Prove that:---=cot(x)

sin(2x)

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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- May 14th 2010, 04:42 PMtallguywhoplaysguitarDouble angle Prove question
**Prove that:**---=cot(x)

sin(2x)

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, 08:06 PMSoroban
Hello, tallguywhoplaysguitar!

We need some double-angle identities:

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

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

Quote:

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 $

- May 15th 2010, 10:48 AMtallguywhoplaysguitar
thank you