Answer Page!!
 
 
1)  a)  q = cos-1 .48  Using your calculator, the reference angle is 61.3o
            The cosine is negative in  the II and III quadrants.  Thus, the answers are:
180 - 61.3 = 118.7o
and
180 + 61.3 = 241.3o
      b)  q = sec-11.5 = cos-11.5-1 = 48.2o   Since the secant is positive in quadrants I and IV the
answers are:
48.2o
and
360 - 48.2 = 311.8o
      c) Since tangent is negative, the answers are in quadrants II and IV.  Use your calculator
          to find the reference angle.  It is .93
In quadrant II, 3.14 - .93 = 2.21
In quadrant IV, 6.28 - .93 = 5.35
 
      d)  The csc is positive in quadrants I and II.  Find x using your calculator by typing
sin-12.4-1 = .43
The answers are .43 in the first quadrant and 3.14 - .43 = 2.71 in the second!

2)  a)  Foil the product to get:  sec2x - 1 = tan2x  by the pythagorean relationship!!
 
     b)  Use the pythagorean relationship and the answer is -1
 
     c)  (sin A)(sin A/ cos A) + cos A    Use the cofunction and basic trig relationship
get a common denominator
(sin2A + cos2A)/ cos A
Use a pythagorean relationship
1/cos A
Use a reciprocal relationship
sec A
 

  3)  amplitude = 5 and period length = 2p/3
zeros at (0, 0), (p/3, 0), (2p /3, 0)
Maximum value at ( p /2, 5)
 Minimum value at (p /6, -5)
 
4) a)  sin x = 1/3    Answers are in quadrant I and II.  reference angle is .34
Answers: .34 and 3.14 - .34 = 2.80
    b)  sin 2x = 1/3 means 2x = .34, 2.80, 6.62, 9.08 because 0 < 2x < 12.56
              Thus x = .17, 1.40, 3.31, 4.54

5)  amplitude = 4
    Period length = 2p /2 = p  
    Translation:  4 down and p right
    Translated points:  zeros translated to (5p /4, -4), (7p /4, -4)
                                  Max translated to  (p , 0), (2p , 0)
                                  Min translated to (3p /2, -8)
6) a)  (sin x)(sec x) = (sin x)(1/cos x) = sin x/cos x = tan x
 
    b)  (1 - sin2 x)(1 + tan2 x) = (cos2 x)(sec2 x)     pythagorean relationships!
                                            = 1                           reciprocal relationship!

7) 

8)  (3cos x + 1)(2cos x + 1) = 0        factor
     cos x = -1/3 or  cos x = -1/2
     Reference angles are 1.23 and 1.05
     Answers are in quadrant II and III
     1.91, 4.37, 2.09, 4.19

9)  

10)  

11) a)  This is not a translation.  It is the normal cycle for the cosine graph.
        Amplitude = 3 with A = 3 because the maximum is at (0, 3)
        Period length = 4 which makes 4 = 2p /B,   4B = 2p ,  B = p/2

        Thus the equation is:  y = 3cos (px/2)
 
 
     b)  This is a translation.  One way to do the problem, certainly not the only way, is to think of it as a sine curved shifted 2 units up and 1 unit to the right.
        Amplitude = (4 - 0)/2 = 2  with A = 2
        Period length = 8  (count from maximum to the next maximum point)
        To find B,  8 = 2p /B,  8B = 2p ,    B = p/4

Putting it all together, gives us the equation     y - 2 = 2sin ( p/4)(x - 3)

Hopefully you did well on this test!  If not, go back and hit the books a little harder
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