# Verify Cos^2x-cos^4x=sin^2x-sin^4x?

May 10, 2018

Please see the proof below

#### Explanation:

We need

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

Therefore,

$L H S = {\cos}^{2} x - {\cos}^{4} x$

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

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

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

$= {\sin}^{2} x - {\sin}^{4} x$

$= R H S$

$Q E D$

May 10, 2018

Kindly refer to Explanation.

#### Explanation:

${\cos}^{2} x - {\cos}^{4} x$,

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

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

$= {\sin}^{2} x - {\sin}^{4} x$, as desired!

Aliter :

We have, ${\cos}^{4} x - {\sin}^{4} x$,

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

$= {\cos}^{2} x - {\sin}^{2} x , i . e . ,$

${\cos}^{4} x - {\sin}^{4} x = {\cos}^{2} x - {\sin}^{2} x , \text{ equivalently, }$

${\sin}^{2} x - {\sin}^{4} x = {\cos}^{2} x - {\cos}^{4} x$.

May 10, 2018

${\cos}^{2} x = 1 - {\sin}^{2} x$
${\cos}^{4} x = {\left(1 - {\sin}^{2} x\right)}^{2} = 1 - 2 {\sin}^{2} x + {\sin}^{4} x$
${\cos}^{2} x - {\cos}^{4} x = 1 - {\sin}^{2} x - 1 + 2 {\sin}^{2} x - {\sin}^{4} x$
$= \left(1 - 1\right) + \left(2 {\sin}^{2} x - {\sin}^{2} \left(x\right)\right) - {\sin}^{4} x$