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

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

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

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

= cos 90° = 0

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

= cos 60° cos 30° - sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt3}{2} - \dfrac{\sqrt3}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{\sqrt3}{4} - \dfrac{\sqrt3}{4}\\[1em] = 0$

∴ L.H.S. = R.H.S.

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