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Math Help - Somewhat Advanced Trig Identity Problem...

  1. #1
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    Somewhat Advanced Trig Identity Problem...

    Show: \frac{1-3\sin x - 4\sin^2 x}{\cos^2 x} = \frac{1 - 4\sin x}{1 - \sin x}

    Your help is greatly appreciated. Thanks.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Savior_Self View Post
    Show: \frac{1-3\sin x - 4\sin^2 x}{\cos^2 x} = \frac{1 - 4\sin x}{1 - \sin x}

    Your help is greatly appreciated. Thanks.
    Hi Savior_Self,

    Take the right side and multiply by \frac{1+\sin x}{1+\sin x}:

    \frac{1-4\sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}=\frac{1-3 \sin x-4 \sin^2 x}{1-\sin^2 x}=\frac{1-3 \sin x - 4 \sin^2 x}{\cos^2 x}
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  3. #3
    Senior Member pacman's Avatar
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    or try factoring the Left Hand Side (LHS) of the equation?

    it is a simple factoring:



    LHS = (1 - 4sin x)(1 + sin x) =
    (1 + sin x)(1 - sin x)

    the (1 + sin x) cancels, you are left with,

    (1 - 4sin x)/(1 - sin x).

    (1 - 4sin 4x)(1 + sin x) = RHS

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