If A = B = 45°, show 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 (45° + 45°)

= cos 90°

= 0

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

= cos 45° . cos 45° - sin 45°. sin 45°

$= \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2} - \dfrac {1}{\sqrt2} \times \dfrac {1}{\sqrt2}\\[1em] = \dfrac {1}{2} - \dfrac {1}{2}\\[1em] = 0$

∴ L.H.S. = R.H.S.

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