# Proving double angles

#### tallguywhoplaysguitar

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

MHF Hall of Honor
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: . $$\displaystyle \frac{\sin2x}{1-\cos2x} \;\;=\;\;\frac{2\sin x\cos x}{2\sin^2x} \;\;=\;\; \frac{\cos x}{\sin x} \;\;=\;\;\cot x$$

#### veljko

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

Are you good write question?

p.s. (22: 59 h) Sory, I now see answer of Soroban.

#### tallguywhoplaysguitar

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