# Question #d3cc2

Jul 10, 2016

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

#### Explanation:

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

$\therefore {\sin}^{2} \frac{x}{1 + \cos x} = 1 - \cos x .$

If we use, the formula, $: 1 - \cos 2 \theta = 2 {\sin}^{2} \left(\frac{\theta}{2}\right)$, the above Exp. can still be expressed as $2 {\sin}^{2} \left(\frac{x}{2}\right) .$

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

Jul 10, 2016

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

#### Explanation:

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

$\therefore {\sin}^{2} \frac{x}{1 + \cos x} = 1 - \cos x .$

If we use, the formula, $: 1 - \cos 2 \theta = 2 {\sin}^{2} \left(\frac{\theta}{2}\right)$, the above Exp. can still be expressed as $2 {\sin}^{2} \left(\frac{x}{2}\right) .$

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