancient astronomers had a difficult time reconciling the data with their simler models of the universe. Section 5.5 The Law of Sines Exloration. If BC AB, the segment will not reach from oint B to the dotted line. On the other hand, if BC 7 AB, then a circle of radius BC will intersect the dotted line in a unique oint. (Note that the line only extends to the left of oint A.). A circle of radius BC will be tangent to the dotted line at C if BCh, thus determining a unique triangle. It will miss the dotted line entirely if BC 6 h, thus determining zero triangles. 3. The second oint (C ) is the reflection of the first oint (C ) on the other side of the altitude.. sin C sin (-C )sin cos C -cos sin C sin C. 5. If BC Ú AB, then BC can only extend to the right of the altitude, thus determining a unique triangle. Quick Review 5.5. abc/d 3. cad/b 7 sin 8 5. L 3.3 sin 3 7. xsin x80 -sin ( 0.7).7 Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

10 Chater 5 Analytic Trigonometry Section 5.5 Exercises. Given: b3.7, B5, A60 an AAS case. C80 -(A+B)75 ; a b sin B b 3.7 sin 60 a L.5; sin B sin 5 3. Given: A00, C35, a an AAS case. B80 -(A+C)5 ; b a sin B sin 5 L 5.8; sin 00 c a sin C sin 35 L.8 sin Given: A0, B30, b0 an AAS case. C80 -(A+B)0 ; a b sin B c b sin C sin B L.9; 7. Given: A33, B70, b7 an AAS case. C80 -(A+B)77 ; a b sin B b sin B Q c sin C Q c b sin C sin B 0 sin 0 sin 30 0 sin 0 sin 30 7 sin 33 sin 70 L 8.8 L.; 3.7 sin 75 sin 5 c b sin C 7 sin 77 L 7.3 sin B sin Given: A3, a7, b an SSA case. hb 5.8; h<b<a, so there is one triangle. B sin - a b b sin - (0.3 Á ) L 0. a C80 -(A+B) 7.9 ; c a sin C 7 sin 7.9 L 5.3 sin 3. Given: B70, b, c9 an SSA case. hc sin B 8.5; h<c<b, so there is one triangle. C sin - a c sin B b sin - (0.60 Á ) L 37. b A80 -(B+C) 7.8 ; a b sin 7.8 L. sin B sin Given: A36, a, b7. hb.; a<h, so no triangle is formed. L Given: C36, a7, c6. ha sin C 0.0; h<c<a, so there are two triangles. 7. Given: C30, a8, c9. ha sin C9; hc, so there is one triangle. 9. Given: A6, a6, b7. hb 5.3; h<a<b, so there are two triangles. B sin - a b b sin - (0.95 Á ) L 7.7 a C 80 -(A+B ) 3.3 ; c a 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 a sin C L.7. Given: C68, a9, c8. ha sin C 7.6; h<c<a, so there are two triangles. A sin - a a sin C b sin - (0.978 Á ) 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 h0 sin 6.69, so: (a) b 6 0. (b) b 6.69 or b Ú 0. (c) b (a) No: This is an SAS case. (b) No: Only two ieces of information given. 7. Given: A6, a8, b an SSA case. hb 8.; a<h, so no triangle is formed. 9. Given: A36, a5, b8 an SSA case. hb 9.5; a<h, so no triangle is formed. 3. Given: B, c8, C39 an AAS case. A80 -(B+C)99 ; a c sin C b c sin B sin C L 8.3; 33. Given: C75, b9, c8. an SSA case. hb sin C 7.3; h<c<b, so there are two triangles. B sin - a b sin C b sin - (0.986 Á ) L 80. c A 80 -(B+C).6 ; a 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. ; a c sin C L Cannot be solved by law of sines (an SAS case). 37. Given: cab56, A7, B53 an ASA case, so C80 -(A+B)55 (a) ACb (b) hb a sin B 5.9 ft. 39. Given: c6, C90-68, B an AAS case. A80 -(B+C)7, so a c sin C 8 sin 99 sin 39 8 sin sin 39 c sin B sin C 6 sin 7 sin 8 L sin 53 sin 55 L.9 ft. L 5.6 ft. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

11 Section 5.6 The Law of Cosines 5. ft 8 0 The length of the brace is the leg of the larger triangle. sin 8 x, so x.9 ft. 3. Consider the triangle with vertices at the to of the flagole (A) and the two observers (B and C). Then a600, B9, and C (an ASA case), so A80 -(B+C)0 ; a sin B 600 sin 9 b L 303.9; sin 0 a sin C 600 sin c L 33.5 sin 0 and finally hb sin Cc sin B 08.9 ft. 5. Given: c0, B5, C33 an AAS case. A80 -(B+C)95, so c 0 sin 95 a L 36.6 mi, and sin C sin 33 c sin B 0 sin 5 b L 8.9 mi. sin C sin True. By the law of sines, sin B, a b which is equivalent to (since, sin B Z 0). sin B a b 9. The third angle is 3. By the law of sines, sin 3 sin 53, which can be solved for x..0 x The answer is C. 5. The longest side is oosite the largest angle, while the shortest side is oosite the smallest angle. By the law of sin 50 sin 70 sines,, which can be solved for x. 9.0 x The answer is A. 53. (a) Given any triangle with side lengths a, b, and c, the law of sines says that sin B sin C. a b c But we can also find another triangle (using ASA) with two angles the same as the first (in which case the third angle is also the same) and a different side length say, a. Suose that a ka for some constant k. Then for this new triangle, we have sin B sin C. Since a b c a ka # sin B, we can see that # sin B, k a b k b so that b kb and similarly, c kc. So for any choice of a ositive constant k, we can create a triangle with angles A, B, and C. (b) Possible answers: a, b 3, c (or any set of three numbers roortional to these). (c) Any set of three identical numbers. 55. (a) hab (b) BC 6 AB (c) BC Ú AB or BCAB x (d) AB 6 BC 6 AB 57. Given: c., B5, C An AAS case: A80 -(B+C)3.5, so c sin B. sin 5 ACb L 8.7 mi, and sin C sin.5 c. sin 3.5 BCa L. mi. sin C sin.5 The height is ha sin 5 b sin mi. Section 5.6 The Law of Cosines Exloration. The semierimeters are 5 and 50. A 5(5-5)(5-8)(5 - ) + 50(50 - )(50-0)(50-86) aces.,0.595 square feet square miles acres 5. The estimate of a little over an acre seems questionable, but the roughness of their measurement system does not rovide firm evidence that it is incorrect. If Jim and Barbara wish to make an issue of it with the owner, they would be well advised to get some more reliable data. 6. Yes. In fact, any olygonal region can be subdivided into triangles. Quick Review 5.6. Acos a b 3. Acos ( 0.68) (a) cos A 8 - x - y -xy (b) Acos a x + y - 8 b xy x + y - 8 xy 7. One answer: (x-)(x-)x -3x+. Generally: (x-a)(x-b)x -(a+b)x+ab for any two ositive numbers a and b. 9. One answer: (x-i)(x+i)x +. Section 5.6 Exercises. Given: B3, c8, a3 an SAS case. b a + c - ac cos B L L 9.; C cos - a a + b - c b L cos - (0.99) L 8.3 ; ab A 80 - (B + C) L Given: a7, b9, c an SSS case. A cos - a b + c - a b L cos - (0.8) L 76.8 ; bc B cos - a a + c - b b L cos - (0.78) L 3. ; ac C 80 - (A + B) L 60. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

12 6 Chater 5 Analytic Trigonometry 5. Given: A55, b, c7 an SAS case. a b + c - bc cos A L L 9.8; B cos - a a + c - b b L cos - (0.0) L 89.3 ; ac C 80 - (A + B) L Given: a, b, C95 an SAS case. c a + b - ab cos C L L 5.; A cos - a b + c - a b L cos - (0.879) L 8.5 ; bc B 80 - (A + C) L No triangles ossible (a+cb).. Given: a3., b7.6, c6. an SSS case. A cos - a b + c - a b L cos - (0.909) L.6 ; bc B cos - a a + c - b b L cos - (-0.60) L 99. ; ac C 80 - (A + B) L 56.. Exercises #3 6 are SSA cases, and can be solved with either the law of sines or the law of cosines. The law of cosines solution is shown. 3. Given: A, a7, b0 an SSA case. Solve the quadratic equation 7 0 +c -(0)c cos, or c -(.86 Á )c+50; there are two ositive a + c - b solutions: L 9.87 or Since cos B : ac c 9.87, B cos (0.9) 7.9, and C 80 -(A+B ) 65., or c 5.376, B cos ( 0.9) 07., and C 80 -(A+B ) Given: A63, a8.6, b. an SSA case. Solve the quadratic equation c -(.)c cos 63, or c -(0.079)c+9.50; there are no real solutions, so there is no triangle. 7. Given: A7, b3, c9 an SAS case. a b + c - bc cos A L L 3.573, so Area L L.33 ft. (using Heron s formula). Or, use Area (3)(9) bc sin 7 L.33 ft. 9. Given: B0, a0, c an SAS case. b a + c - ac cos B L L 5.85, so Area L L cm. (using Heron s formula). Or, use Area (0)() ac sin B sin 0 L cm. For # 8, a triangle can be formed if a+b<c, a+c<b, and b+c<a. 7. s ;Area L No triangle is formed (a+bc). 5. a36.;area 6,70.36 L s.; Area 98, L Let a, b5, and c6. The largest angle is oosite the largest side, so we call it C. Since cos C a + b - c, C cos - a ab 8 b L radians. 3. Following the method of Examle 3, divide the hexagon into six triangles. Each has two -inch sides that form a 60 angle * ()()sin L 37. square inches. 30 s a s In the figure, a and so s sec The area of the hexagon is 6 * (83)(83) sin L 98.8 square inches. 35. Given: C5, BCa60, ACb0 an SAS case. AB c a + b - ab cos C 7, L 30. ft. 37. (a) c (0)(60) cos L.5 ft. (b) The home-to-second segment is the hyotenuse of a right triangle, so the distance from the itcher s rubber to second base is 60-0 L.9 ft. (c) B cos - a a + c - b b L cos - (-0.057) ac (a) Using right ACE, mjcae tan - a 6 8 b tan - a 3 b L (b) Using A L 8.35, we have an SAS case, so DF (9)() cos A L ft. (c) EF (8)() cos A L ft.. AB (73)(65) cos L.5 yd. 3. AB c + 3 3, AC b + 3 0, and BC a + 5, so A cos - a b + c - a mjcab b bc cos - 9 a b 5. True. By the law of cosines, b +c -bc cos Aa, which is a ositive number. Since b +c -bc cos A>0, it follows that b +c >bc cos A. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

13 Chater 5 Review 7 7. Following the method of Examle 3, divide the dodecagon into triangles. Each has two -inch sides that form a 30 angle. * ()() sin 30 3 The answer is B. 9. After 30 minutes, the first boat has traveled miles and the second has traveled 6 miles. By the law of cosines, the two boats are ()(6) cos 0 L 3.05 miles aart. The answer is C. 5. Consider that a n-sided regular olygon inscribed within a circle can divide into n equilateral triangles, each with r 360 equal area of sin. (The two equal sides of the n equilateral triangle are of length r, the radius of the circle.) Then, the area of the olygon is exactly nr 360 sin. n (a) Shi A: 5. knots; hr Shi B:. knots hrs (b) cos A b + c - a bc (5.) + (.) - (8.7) (5.)(.) A 35.8 (c) c a +b -ab cos C (9.6) +(60.) -(9.6)(60.) cos (35.8 ).0, so the boats are 3.8 nautical miles aart at noon. 55. Let P be the center of the circle. Then, cos P (5)(5) 0.0, so P The area of the segment is The area of the triangle, however, is (5)(5) sin (88.9).50 in, so the area of the shaded region is aroximately 6.9 in. Chater 5 Review. sin 00 cos 00 sin 00 r # L 5 # (0.7) L 9.39 in. 3. ; the exression simlifies to (cos ) +( sin cos ) (cos ) +(sin ). 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 7. tan x-sin xsin x a - cos b x sin x # sin x tan x cos x 9. csc x- cot x + tan u. Recall that tan cot. - tan u + + cot u - cot u ( + tan u)( - cot u) + ( + cot u)( - tan u) ( - tan u)( - cot u) ( + tan u - cot u - ) + ( + cot u - tan u - ) ( - tan u)( - cot u) 0 ( - tan u)( - cot u) 0 t 3. cos c ; A ( + cos t) d ( + cos t) cos x a + cos t b a sec t sec t b + sec t sec t cosf - tanf + sinf - cotf a cosf sinf a - tanf bacosf cosf b + - cotf basinf sinf b cos f sin f sinf - cosf cos f - sin f cosf - sinf + cosf - sinf cosf + sinf - cos y 7. B + cos y ( - cos y) B( + cos y)( - cos y) ( - cos y) B - cos y ( - cos y) B sin y ƒ - cos y ƒ - cos y since -cos y Ú 0, ƒ sin y ƒ ƒ sin y ƒ we can dro that absolute value. 9. tan au + 3 b tan u + tan (3/) - tan u tan (3/) tan u + (-) - tan u(-) tan u - + tan u tan b - cos b sin b - # sin b - cos b csc b - cot b sin b 3. Yes: sec x- tan x - - sin x. 5. Many answers are ossible, for examle, sin 3x+cos 3x (3 - sin 3 x)+( cos 3 x-3 ) 3(-)-(sin 3 x-cos 3 x) (-)[3-(sin x+ +cos x)] (-) (3-- ) (-)(+ ). Check other answers with a graher. 7. Many answers are ossible, for examle, cos x--sin x- - sin x cos x-. Check other answers with a graher. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

14 8 Chater 5 Analytic Trigonometry In #9 33, n reresents any integer when x +n or x +n, so x +n or x +n. 3. tan x when x - +n 33. If tan x, then xtan. 35. x. or x x.5 [0, ] by [, ] [0, ] by [ 3, ] 5 39., so x or x The left side factors to (-3)(+)0; 3 only is ossible, so x. 3. sin() only if. No choice of + n n gives a value in [, ], so there are no solutions. For #5 8, use grahs to suggest the intervals. To find the endoints of the intervals, treat the inequalities as equations, and solve. 5. has solutions x 6, x 5 6, x 7 6, and x in interval [0, ). The solution 6 set for the inequality is 0 x or or 6 x 6 ; 6 6 x that is, c0,. 6 b h a 5 6, 7 6 b h a 6, b 7. has solutions x and x 5 in the 3 3 interval [0, ). The solution set for the inequality is ; that is, a. 3, x b 9. y5 sin (3x+cos (3/5)) 5 sin (3x+0.973) 5. Given: A79, B33, a7 an AAS case. C 80 - (A + B) 68 b a sin B 7 sin 33 L 3.9; sin 79 c a sin C 7 sin 68 sin 79 L Given: a8, b3, B30 an SSA case. Using the law of sines: ha sin B; b<h, so no triangle is formed. Using the law of cosines: Solve the quadratic equation 3 8 +c -(8)c cos 30, or c -(83)c ; there are no real solutions. 55. Given: A3, B7, c5 an ASA case. C80 -(A+B)7 ; a c sin C b c sin B sin C 5 sin 3 sin 7 5 sin 7 sin 7 L.9; L Given: a5, b7, c6 an SSS case. A cos - a b + c - a b L cos - (0.7) L. ; bc B cos - a a + c - b b L cos - (0.) L 78.5 ; ac C 80 - (A + B) L s (3+5+6)7; Area s(s - a)(s - b)(s - c) 7(7-3)(7-5)(7-6) 56 L h sin 8 5.6, so: (a) 5.6<b<. (b) b 5.6 or b Ú. (c) b< Given: c.75, A33, B37 an ASA case, so C80 -(A+B)0 ; a c.75 sin 33 L.0; sin C sin 0 b c sin B.75 sin 37 L., sin C sin 0 and finally, the height is hb a sin b 0.6 mi. 65. Let a8, b9, and c0. The largest angle is oosite the largest side, so we call it C. a + b - c Since cos C, Ccos a 5 ab 6 b 7.790,.53 rad. 67. (a) The oint (x, y) has coordinates (cos, sin ), so the bottom is b units wide, the to is b x cos units wide, and the height is hysin units. Either use the formula for the area of a traezoid, A (b +b )h, or notice that the traezoid can be slit into two triangles and a rectangle. Either way: A( )sin +sin cos sin (+cos ) sin + sin. (b) The maximizing angle is 60 ; the maximum 3 3 area is.30 square units. 3 L 69. (a) Slit the quadrilateral in half to leave two (identical) right triangles, with one leg 000 mi, hyotenuse 000+h mi, and one acute angle /. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

15 Chater 5 Project 9 u Then cos ; solve for h to leave h 000 h miles. cos(u/) sec u (b) cos u, so u cos - a 0 b L 0.6 L The hexagon is made u of 6 equilateral triangles; using Heron s formula (or some other method), we find that each triangle has area ( - 6) 3,88 63.The hexagon s area is therefore, 383 cm, and the radius of the circle is 6 cm, so the area of the circle is 56 cm, and the area outside the hexagon is L 39.0 cm. 73. The volume of a cylinder with radius r and height h is V r h, so the wheel of cheese has volume (9 )(5)05 cm 3 ; a 5 wedge would have 5 05 fraction of that volume, or L 53.0 cm (a) By the roduct-to-sum formula in Exercise 7 (c), u + v u - v sin cos # asin u + v + u - v u + v - (u - v) + sin b sin u+sin v (b) By the roduct-to-sum formula in Exercise 7 (c), u - v u + v sin cos # asin u - v + u + v u - v - (u + v) + sin b sin u+sin( v) sin u-sin v (c) By the roduct-to-sum formula in Exercise 7 (b), u + v u - v cos cos # u + v - (u - v) acos + cos u + v + u - v b cos v+cos u cos u+cos v (d) By the roduct-to-sum formula in 7 (a), u + v u - v sin sin - # u + v - (u - v) acos -cos u + v + u - v b (cos v-cos u) cos u-cos v 77. (a) Any inscribed angle that intercets an arc of 80 is a right angle. (b) Two inscribed angles that intercet the same arc are congruent. (c) In right A BC, o hy a d. (d) Because ja and ja are congruent, a/d a a a d. (e) Of course. They both equal by the law of sines. a Chater 5 Project. [, 3] by [ 0.,.]. We set the amlitude as half the difference between the maximum value,.00, and the minimum value, 0.00, so a0.5. And we set the average value as the average of the maximum and minimum values, so k0.5. Since cos (b(x-h)) has a eriod of about 30, we set b / Exerimenting with the grah suggests that h should be about 0.5. So the equation is y L 0.5 cos (0.(x +.05)) We set the amlitude as half the difference between the maximum value,.00, and the minimum value, 0.00, so a0.5. And we set the average value as the average of the maximum and minimum values, so k0.5. Since sin (b(x-h)) has a eriod of about 30, we set b /300.. Exerimenting with the grah suggests that h should be about 8.. So the equation is y L 0.5 sin (0.(x + 8.)) The grahs are the same. Coyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.