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Math Help - Trig Identity

  1. #1
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    Trig Identity

    Hello all. I am having trouble with a trig identity, and was wondering if you could help. This is a problem from a take home quiz, and unfortunately I can't find anything in my text about how to deal with the \cos{4\theta}.

    Here is the identity.

    \sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}]
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Deimos View Post
    Hello all. I am having trouble with a trig identity, and was wondering if you could help. This is a problem from a take home quiz, and unfortunately I can't find anything in my text about how to deal with the \cos{4\theta}.

    Here is the identity.

    \sin^2\theta\cos^2\theta=\frac{1}{8}[1-\cos{4\theta}]
    You can write the given expression as:
    8 \sin^2\theta\cos^2\theta=[1-\cos{4\theta}]

    Firstly, I assume you know that:

    1 - cos2x = 2sin^{2}x

    \therefore RHS = 1 - cos(4\theta) = 1 - cos2(2\theta) = 2 sin^{2}(2\theta) = 2 \times [sin(2\theta)]^2 = 2 \times [ 2 \sin(\theta) \cos(\theta)]^2 = 2 \times 4 \sin^{2}(\theta) \cos^{2}(\theta) = 8  \sin^{2}(\theta) \cos^{2}(\theta) = LHS
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  3. #3
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    Thank you very much!
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