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Math Help - Proving double angles

  1. #1
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    Proving double angles

    Apparently I'm supposed to put these questions in this subforum, so here it goes.
    I've tried this problem several times in different ways and never managed to prove it.

    Prove:
    --sin2x--= cot2x
    1-cos2x

    These are double angles, not squared.
    Thanks for helping.
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  2. #2
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    Hello, tallguywhoplaysguitar!

    There must be a typo.
    As written, the statement is not an identity.


    We need these two identities:

    . . \sin2x \:=\:2\sin x\cos x

    . . \sin^2\!x \:=\:\frac{1-\cos2x}{2} \quad\Rightarrow\quad 1 - \cos2x \:=\:2\sin^2\!x



    Prove: . \frac{\sin2x}{1 -\cos2x} \: =\:{\color{red}\cot x}

    We have: . \frac{\sin2x}{1-\cos2x} \;\;=\;\;\frac{2\sin x\cos x}{2\sin^2x} \;\;=\;\; \frac{\cos x}{\sin x} \;\;=\;\;\cot x

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  3. #3
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    Quote Originally Posted by tallguywhoplaysguitar View Post

    Prove:
    --sin2x--= cot2x
    1-cos2x
    \frac{2\sin x \cos x}{1-\cos^2x + \sin^2x}= \frac{2\sin x \cos x}{2 \sin^2x}  = \frac{\cos x}{\sin x}= \cot x \neq \cot 2x

    Are you good write question?

    p.s. (22: 59 h) Sory, I now see answer of Soroban.
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  4. #4
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    yah, my teacher either made a mistake or wanted us to prove that it doesn't equal cot2x. Thank you very much
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