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Math Help - Difficult-ish Trig Identity

  1. #1
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    Difficult-ish Trig Identity

    (cos^2x)/(1+2sinx-3sinx)=(1+sinx)/(1+3sinx)

    The above is the trig identity, I'm sorry for not having in the proper fraction way but I couldn't figure out how to do it. And I'm dead tired for going away at this question for an hour. If anyone can help me out?
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  2. #2
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    Re: Difficult-ish Trig Identity

    \frac{\cos ^2 (x)}{1+ 2\sin (x) - 3 \sin (x)} = \frac{1 - \sin ^2 (x)}{1 - \sin (x) }

    I used the identity  \sin ^2 (x) + \cos ^2 (x) = 1
    note that
     1 - \sin ^2 (x) = (1 - \sin (x)) ( 1 + \sin (x))

    the left hand side is not equal to the right hand side
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  3. #3
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    Re: Difficult-ish Trig Identity

    Quote Originally Posted by astrophysicist View Post
    (cos^2x)/(1+2sinx-3sinx)=(1+sinx)/(1+3sinx)

    The above is the trig identity, I'm sorry for not having in the proper fraction way but I couldn't figure out how to do it. And I'm dead tired for going away at this question for an hour. If anyone can help me out?
    Assuming that your identity is actually \displaystyle \begin{align*} \frac{\cos^2{x}}{1 + 2\sin{x} - 3\sin^2{x}} \equiv \frac{1 + \sin{x}}{1 + 3\sin{x}} \end{align*} ...

    \displaystyle \begin{align*} \frac{\cos^2{x}}{1 + 2\sin{x} - 3\sin^2{x}} &\equiv \frac{1 - \sin^2{x}}{1 + 3\sin{x} - \sin{x} - 3\sin^2{x}} \\ &\equiv \frac{(1 - \sin{x})(1 + \sin{x})}{1(1 + 3\sin{x}) - \sin{x}(1 + 3\sin{x})} \\ &\equiv \frac{(1 - \sin{x})(1 + \sin{x})}{(1 + 3\sin{x})(1 - \sin{x})} \\ &\equiv \frac{1 + \sin{x}}{1 + 3\sin{x}}  \end{align*}
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