Question

f A = B = 45°, show 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

Step-by-step explanation:

(i) L.H.S= sin (A - B) = sin ( 45 - 45) = 0

R.H.S = Sin 45° cos 45° - cos 45° sin 45°

=

$\frac{1}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} }$

= 0 = L.H.S, Proved

(ii) L.H.S= cos (A + B) = Cos( 45 + 45) =cos(90°)

= 0

R.H.S = cos 45° cos 45° - sin45° sin 45°

=

$\frac{1}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } \times \frac{1}{ \sqrt{2} }$

= 0 = L.H.S ,

proved.