Question

State true or false

Given $A={60}^0$ and $B={30}^0$, then

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

If true enter 1 else 0

Answer 1

$A={60}^{\circ }$ and $B={30}^{\circ }$
Now, $\sin (A+B)=\sin A\cos B+\cos A\sin B$
L.H.S: $\sin (A+B)=\sin ({60}^0+{30}^0)=\sin {90}^0=1$
R.H.S: $\sin A\cos B+\cos A\sin B$
= $\sin {60}^0\cos {30}^0+\cos {60}^0\sin {30}^0$
= $\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
= $\frac{3}{4}+\frac{1}{4}$
= $1$