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.
 

Soroban

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May 2006
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Hello, tallguywhoplaysguitar!

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


We need these two identities:

. . \(\displaystyle \sin2x \:=\:2\sin x\cos x\)

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



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

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

 
yah, my teacher either made a mistake or wanted us to prove that it doesn't equal cot2x. Thank you very much