Solution to Chapter 11 Problems

1) sin15° = sin(45°-30°) = sin45°cos30° - cos45°sin30° =

(2/2)(3/2) - (2/2)(1/2) = (6 - 2)/4.

cos15° = cos(45°-30°) = cos45°cos30° + sin45°sin30° =

(2/2)(3/2) + (2/2)(1/2) = (6 + 2)/4.

tan 15° = sin15° / cos15° = (6 - 2)/(6 + 2).

cot 15° = cos15° / sin15° = (6 + 2)/(6 - 2).

sin75° = cos15° = (6 + 2)/4.

cos75°  = sin15°  = (6 - 2)/4.

tan 75° = 1/tan15° =  (6 + 2)/(6 - 2).

cot 75° = 1/tan75° =  (6 - 2)/(6 + 2).

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2) sin20°cos10°  + cos20°sin10°  =  sin(20°+10°=  sin30° = 1/2.

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3) sin80°cos20° - cos80°sin20°  =  sin(80°-20°)  =  sin60°  = 3/2..

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4) cos55°cos10° + sin55°sin10° = cos(55°-10°) = cos 45° = 2/2..

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5) cos85°cos35° - sin85°sin35° = cos(85°+35°) = cos 120° = -1/2.

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13) Similar to (12).   To be done by students.  Verify that you get:

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15) cos(π/2 - a) = cos(π/2 )cosa + sin( π/2)sina = (0)cosa +(1)sina = sina

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16) cos(π/2 + a) = cos(π/2 )cosa - sin( π/2)sina = (0)cosa -(1)sina = -sina

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17) sin(π - a) + sin(π + a) = sinπ cosa - cosπ sina   +  sinπ  cosa + cosπ sina =

=  (0)cosa   -  (-1)sina   +     (0)cosa   + (-1)sina = 0

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18) cos(2π - a) + cos(2π + a) =

cos(2π )cosa + sin(2π)sina + cos(2π )cosa - sin(2π)sina=

(+1)cosa     +     (0)sina   +       (+1)cosa  -    (0)sina  = 2cosa

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20) sin3x cos2x + cos3x sin2x = sin(3x + 2x) = sin5x

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21) cos(a + 10° )cos(a -10° ) - sin(a + 10° )sin(a -10° ) = cos(a+10° +a - 10° )

= cos2a

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26) sin(a+b) sin(a-b) = [sina cosb + cosa sinb][sina cosb - cosa sinb] =

sin2a cos2b - cos2a sin2b =

sin2a ( 1 - sin2b) - (1 - sin2a) sin2b = sin2a - sin2a sin2b - sin2b + sin2a sin2b =

= sin2a - sin2b

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27)  cos(a+b) cos(a-b) = [cosa cosb - sina sinb][cosa cosb + sina sinb] =

cos2a cos2b - sin2a sin2b =

cos2a ( 1 - sin2b) - (1 - cos2a) sin2b = cos2a - cos2a sin2b - sin2b + cos2a sin2b =

= cos2a - sin2b

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35) {sina sin(b-c) + sinb sin(c-a) + sinc sin(a-b) = 0 Expanding results in:

sina(sinbcosc - cosbsinc) + sinb(sinc cosa - cosc sina)

+ sinc(sina cosb - cosa sinb) =

sinasinbcosc - sinacosbsinc + sinb sinc cosa - sinb cosc sina

+ sinc sina cosb - sinc cosa sinb = 0

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36) {cosa sin(b-c) + cosb sin(c-a) + cosc sin(a-b) = 0 Expanding results in:

cosa(sinbcosc - cosbsinc) + cosb(sinc cosa - cosc sina)

+ cosc(sina cosb - cosa sinb) =

coasinbcosc - cosacosbsinc + cosb sinc cosa - cosb cosc sina

+ cosc sina cosb - cosc cosa sinb = 0

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39) {(cosx + cosy)2 + (sinx + siny)2 -2 = 2cos(x-y)}  Expanding the left side,

cos2x + cos2y + 2cosxcosy + sin2x + sin2y +2sinxsiny -2 =

(cos2x + sin2x) + (cos2y + sin2y) -2 + 2(cosx cosy + sinx siny) =

1            +            1           -2  +2cos(x-y) =  2cos(x-y)

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40) {cos(x-y) + sin(x+y) = (sinx + cosx)(siny + cosy)}  Expanding the left side,

cosx cosy + sinx siny + sinx cosy +cosx siny  =    Grouping and factoring,

cosx( siny + cosy )  + sinx( siny + cosy) =

( sinx + cosx )( siny + cosy).

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45) { sin(a+b)sin(a-b) + sin(b+c)sin(b-c) + sin(c+a)sin(c-a) = 0 };  Expanding:

(sinacosb+cosasinb)(sinacosb- cosasinb) +(sinbcosc+cosbsinc)(sinbcosc- cosbsinc)

+  (sinccosa + coscsina)(sinccosa - coscsina) =

sin2a cos2b - cos2a sin2b+ sin2b cos2c - cos2b sin2c+ sin2c cos2a - cos2c sin2a =

sin2a (1-sin2b) - (1-sin2a) sin2+  sin2b (1-sin2c) - (1-sin2b) sin2+

sin2c (1-sin2a) - (1-sin2c) sin2a =

sin2a -sin2a sin2b -sin2b + sin2a sin2b + sin2b -sin2b sin2c -sin2c +sin2b sin2c +

sin2c -sin2c sin2a -sin2a + sin2c sin2a = 0.    All terms cancel.

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61) To show that A = cos2(x+y) + cos2(x-y) - cos2x cos2y  does not depend

on x and y, expand the terms while writing 2x as x+x and 2y as y+y.

A = cos2(x+y) + cos2(x-y) - cos(x+x) cos(y+y). Now expanding,

A = (cosx cosy - sinx siny)2 +  (cosx cosy + sinx siny)2

- [cosxcosx - sinx sinx][ cosy cosy - siny siny]

= (cos2x cos2y + sin2x sin2y + 2cosx cosy sinx siny)

+ (cos2x cos2y + sin2x sin2y - 2cosx cosy sinx siny)

- [cos2x - sin2x][ cos2y - sin2y]

= 2cos2x cos2y + 2sin2x sin2y

-cos2x cos2y + cos2x sin2y + sin2x cos2y - sin2x sin2y

= cos2x cos2y + cos2x sin2y + sin2x cos2y + sin2x sin2y

= cos2x(cos2y + sin2y) + sin2x(cos2y + sin2y)

= (cos2x + sin2x)(cos2y + sin2y) = 1 (Independent of x & y)

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62) a) From sin(2x+y) = 5siny deduce the relation: 2tan(x+y) = 3tanx.

sin(2x+y) = 5siny may be written as:

sin[ (x+y) +x] = 5sin[ (x+y) -x].  Expanding both sides, yields:

sin(x+y) cosx  + cos(x+y) sinx  =  5sin(x+y) cosx - 5cos(x+y) sinx , or

6cos(x+y) sinx  = 4sin(x+y) cosx,   or,

3cos(x+y) sinx  = 2sin(x+y) cosx,

Dividing both sides once by cosx and once by cos(x+y), we get:

3tanx= 2tan(x+y).

b) From 2tan(x+y) = 3tanxsiny deduce the relation: sin(2x+y) = 5.

Write the tangent on each side as sine/cosine, and then multiply

once by cosx and once by cos(x+y).  The reverse process is left

for students.

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63) Determine (K) and (tana) such that the equality

3sinx + 4cosx = Ksin(x+a) is an identity that means it holds true for all x.

Expanding the right side, we get:

3sinx + 4cosx = Ksinx cosa + Kcosx sina.

By convention, x is usually set to be the variable, and therefore, the

coefficients of sinx and cosx must equal on both sides.  We must have:

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