Solution to Chapter 11 Problems

1) sina = 5/13,    cos2a = 1-sin2a = 1 -25/169 = 144/169,     cosa = 12/13

    (In the 1st Quad., cosa is positive); thus,          cosa = +12/13.

    tana = (sina)/(cosa) = (5/13)/(12/13) = 5/12,  or,  tana =5/12.

    cota = 1/(tana) = 1/( 5/12) = 12/5,                  or, cota = 12/5. 

    Now,             sin2a = 2sina cosa = 2(5/13)(12/13) = 120/169.

               cos2a = cos2a - sin2a = (12/13)2 - (5/13)2 = 119/169.

                                            tan2a = (sin2a)/(cos2a) =  120/119.

                                                     cot2a = 1/(tan2a) = 119/120.

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5) We have:     sinx + cosx = 0.2    Squaring and replacing   sin2x + cos2x = 1,

  sin2x + cos2x + 2sinx cosx = 0.04  or   1 + sin2x = 0.04  or {sin2x = -0.96}

  Since  sin22x + cos22x = 1  we get:  cos22x = 1 - sin22x   or 

  cos22x = 1 - (-0.96)2    or,    {cos2x = 0.28}

  tan2x = (sin2x)/(cos2x) = -0.96/( 0.28) = 3.43    or    {tan2x = 3.43}

  Finally, cot2x = 1/tan2x = 1/( 3.43) = 0.292        or    {cot2x = 0.292}

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8)  sin2 b + (1/2)cos2b = {1/2}

     Working on the left side,

    sin2b +(1/2)[1 - 2sin2b] =  sin2b + 1/2  - sin2b = {1/2}

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12)

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14)  cos2(a+b) + cos2(a-b) - cos2a cos2b = 1

Since {cos2x = 1/2 + (1/2) cos2x }, the left side may be written as:

[1/2 + (1/2)cos(2a+2b)] + [1/2 + (1/2)cos(2a -2b) ] - cos2a cos2b =

[1/2 + (1/2)(cos2a cos2b - sin2a sin2b)] +

[1/2 + (1/2)(cos2a cos2b + sin2a sin2b)]  - cos2a cos2b = 1

The (1/2)s add up to 1, the rest goes to zero after doing the algebra.

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29)  {cos22a - sin2a = cosa cos3a} ;  Since cos2a = cos2a - sin2a; therefore,

(cos2a - sin2a)2 - sin2a = cos4a + sin4a -2sin2a cos2a - sin2a =

Adding and subtraction 2sin2a cos2a to the cos4a + sin4a terms results in

[cos4a + sin4a + 2sin2a cos2a -2sin2a cos2a] + (-2sin2a cos2a - sin2a) =

[(sin2a + cos2a)2 -2sin2a cos2a ] + (-2sin2a cos2a - sin2a) =

[(1 -2sin2a cos2a ] + (-2sin2a cos2a - sin2a) = 1 - sin2a - 4sin2a cos2=

cos2a - 4sin2a cos2= cos2a - 4(1 - cos2a)cos2 = cos2a - 4cos2a + 4cos4a

= - 3cos2a + 4cos4a = cosa (- 3cosa +4cos3a) = cosa cos3a

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32)     {sina ( sina + sin3a) = cosa ( cosa - cos3a)}    ;

Working on the right side, multiplying through results in:

sin2a + sina sin3a = sin2a + sina (3sina - 4sin3a) =  sin2a + 3sin2a - 4sin4a =

4sin2a - 4sin4a = 4sin2a (1 - sin2a) = 4sin2acos2a = 4( 1 - cos2a) cos2a =

4 cos2a - 4cos4a = cosa (4cosa - 4cos3a) = cosa (cosa  + 3cosa - 4cos3a) =

cosa (cosa - cos3a)

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 34)          {sin3x sin3x + cos3x cos3x = cos32x} ;

Replacing sin3x and cos3x with their equivalents, we get:

 (3sinx - 4sin3x) sin3+  (4cos3x - 3cosx) cos3x =

 3sin4x - 4sin6x - 3cos4x + 4cos6x = 3(sin4x - cos4x)  - 4(sin6x - cos6x) =

 3(sin2x - cos2x)(sin2x + cos2x)  - 4(sin2x - cos2x)(sin4x + cos4x + sin2x cos2x) =

 3(-cos2x)( 1 )  -4(-cos2x)(sin4x + cos4x + sin2x cos2x ) =

cos2x { -3 + 4[ sin2x (sin2x + cos2x) + cos4x ] }=

cos2x { -3 + 4[ sin2x + cos4x ] }= cos2x { -3 + 4[ 1 - cos2x + cos4x ] }=

cos2x { -3 + 4 - 4cos2x + 4cos4x } = cos2x {4cos4x - 4cos2x + 1}=

cos2x (2cos2x - 1)2 = cos2x ( cos22x) = cos32x

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