Hence proved.
(iv) sin2 B = sin2 A + sin2 (A-B)– 2sin A cos B sin (A – B)
Let us consider RHS:
sin2A + sin2 (A -B) – 2 sinA cos B sin (A – B)
sin2A + sin (A -B) [sin (A –B) – 2 sin Acos B]
We know that, sin (A –B) = sin A cos B – cos A sin B
So,
sin2A + sin (A -B) [sin A cos B – cos A sinB – 2 sin A cos B]
sin2A + sin (A -B) [-sin A cos B – cos Asin B]
sin2A – sin (A -B) [sin A cos B + cos A sinB]
We know that, sin (A +B) = sin A cos B + cos A sin B
So,
sin2A – sin (A – B) sin (A + B)
sin2 A – sin2 A + sin2 B
sin2 B
= LHS
∴ LHS = RHS
Hence proved.
(v) cos2 A + cos2 B – 2cos A cos B cos (A + B) = sin2 (A + B)
Let us consider LHS:
cos2A + cos2B – 2 cos A cos Bcos (A +B)
cos2A + 1 – sin2B – 2 cos A cosB cos (A +B)
1 + cos2A – sin2B – 2 cos Acos B cos (A +B)
We know that, cos2A – sin2B =cos (A +B) cos (A –B)
So,
1 + cos (A +B) cos (A –B) – 2 cos A cos B cos (A+B)
1 + cos (A +B) [cos (A –B) – 2 cos A cos B]
We know that, cos (A – B) = cos A cos B + sin A sin B.
So,
1 + cos (A +B) [cos A cos B + sin A sin B – 2 cosA cos B]
1 + cos (A +B) [-cos A cos B + sin A sin B]
1 – cos (A +B) [cos A cos B – sin A sin B]
We know that, cos (A +B) = cos A cos B – sin A sin B.
So,
1 – cos2 (A + B)
sin2 (A + B)
= RHS
∴ LHS = RHS
Hence proved.
(vi)
∴ LHS = RHS
Hence proved.