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Math Help - Prove the identity sin 3x = 3 sin - 4 sin^3 x

  1. #1
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    Prove the identity sin 3x = 3 sin - 4 sin^3 x

    This as far as I get to making the left side the same as the right. I could use the steps I am missing

    sin\, 3x = sin(x+2x)

    = sin\, x \cos 2x + \cos x \sin 2x

    =\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x)

    but I get lost from here to get the left and right sides the same.


    thank you!
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  2. #2
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    Re: Prove the identity sin 3x = 3 sin - 4 sin^3 x

    Quote Originally Posted by Jonroberts74 View Post
    This as far as I get to making the left side the same as the right. I could use the steps I am missing

    sin\, 3x = sin(x+2x)

    = sin\, x \cos 2x + \cos x \sin 2x

    =\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x)

    but I get lost from here to get the left and right sides the same.


    thank you!
    Let's multiply out what you have:

    \begin{align*}\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x) & = \sin x \cos^2 x - \sin^3 x + 2\sin x \cos^2 x \\ & = 3\sin x \cos^2 x - \sin^3 x \\ & = 3\sin x(1-\sin^2 x) - \sin^3 x \\ & = 3\sin x - 3\sin^3 x - \sin^3 x \\ & = 3\sin x - 4\sin^3 x\end{align*}
    Thanks from Jonroberts74
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