In any AABC, prove that sin2A + sin2B – sin2C = 4 Cos A Cos B sin C

1.	In any AABC, prove that sin2A + sin2B – sin2C = 4 Cos A Cos B sin C
Answer» LHS = sin2A + sin 2B -sin 20 = 2sin (A+ B). Cos (A – B) – 2 sin C Cos C[ using transformation formula] = 2sin C. Cos (A – B) – 2sin C. Cos C = 2sin C[Cos (A – B) – Cos C] = 2 sin C [Cos (A – B)+ Cos (A + B)] ∵ Cos (A + B) = – Cos C = 2 sin C. 2.cos A. Cos B = 4 Cos A. sin C. Cos B [∵ CosC + CosD = 2cos\(\frac{C + D}{2}cos\frac{C - D}{2}]\) = 4 cos A. Cos B. sin C = RHS ∴ sin 2A + sin 2 B sin 2 C = 4 Cos A. Cos B. sin C.

Answer»

LHS = sin2A + sin 2B -sin 20

= 2sin (A+ B). Cos (A – B) – 2 sin C Cos C[ using transformation formula]

= 2sin C. Cos (A – B) – 2sin C. Cos C

= 2sin C[Cos (A – B) – Cos C]

= 2 sin C [Cos (A – B)+ Cos (A + B)]

∵ Cos (A + B) = – Cos C

= 2 sin C. 2.cos A. Cos B

= 4 Cos A. sin C. Cos B [∵ CosC + CosD = 2cos\(\frac{C + D}{2}cos\frac{C - D}{2}]\)

= 4 cos A. Cos B. sin C = RHS

∴ sin 2A + sin 2 B sin 2 C = 4 Cos A. Cos B. sin C.

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In any AABC, prove that sin2A + sin2B – sin2C = 4 Cos A Cos B sin C

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