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

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59
Alan P. Share
Jun 2, 2017

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

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

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

Then teach the underlying concepts
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16
Jim H Share
Apr 22, 2015

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

Here's one:

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

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

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

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

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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15
Tom Share
Apr 22, 2015

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

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

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