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

Apr 22, 2015

From the basic definitions of $\sin$ and $\cos$ and the Pythagorean Theorem
${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$

or
${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

So
${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

$= \left(1 - {\sin}^{2} \left(x\right)\right) - {\sin}^{2} \left(x\right)$

$= 1 - 2 {\sin}^{2} \left(x\right)$

Apr 22, 2015

There are many acceptable (correct) ways to do this.

Here's one:

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

$1 - 2 {\sin}^{2} x = \left({\sin}^{2} x + {\cos}^{2} x\right) - 2 {\sin}^{2} x$

$\textcolor{w h i t e}{\text{ssssssssssss}}$ $= {\cos}^{2} x + {\sin}^{2} x - {\sin}^{2} x - {\sin}^{2} x$

$\textcolor{w h i t e}{\text{ssssssssssss}}$ $= {\cos}^{2} x - {\sin}^{2} x$

Apr 22, 2015

${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

$= \left({\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)\right) - \left({\sin}^{2} \left(x\right) + {\sin}^{2} \left(x\right)\right) = 1 - 2 {\sin}^{2} \left(x\right)$

Jul 22, 2018

See below:

#### Explanation:

Recall the Pythagorean Identity

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

Which can be manipulated into this form:

$\textcolor{b l u e}{{\cos}^{2} x = 1 - {\sin}^{2} x}$

In our equation, we can replace ${\cos}^{2} x$ with this to get

$\textcolor{b l u e}{1 - {\sin}^{2} x} - {\sin}^{2} x$, which simplifies to

$1 - 2 {\sin}^{2} x$. We have just verified the identity

$\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} {\cos}^{2} x - {\sin}^{2} x = 1 - 2 {\sin}^{2} x \textcolor{w h i t e}{\frac{2}{2}} |}}$

With the use of the Pythagorean Identity.

Hope this helps!