Question

Prove that
$\sin ({45}^o+A)+\sin ({45}^o-A)=\sqrt{2}\cos A$

Answer 1

L.H.S

$=\sin ({45}^0+A)+\sin ({45}^0-A)$

$=\sin {45}^0\cos A+\sin A\cos {45}^0+\sin {45}^0\cos A-\sin A\cos {45}^0$

$=2\sin {45}^0\cos A$

$=\frac{2}{\sqrt{2}}\cos A$

$=\sqrt{2}\cos A$

R.H.S

Hence, proved.