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Math Help - Please help with Using Period to Evaluate Sin and Cosine

  1. #1
    Junior Member
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    Jun 2005
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    Please help with Using Period to Evaluate Sin and Cosine

    Hi, I am trying to follow the logic of tis problem and the solution.

    cos (- 7pi/2) = cos(-4pi + pi/2) = cos pi/2 = 0.

    Why do we use -4pi + pi/2 to get pi/2? How do we know to add -4pi + pi/2?

    Please help me to understand what's going on with this one.

    Thanks a lot for any help.
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  2. #2
    MHF Contributor
    Joined
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    Period of sine or cosine has nothing to do with this.

    This is just to explain the posted
    "cos (- 7pi/2) = cos(-4pi + pi/2) = cos pi/2 = 0."

    -7pi/2
    = ((-8 +1)pi)/2
    = -8pi/2 +pi/2
    = -4pi +pi/2
    That is easy.

    --------------
    The cos(-4pi +pi/2) = cos(pi/2) is not that easy.

    1) The author of the answer could have meant that by going all the way to -4pi then going back pi/2, the resulting angle is the same as pi/2.
    (In measuring the -4pi, we go two revolutions clockwise. Then, pi/2 is pi/2 by going the normal way, counterclockwise.)

    2)Another way is by using the trig identity cos(A+B) = cosAcosB -sinAsinB. (I assume that the author did not mean to use this, though.)
    cos(-4pi +pi/2)
    = cos(-4pi)cos(pi/2) -sin(-4pi)sin(pi/2)
    = (1)cos(pi/2) -(0)sin(pi/2)
    = cos(pi/2) -0
    = cos(pi/2)

    I am sure you know that cos(-4pi) = 1, and sin(-4pi) = 0.
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  3. #3
    Junior Member
    Joined
    Jun 2005
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    Tanks for your help ticbol. I couldn't have solved it without your way of expleaning it.
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