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.
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 $