# Double angle Prove question

• May 14th 2010, 05:42 PM
tallguywhoplaysguitar
Double angle Prove question
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.
• May 14th 2010, 09:06 PM
Soroban
Hello, tallguywhoplaysguitar!

We need some double-angle identities:

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

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

Quote:

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

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

• May 15th 2010, 11:48 AM
tallguywhoplaysguitar
thank you