# Verify this is an identity? 2 sin^2 (u/2)=sin^2u/(1+cosu)

Apr 3, 2018

$L H S = 2 {\sin}^{2} \left(\frac{u}{2}\right)$

$= \frac{2 {\sin}^{2} \left(\frac{u}{2}\right) \times 2 {\cos}^{2} \left(\frac{u}{2}\right)}{2 {\cos}^{2} \left(\frac{u}{2}\right)}$

$= {\left(2 \sin \left(\frac{u}{2}\right) \cos \left(\frac{u}{2}\right)\right)}^{2} / \left(1 + \cos u\right)$

$= {\sin}^{2} \frac{u}{1 + \cos u} = R H S$

Apr 3, 2018

Proof

#### Explanation:

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

$= \frac{2 {\sin}^{2} \left(\frac{u}{2}\right) \setminus \times 2 {\cos}^{2} \left(\frac{u}{2}\right)}{2 {\cos}^{2} \left(\frac{u}{2}\right)}$

$= {\left(2 \sin \left(\frac{u}{2}\right) \cos \left(\frac{u}{2}\right)\right)}^{2} / \left(2 {\cos}^{2} \left(\frac{u}{2}\right)\right)$

$\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$ and $\cos \left(2 \theta\right) = 2 {\cos}^{2} \theta - 1$. So,

$= {\sin}^{2} \frac{u}{1 + \cos u}$

You can prove the same starting from RHS too.....