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Math Help - quick question

  1. #1
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    quick question

    Determine an equivalent sine function for each cosine function:
    y=-cos2(x-pi/12) y=-1/2cos1/4x

    Thank you
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  2. #2
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    Hello, johett!

    Use the identity: . \sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}\quad\Rightarrow\quad\cos2\theta \:=\:1-2\sin^2\!\theta


    Determine an equivalent sine function for each cosine function:

    1)\;\;y\:=\:-\cos\left[2\left(x-\frac{\pi}{12}\right)\right]
    Use the identity: . y \;=\;-\left[1 - 2\sin^2\!\left(x-\frac{\pi}{12}\right)\right] \;=\;2\sin^2\!\left(x-\frac{\pi}{12}\right) - 1


    2)\;\;y\:=\:-\frac{1}{2}\cos\left(\frac{x}{4}\right)
    We have: . y \;=\;-\frac{1}{2}\cos\left(2\cdot\frac{x}{8}\right)

    Use the identity: . y \;=\;-\frac{1}{2}\left[1 - 2\sin^2\!\left(\frac{x}{8}\right)\right] \;=\;\sin^2\!\left(\frac{x}{8}\right) - \frac{1}{2}

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