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Math Help - Proving trig identity with exponential form of complex numbers

  1. #1
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    Proving trig identity with exponential form of complex numbers

    Express \cos\theta and \sin\theta in terms of e^{i\theta} and e^{-i\theta}.
    Use your results to prove that:
    16\cos^3\theta\sin^2\theta\equiv 2\cos\theta-\cos 3\theta-\cos 5\theta

    I have done:
    \cos\theta=Re(e^{i\theta}) and Re(e^{-i\theta})
    \sin\theta=Im(e^{i\theta}) and -Im(e^{-i\theta})

    The proving part:
    LHS\equiv 16[Re(e^{i\theta})]^3[Im(e^{i\theta})]^2
    I'm stuck with the [Im(e^{i\theta})]^2 part. The power of 2 indicates its real, but I haven't figured out how to express it in terms of Re(e^{i\theta}).
    Thanks!
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  2. #2
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    You should know that

    e^{i\theta} = \cos{\theta} + i\sin{\theta} and e^{-i\theta} = \cos{\theta} - i\sin{\theta}.


    Then e^{i\theta} + e^{-i\theta} = \cos{\theta} + i\sin{\theta} + \cos{\theta} - i\sin{\theta}

     = 2\cos{\theta}.

    Thus \cos{\theta} = \frac{1}{2}(e^{i\theta} + e^{-i\theta}).


    e^{i\theta} - e^{-i\theta} = \cos{\theta} + i\sin{\theta} - (\cos{\theta} - i\sin{\theta})

     = 2i\sin{\theta}.


    Thus \sin{\theta} = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})

     = -\frac{i}{2}(e^{i\theta} - e^{-i\theta}).


    Now try to go from here...
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