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Math Help - Sin and Cos Graph

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    Sin and Cos Graph

    Sin and Cos Graph-untitled565464.jpg
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mr_Green View Post
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    The general form will be
    y = A sin(kx + \phi)

    Since A is the amplitude, this part is easy: A = 3.

    y = 3 sin(kx + \phi)

    Now, we want a 0 of the sine function to be positioned at x = - \frac{\pi}{2}. We can do this by setting kx + \phi = 0 at x = - \frac{\pi}{2}, so we have the equation:
    k \left ( - \frac{\pi}{2} \right ) + \phi = 0

    Again, we want a 0 of the sine function to be positioned at x = \frac{3 \pi}{2}. We can do this by setting kx + \phi = \pi ( \pi is the next zero for the sine function after 0) at x = \frac{3 \pi}{2}, so we have the equation:
    k \left (  \frac{3 \pi}{2} \right ) + \phi = 0

    Solving the first equation for \phi gives:
    \phi = \frac{\pi}{2}k

    Inserting this into the second equation gives:
    \frac{3\pi}{2}k + \frac{\pi}{2}k = \pi
    or
    k = \frac{1}{2}

    This gives
    \phi = \frac{\pi}{4}.

    Thus
    y = 3 sin \left ( \frac{1}{2}x + \frac{\pi}{4} \right ).

    -Dan
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    Thanks!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mr_Green View Post
    THANKS!



    FOR COS WOULD IT BE:


    3cos(.5x+pi/4)

    ???
    Close. You can do that in two ways. The first is to do it the same way I did below, except to use
    -\frac{\pi}{2}k + \phi = -\frac{\pi}{2} ( -\frac{\pi}{2} is a convenient first zero for cos)

    \frac{3\pi}{2}k + \phi = \frac{\pi}{2} (the next zero for cos after -\frac{\pi}{2})

    The other way is to recall that sin(\theta) = cos(\theta - \pi/2) so to switch to cosine we need to "shave" a \pi/2 off the argument.

    Either way I got:
    y = 3 cos \left ( \frac{1}{2}k - \frac{\pi}{4} \right )

    -Dan
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