Question

If $A={30}^0$, then the value of

$\sin ({60}^0+A)-\sin ({60}^0-A)=\sin A$

If true then enter $1$ and if false then write $0$

Answer 1

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