Double angle Prove question

**tallguywhoplaysguitar** · May 14th 2010, 04:42 PM

Prove that:
sin(2x) ---=cot(x)
1-cos(2x)

I've worked it down to simply cosx on the left side, but the other side is very annoying and confusing.

**Soroban** · May 14th 2010, 08:06 PM

Hello, tallguywhoplaysguitar!

We need some double-angle identities:

. . $\sin2x \;=\;2\sin x\cos x$

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

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

We have: . $\frac{\sin2x}{1-\cos2x}\;\;\begin{array}{c}\to \\ \to \end{array} \;\; \frac{2\sin x\cos x}{2\sin^2\!x} \;\;=\;\;\frac{\cos x}{\sin x} \;\;=\;\;\cot x$

**tallguywhoplaysguitar** · May 15th 2010, 10:48 AM

thank you

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