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Math Help - Trig. Identities (cos3x)

  1. #1
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    Trig. Identities (cos3x)

    Alright, I'll write what I have so far..

    Show: \cos3x = 4\cos^3x - 3\cos x (changing only \cos3x)

    ---

    considering that \cos3x = \cos(2x + x) and \cos(A+B) = \cos A\cos B - \sin A\sin B

    we get...
    \cos2x\cos x - \sin2x\sin x

    using the double angle identities...

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

    \cos^3 x - (\sin^2 x)\cos x - 2\sin^2 x\cos x


    Using the Pythagorean identity..  2\sin^2 x = 2(1 -\cos^2 x)

    \cos^3 x - (1 - \cos^2 x)\cos x  - (2 - 2\cos x)\cos x

    \cos^3 x - (\cos x - \cos^3 x) - 2\cos x + 2\cos^2 x

    \cos^3 x - \cos x + \cos^3x - 2\cos x + 2cos^2 x

    2\cos^3x - 3\cos x + 2\cos^2x


    So I don't really know where to go from here... and I'm sure I made mistakes thus far. So could anyone help me out here? I'd really appreciate it. Thanks guys.
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  2. #2
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    Quote Originally Posted by Savior_Self View Post
    Alright, I'll write what I have so far..

    Show: \cos3x = 4\cos^3x - 3\cos x (changing only \cos3x)

    ---

    considering that \cos3x = \cos(2x + x) and \cos(A+B) = \cos A\cos B - \sin A\sin B

    we get...
    \cos2x\cos x - \sin2x\sin x

    using the double angle identities...

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

    \cos^3 x - (\sin^2 x)\cos x - 2\sin^2 x\cos x


    Using the Pythagorean identity..  2\sin^2 x = 2(1 -\cos^2 x)

    \cos^3 x - (1 - \cos^2 x)\cos x  - (2 - 2\cos\textcolor{red}{^2}{x})\cos x
    ...
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  3. #3
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    hahaha

    thanks, man.
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  4. #4
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    Cos3x=cos(x+2x) and

    cos3x=cosx.cos2x-sinxsin2x
    =cosx(2cos^2 x-1)-sinx(2sinxcosx)
    =2cos^3x-cosx-2sin^2xcosx
    =2cos^3x-cosx-2(1-cos^2x)cosx
    =2cos^3x-cosx-2cosx+2cos^3x
    =4cos^3x-3cosx
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