Question

If $A={45}^{\circ }$, verify that:
$\sin 2A=2\sin A\cos A$

Answer 1

$A={45}^{\circ }$ then $2A={90}^{\circ }$

$\sin 2A=\sin {90}^{\circ }$

$RHS$

$2\sin A\cos B=2\sin {45}^{\circ }\cos {45}^{\circ }=2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}$

$2\sin A\cos B=1$

$LHS$:

$\sin {90}^{\circ }=1$

$L.H.S.=R.H.S.$