# How do you verify 8sin²xcos²x = 1-cos(4x)?

Mar 30, 2018

See below:

#### Explanation:

We know

$\sin \left(2 x\right) = 2 \sin x \cos x$

and

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

So

$8 {\sin}^{2} x {\cos}^{2} x = 2 \left(4 {\sin}^{2} x {\cos}^{2} x\right) = 2 {\left(2 \sin x \cos x\right)}^{2}$
$q \quad = 2 {\sin}^{2} \left(2 x\right) = 1 - \cos \left(2 \times 2 x\right) = 1 - \cos \left(4 x\right)$

Mar 30, 2018

Please see below.

#### Explanation:

We have:

$8 {\sin}^{2} {\cos}^{2} x = 1 - \cos \left(2 x + 2 x\right)$

$8 {\sin}^{2} x {\cos}^{2} x = 1 - \left(\cos \left(2 x\right) \cos \left(2 x\right) - \sin \left(2 x\right) \sin \left(2 x\right)\right)$

$8 {\sin}^{2} x {\cos}^{2} x = 1 - \left({\cos}^{2} \left(2 x\right) - {\sin}^{2} \left(2 x\right)\right)$

$8 {\sin}^{2} x {\cos}^{2} x = 1 - {\cos}^{2} \left(2 x\right) + {\sin}^{2} \left(2 x\right)$

$8 {\sin}^{2} x {\cos}^{2} x = {\sin}^{2} \left(2 x\right) + {\sin}^{2} \left(2 x\right)$

$8 {\sin}^{2} x {\cos}^{2} x = 2 {\sin}^{2} \left(2 x\right)$

Now recall that $\sin \left(2 x\right) = 2 \sin x \cos x$, therefore ${\sin}^{2} \left(2 x\right) = 4 {\sin}^{2} x {\cos}^{2} x$.

$8 {\sin}^{2} x {\cos}^{2} x = 2 \left(4 {\sin}^{2} x {\cos}^{2} x\right)$

$8 {\sin}^{2} x {\cos}^{2} x = 8 {\sin}^{2} x {\cos}^{2} x$

$L H S = R H S$

As required.

Hopefully this helps!