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  1. #1
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    trig help

    express in terms of the first power of cosine: (cos^2(x))(sin^2(x))

    how do i solve??
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  2. #2
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    Hello, jrose011!

    We're expected to know these Half-angle Identities:

    . . \begin{array}{ccc}\cos^2\!\theta &=& \dfrac{1+\cos2\theta}{2} \\ \\[-4mm] \sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} \end{array}


    Express in the first power of cosine: . \cos^2\!x\sin^2\!x

    We have: . \cos^2\!x\sin^2\!x \;=\; \frac{1+\cos 2x}{2}\cdot\frac{1-\cos2x}{2} \;=\;\frac{1}{4}\left[1 - \cos^2\!2x\right]

    . . . . . . . = \;\frac{1}{4}\bigg[1 - \frac{1+\cos4x}{2}\bigg] \;=\; \frac{1}{8}\left(1 - \cos 4x\right)

    Last edited by mr fantastic; March 17th 2009 at 03:35 AM. Reason: No edit, just flagging the reply as having been moved from a thread where the question was repeat posted.
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  3. #3
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    Trig Identities

    Hello jrose011
    Quote Originally Posted by jrose011 View Post
    express in terms of the first power of cosine: (cos^2(x))(sin^2(x))

    how do i solve??
    I assume that this means that you shouldn't have any power higher than 1, so you'll have to use the double-angle identities:

    \cos^2x = \tfrac{1}{2}(1 + \cos 2x) and \sin^2x = \tfrac{1}{2}(1-\cos 2x) to get:

    \cos^2x \sin^2x= \tfrac{1}{2}(1 + \cos 2x)\cdot \tfrac{1}{2}(1-\cos 2x)

    = \tfrac{1}{4}(1-\cos^22x)

    Then do the same again by expressing \cos^22x in terms of \cos 4x.

    Can you complete it now?

    Grandad
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