# How do you prove cos^2x - sin^2x = 2cos^2x - 1?

Jun 13, 2015

Have a look:

#### Explanation:

Given:

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

we can write it as (taking $- 1$ to the left and ${\cos}^{2} x$ to the right):

$1 - {\sin}^{2} x = - {\cos}^{2} x + 2 {\cos}^{2} x$

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

But ${\sin}^{2} x + {\cos}^{2} x = 1$;
then: $1 - {\sin}^{2} x = {\cos}^{2} x$;
so:

${\cos}^{2} x = {\cos}^{2} x$