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Math Help - Proof of trig identity

  1. #1
    Newbie Chief65's Avatar
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    Proof of trig identity

    Establish the following identity: \cos (\theta)= 4 \cos^3(\theta/3) - 3\cos(\theta/3) Our hint is to use DeMoivre's formula, \left(\cos x + i \sin x\right)^k = \cos\left(kx\right) + i \sin\left(kx\right)  \, I have no idea where to begin and any help would be appreciated!
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  2. #2
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    Quote Originally Posted by Chief65 View Post
    Establish the following identity: \cos (\theta)= 4 \cos^3(\theta/3) - 3\cos(\theta/3) Our hint is to use DeMoivre's formula, \left(\cos x + i \sin x\right)^k = \cos\left(kx\right) + i \sin\left(kx\right) \, I have no idea where to begin and any help would be appreciated!
    (\cos{x} + i\sin{x})^k = \cos{(kx)} + i\sin{(kx)}

    (\cos{x} + i\sin{x})^3 = \cos{(3x)} + i\sin{(3x)}

    \cos^3{x} + 3i\cos^2{x}\sin{x} + 3i^2\cos{x}\sin^2{x} + i^3\sin^3{x} = \cos{(3x)} + i\sin{(3x)}

    \cos^3{x} + 3i\cos^2{x}\sin{x} - 3\cos{x}\sin^2{x} - i\sin^3{x} = \cos{(3x)} + i\sin{(3x)}

    \cos^3{x} + 3i(1 - \sin^2{x})\sin{x} - 3\cos{x}(1 - \cos^2{x}) - i\sin^3{x} = \cos{(3x)} + i(3\sin{x} - 4\sin^3{x})

    \cos^3{x} + 3i\sin{x} - 3i\sin^3{x} - 3\cos{x} + 3\cos^3{x} - i\sin^3{x} = \cos{(3x)} + 3i\sin{x} - 4i\sin^3{x}

    4\cos^3{x} - 3\cos{x} = \cos{(3x)}.


    Now let x = \frac{\theta}{3}.
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  3. #3
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    A shorter method:

    Using De Moivre's Theorem
     (\cos x + i \sin x)^3 = (\cos 3x + i \sin 3x)

    Using binomial expansion:
     (\cos x + i \sin x)^3 = \cos^3x + 3i \cos^2x \sin x - 3\cos x \sin^2x - i\sin^3x

    Equating the real parts:

     \cos 3x =  \cos^3x - 3\cos x \sin^2x
     \cos 3x =  \cos^3x - 3\cos x (1 - \cos^2x)
     \cos 3x =  4\cos^3x - 3\cos x

    Substituting  x = \frac{\theta}{3}
     \cos \theta = 4\cos^3 \frac{\theta}{3} - 3\cos \frac{\theta}{3} as required
    Last edited by Gusbob; December 12th 2009 at 05:12 PM. Reason: Stupid mistake
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  4. #4
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    Quote Originally Posted by Gusbob View Post
    A shorter method:

    Using De Moivre's Theorem
     (\cos x + i \sin x)^3 = (\cos 3x + i \sin 3x)

    Using binomial expansion:
     (\cos x + i \sin x)^3 = \cos^3x + 3i \cos^2x \sin x - 3\cos x \sin^2x - i\sin^3x

    Equating the real parts:

     \cos 3x = \cos^3x - 3\cos x \sin^2x
     \cos 3x = \cos^3x - 3\cos x (1 - \cos^2x)
     \cos 3x = 4\cos^3x - 3\cos x

    Substituting  x = \frac{\theta}{3}
     \cos \theta = 4\cos^4 \frac{\theta}{3} - 3\cos \frac{\theta}{3} as required
    That's probably the preferred method, as I'm sure the next part of the question is to prove the triple/third-angle identity for sine, which I already used in my proof...
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