Solution to Chapter 11 Problems

1) sin15 = sin(45-30) = sin45cos30 - cos45sin30 =

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

cos15 = cos(45-30) = cos45cos30 + sin45sin30 =

                                     (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) sin20cos10  + cos20sin10  =  sin(20+10=  sin30 = 1/2.

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3) sin80cos20 - cos80sin20  =  sin(80-20)  =  sin60  = 3/2..

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4) cos55cos10 + sin55sin10 = cos(55-10) = cos 45 = 2/2..

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5) cos85cos35 - sin85sin35 = 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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