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Math Help - Proving of trigonometric ID

  1. #1
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    Proving of trigonometric ID (solved)

    Good Day,

    I'm stuck on proving the following trigonometric identity:

    (sin 2x + sin 4x + sin 6x) / (cos 2x + cos 4x + cos 6x) = tan 4x

    I do hope that some one can give me some advice on how to prove this as I've been going at this for the past 2 days and I feel that I'm not getting anywhere near.

    Thanks in advance.
    Last edited by dd86; July 22nd 2012 at 05:23 AM.
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Re: Proving of trigonometric ID

    Quote Originally Posted by dd86 View Post
    Good Day,

    I'm stuck on proving the following trigonometric identity:

    (sin 2x + sin 4x + sin 6x) / (cos 2x + cos 4x + cos 6x) = tan 4x

    I do hope that some one can give me some advice on how to prove this as I've been going at this for the past 2 days and I feel that I'm not getting anywhere near.
    Use the sum-to-product identities,

    \sin\theta+\sin\phi = 2\sin\left(\frac{\theta+\phi}2\right)\cos\left( \frac{\theta-\phi}2\right)

    \cos\theta+\cos\phi = 2\cos\left(\frac{\theta+\phi}2\right)\cos\left( \frac{\theta-\phi}2\right)

    We have

    \frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}

    =\frac{\sin4x+\left(\sin2x + \sin6x\right)}{\cos4x + \left(\cos2x+\cos6x\right)}

    =\frac{\sin4x+2\sin4x\cos2x}{\cos4x + 2\cos4x\cos2x}

    =\frac{\sin4x\left(1+2\cos2x\right)}{\cos4x\left(1 + 2\cos2x\right)}

    =\frac{\sin4x}{\cos4x} = \tan4x.


    Please note, however, that the cancellation step is not valid when 1+2\cos2x = 0. The identity actually does not hold for these values of x, as the original expression is undefined at such values. So the equation is only identically true for

    x\neq\pi\left(k\pm\frac13\right)\!,\ \ k\in\mathbb{Z}.
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  3. #3
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    Re: Proving of trigonometric ID

    Hi Reckoner,

    Thanks so much. Now it's clear where I went wrong.
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