Question

Prove $\sqrt{2}.\sin (\frac{\pi }{4}-\theta )=\cos \theta -\sin \theta$

Answer 1

We know that

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\sin (\frac{\pi }{4}-\theta )=\sin \frac{\pi }{4}\cos \theta -\cos \frac{\pi }{4}\sin \theta$

$\sin (\frac{\pi }{4}-\theta )=\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta$

$\sqrt{2}\sin (\frac{\pi }{4}-\theta )=\cos \theta -\sin \theta$

Hence proved