Given A = 60° and B = 30°, prove that :

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

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

L.H.S. = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

∴ L.H.S. = R.H.S.

Hence, sin (A + B) = sin A cos B + cos A sin B.