# Double angle Prove question

#### tallguywhoplaysguitar

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.

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#### Soroban

MHF Hall of Honor
Hello, tallguywhoplaysguitar!

We need some double-angle 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 that: .$$\displaystyle \frac{\sin2x}{1-\cos2x} \:=\:{\color{red}\cot x}$$

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

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thank you