How do you verify (1-cos2x)/ tanx = sin2x?

Aug 6, 2015

$\setminus \frac{1 - \cos 2 x}{\tan x} = \setminus \frac{2 {\sin}^{2} x}{\tan x} = 2 \sin x \cos x = \sin 2 x$

Explanation:

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x = 1 - 2 {\sin}^{2} x$
$\sin 2 x = 2 \sin x \cos x$
$\tan x = \setminus \frac{\sin x}{\cos x}$

$\setminus \frac{1 - \cos 2 x}{\tan x} = \setminus \frac{2 {\sin}^{2} x}{\tan x} = 2 \sin x \cos x = \sin 2 x$

Aug 7, 2015

Starting with $\frac{1 - \cos 2 x}{\tan x} = \sin 2 x$ and using the identity $\frac{1 - \cos 2 x}{2} = {\sin}^{2} x$:

$= \frac{2 {\sin}^{2} x}{\tan x} = \sin 2 x$

Of course, $\tan x = \frac{\sin x}{\cos x}$:

$= \frac{2 \cos x {\sin}^{\cancel{2}} x}{\cancel{\sin x}} = \sin 2 x$

and:

$2 \sin x \cos x = \sin x \cos x + \cos x \sin x$
$= \sin \left(x + x\right) = \sin 2 x$:

$\implies \textcolor{b l u e}{\sin 2 x = \sin 2 x}$