$1 - 2sin^2 \left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right ) = sin \theta$

Solution:

LHS

$= 1 - 2sin^2 \left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right )$

[Using formula, cos2A = 1 - 2sin²A]

$= cos 2\left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right )$

$= cos \left ( \dfrac{\pi}{4} × 2 - \dfrac{\theta}{2} × 2 \right )$

$= cos \left ( \dfrac{\pi}{2} - \theta \right )$

$= cos (90° - \theta)$

$= sin \theta$

RHS