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Math Help - Tough Trig Identity

  1. #1
    Senior Member sakonpure6's Avatar
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    Tough Trig Identity

    cot(2x) = csc(2x) - tanx

    I tried working with either side, but I couldn't get anywhere. Any help is appreciated!!
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    Re: Tough Trig Identity

    Quote Originally Posted by sakonpure6 View Post
    cot(2x) = csc(2x) - tanx

    I tried working with either side, but I couldn't get anywhere. Any help is appreciated!!
    couldn't get anywhere? why?

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

    \csc(2x)=\frac{1}{\sin(2x)}=\frac{1}{2\sin(x) \cos(x)}

    you can finish it from here
    Last edited by romsek; January 27th 2014 at 05:54 PM. Reason: correcting mistake caught by SlipEternal
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  3. #3
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    Re: Tough Trig Identity

    Start with the LHS.
    Use the following identities:

    \begin{align*}\cot x & = \dfrac{\cos x}{\sin x} \\ \cos (2x) & = 1-2\sin^2 x \\ \dfrac{a-b}{c} & = \dfrac{a}{c} - \dfrac{b}{c}\\ \dfrac{1}{\sin x} & = \csc x \\ \sin(2x) & = 2\sin x \cos x \\ \dfrac{\sin x}{\cos x} & = \tan x\end{align*}
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  4. #4
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    Re: Tough Trig Identity

    Quote Originally Posted by romsek View Post
    couldn't get anywhere? why?

    \cot(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin(x) \cos(x)}{2\cos^2(x)-1}
    Should be: \cot(2x) = \dfrac{\cos(2x)}{\sin (2x)} = \dfrac{2\cos^2 (x)-1}{2\sin (x) \cos (x)}
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    Re: Tough Trig Identity

    Tough Trig Identity-28-jan-2014.png
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