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Math Help - trig identity

  1. #1
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    trig identity

    I have problems with solving this identity, this is how I started

    \frac{cos4x+1}{ctgx+tgx} = \frac{1}{2}\cdot\sin4x  (1)

    \frac{cosx(2x+2x)+1}{\dfrac{cosx}{sinx} - \dfrac{sinx}{cosx}} = \frac{1}{2}\cdot\sin4x  (2)

    \frac{cos^2{2x} - sin^2{2x} +1}{\dfrac{cos^2x-sin^2x}{sinx \cdot cosx}} =\frac{1}{2}\cdot\sin4x   (3)

    now I don't now how to finish it

    any help appreciated
    Last edited by livmed; November 24th 2009 at 12:56 PM.
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  2. #2
    MHF Contributor
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    In the third line you dropped 1 from the numerator. Use \sin^2(2x)+\cos^2(2x)=1 to substitute for 1. Also notice that \cos^2(x)-\sin^2(x)=\cos(2x)

    That should keep you going.
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