Question

If A = ${60}^{\circ }$ and B = ${30}^{\circ }$ then cos (A + B) = cos A cos B - sin A sin B.

If above statement is true then write 1 and if false then write 0.

Answer 1

$A={60}^{\circ }$ and $B={30}^{\circ }$
Now, $\cos (A+B)=\cos A\cos B-\sin A\sin B$
L.H.S.: $\cos (A+B)=\cos ({60}^0+{30}^0)=\cos {90}^0=0$
R.H,S $\cos A\cos B-\sin A\sin B$

= $\cos {60}^0\cos {30}^0-\sin {60}^0\sin {30}^0$

= $\frac{\sqrt{3}}{2}\frac{1}{2}-\frac{1}{2}\frac{\sqrt{3}}{2}$

= $\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

= $0$
Hence, $L.H.S=R.H.S$

The given statement is true.