If A = B = 60°, then prove that :(i) cos(A – B) = cos A cos B + s

1.	If A = B = 60°, then prove that :(i) cos(A – B) = cos A cos B + sin A sin B(ii) sin(A – B) = sin A cos B – cos A sin B
Answer» (i) A = B = 60° (Given) LH.S. cos(A – B) = cos(60° – 60°) = cos 0° = 1 R.HS. cosA cosB + sinA sinB = cos 60° cos 60° + sin 60° sin 60° cos² 60° + sin² 60° = 1 [ sin² θ + cos² θ = 1] Thus, L.H.S. R.H.S. (ii) sin(A – B) = sin A cos B – cos A sin B L.H.S. sin(A – B) = sin(60° – 60°) = sin 0° = 0 R.H.S. sin A cos B – cos A sin B = sin 60° cos 60° – cos 60° sin 60° = 0 Thus, L.H.S. = R.H.S.

If A = B = 60°, then prove that :(i) cos(A – B) = cos A cos B + sin A sin B(ii) sin(A – B) = sin A cos B – cos A sin B

Answer»

(i) A = B = 60° (Given)

LH.S. cos(A – B) = cos(60° – 60°)

= cos 0° = 1

R.HS. cosA cosB + sinA sinB

= cos 60° cos 60° + sin 60° sin 60°

cos² 60° + sin² 60° = 1 [ sin² θ + cos² θ = 1]

Thus, L.H.S. R.H.S.

(ii) sin(A – B) = sin A cos B – cos A sin B

L.H.S. sin(A – B) = sin(60° – 60°)

= sin 0° = 0

R.H.S. sin A cos B – cos A sin B

= sin 60° cos 60° – cos 60° sin 60°

= 0

Thus, L.H.S. = R.H.S.