Question

If $A={30}^0$, then
$\sin 3A=4\sin A\sin ({60}^0-A)\sin ({60}^0+A)$

If true enter 1 else 0

Answer 1

Given, $A={30}^{\circ }$
To Prove: $\sin 3A=4\sin A\sin ({60}^0-A)\sin ({60}^0+A)$
L.H.S: $sin3A=sin{90}^0=1$
R.H.S: $4\sin A\sin ({60}^0-A)\sin ({60}^0+A)$
= $4\sin {30}^0\sin ({60}^0-{30}^0)\sin ({60}^0+{30}^0)$
= $4\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(1\right)$
= $1$