Question

$\sqrt{2}\cdot \cos (\pi /4+A)=\cos A-\sin A$.

Answer 1

$\sqrt{2}\cos (\pi /4+A)=\cos A-\sin A$

$\sqrt{2}(\cos \pi /4\cos A-\sin \frac{\pi }{4}\sin A)$

$\sqrt{2}[\frac{\cos A}{\sqrt{2}}-\frac{\sin A}{\sqrt{2}}]=[\cos A-\sin A]$