Question

If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B

Answer 1

A = 60^o and B = 30^o
Now, A + B = 60^o + 30^o​ = 90^o
Also, A − B = 60^o − 30^o = 30^o

(i) sin (A + B) = sin 90^o = 1
sin A cos B + cos A sin B = sin 60^o cos 30^o + cos 60^o sin 30^o
= $\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=1$
∴ sin (A + B) = sin A cos B + cos A sin B


(ii) cos (A + B) = cos 90^o = 0
cos A cos B − sin A sin B =cos 60^o cos 30^o − sin 60^o sin 30^o $=\left(\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$
∴​ cos (A + B) = cos A cos B − sin A sin B