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Thread: verifying trig identities

  1. #1
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    verifying trig identities

    Could someone please explain to me how to get through to the next step of verifying the identity?

    $\displaystyle
    cos^3 x sin^2 x = (sin^2 x - sin^4 x) cos x
    $

    $\displaystyle
    cos x(sin^2 x - sin^4 x) =
    $

    $\displaystyle
    cos x sin^2 x - sin^4 x cos x =
    $


    What Next?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by robasc View Post
    I am tyring to prove that:

    $\displaystyle
    cos^3 x sin^2 x
    $


    Equals:
    $\displaystyle
    (sin^2 x - sin^4 x) cos x
    $
    Prove $\displaystyle \cos^3 x \sin^2 x \equiv (\sin^2 x - \sin^4 x) \cos x$

    Consider the LHS

    $\displaystyle \cos^3 x \sin^2 x = \cos x \cos^2 x \sin^2 x$

    $\displaystyle = \cos x (1 - \sin^2 x ) \sin^2 x$

    $\displaystyle = \cdots $
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  3. #3
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    Prove

    Consider the LHS





    __________________

    So the next move would be:

    $\displaystyle
    cos x(1 - sin^2 x) sin^2 x =
    $

    $\displaystyle
    cos x(sin^2 x - sin^4 x)
    $

    and that's it!
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