How do you verify the identity: 1 - cos 2x = tan x sin 2x?

Apr 16, 2018

Here's how I proved it:

Explanation:

$1 - \cos 2 x = \tan x \sin 2 x$

I'll prove using the right hand side of the equation.

From the double angle identities, $\sin 2 x = 2 \sin x \cos x$:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \sin \frac{x}{\cos} x \cdot 2 \sin x \cos x$

Combine by multiplying:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{2 {\sin}^{2} x \cancel{\cos x}}{\cancel{\cos x}}$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad = 2 {\sin}^{2} x$

From the Pythagorean Identities, ${\sin}^{2} x = 1 - {\cos}^{2} x$:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = 2 \left(1 - {\cos}^{2} x\right)$

Simplify:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = 2 - 2 {\cos}^{2} x$

Factor out a $- 1$:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = - 1 \left(- 2 + 2 {\cos}^{2} x\right)$

Since we need $2 {\cos}^{2} x - 1$ to get $\cos 2 x$, let's rewrite it so that we can get that:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = - 1 \left(- 1 + 2 {\cos}^{2} x - 1\right)$

From the double angle identities, ${\cos}^{2} x - 1 = \cos 2 x$:
$\quad \quad \quad \quad \quad \quad \quad \quad \quad = - 1 \left(- 1 + \cos 2 x\right)$

Finally:
$1 - \cos 2 x = 1 - \cos 2 x$

We have proved that $1 - \cos 2 x = \tan x \sin 2 x$.

Hope this helps!