Question

if A +B + C = Π then prove that, sin^2A + sin^2B - sin^2C = 2sinA.sinB.cosC

Answer 1

Answer:-

Given:-

A + B + C = π

⟹ A + B = π - C -- equation (1).

Applying cos both sides we get,

⟹ cos (A + B) = cos (π - C)

using cos (A + B) = cos A cos B - sin A sin B and cos (π - θ) = - cos θ we get,

⟹ cos A cos B - sin A sin B = - cos C -- equation (2)

Again from equation (1),

Applying sin both sides we get,

⟹ sin (A + B) = sin (π - C)

using sin (A + B) = sin A cos B + cos A sin B and sin (π - θ) = sin θ we get,

⟹ sin A cos B + cos A sin B = sin C

Squaring both sides we get,

⟹ (sin A cos B + cos A sin B)² = sin² C

using (a + b)² = a² + b² + 2ab in LHS we get,

⟹ sin² A cos² B + cos² A sin² B + 2 sin A cos B * cos A sin B = sin² C

using the identity cos² θ = 1 - sin² θ in LHS we get,

⟹ sin² A (1 - sin² B) + (1 - sin² A) sin² B + 2 sin A cos A sin B cos B = sin² C

⟹ sin² A - sin² A sin² B + sin² B - sin² A sin² B + 2 sin A cos A sin B cos B = sin² C

⟹ sin² A + sin² B - 2 sin² A sin² B + 2 sin A sin B cos A cos B = sin² C

⟹ sin² A + sin² B - sin² C = 2 sin² A sin² B - 2 sin A sin B cos A cos B

⟹ sin² A + sin² B - sin² C = 2 sin A sin B (sin A sin B - cos A cos B)

⟹ sin² A + sin² B - sin² C = - 2 sin A sin B (cos A cos B - sin A sin B)

⟹ sin² A + sin² B - sin² C = - 2 sin A sin B ( - cos C)

[ ∵ From equation (2) ]

⟹ sin² A + sin² B - sin² C = 2 sin A sin B cos C

Hence, Proved.