# How do you prove that  1- cos^2(x/2) = (sin^2x)/(2(1+cosx)) ?

Jun 5, 2018

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

#### Explanation:

so we have
$2 \left(1 + \cos \left(x\right)\right) = 2 \left(2 {\cos}^{2} \left(\frac{x}{2}\right) - 1\right) = 4 {\cos}^{2} \left(\frac{x}{2}\right)$
so we get
$4 {\cos}^{2} \left(\frac{x}{2}\right) \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right)$
and
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
$1 - {\left(2 {\cos}^{2} \left(\frac{x}{2}\right) - 1\right)}^{2}$
$1 - 4 {\cos}^{2} \left(\frac{x}{2}\right) - 1 + 4 {\cos}^{2} \left(\frac{x}{2}\right)$
$4 {\cos}^{2} \left(\frac{x}{2}\right) \left(1 - {\cos}^{2} \left(\frac{x}{2}\right)\right)$

Jun 5, 2018

Please find a Proof in Explanation.

#### Explanation:

Since, $1 - \cos x = 2 {\sin}^{2} \left(\frac{x}{2}\right)$, we have,

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

={cancel((1+cosx))(1-cosx)}/(2cancel((1+cosx)),

$= \frac{1 - \cos x}{2}$,

$= \frac{\cancel{2} {\sin}^{2} \left(\frac{x}{2}\right)}{\cancel{2}}$,

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

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