If A = B = 45°, show that :

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

(ii) cos(A + B) = cosA cosB - sin A sin B

(i) Left Hand Side :

sin(A - B) = sin (45° - 45°) = sin 0

= 0

Right Hand Side :

sin A cos B - cos A sin B

= sin 45° cos 45° - cos 45° sin 45°

= $\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}$

= 0.

Hence, proved that sin(A - B) = sin A cos B - cos A sin B.

(ii) Left Hand Side :

cos(A + B) = cos (45° + 45°) = cos 90°

= 0.

Right Hand Side :

cos A cos B - sin A sin B

= cos 45° cos 45° - sin 45° sin 45°

= $\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}$

= $\dfrac{1}{2} - \dfrac{1}{2}$

= 0.

Hence, proved that cos(A + B) = cosA cosB - sin A sin B.