Question

sin2a+sin2B+sin2C=4cosAsinBcosC​

Answer 1

Step-by-step explanation:

A+B+C= π

• A+B= π-C

• sin 2C = 2 sin C cos C

• sin A + sin B = 2 sin (A+B / 2 ) + cos (A - B / 2 )

LHS = 2 sin (2A + 2B / 2 ) cos ( 2A - 2B / 2) + 2 sin C cos C

2 sin (A + B ) cos ( A - B ) + 2 sin C cos C

2 sin ( π - C ) cos ( A - B ) + 2 sin C cos C

2 sin C cos ( A - B ) + 2 sin C cos C

2 sin C ( cos (A - B ) + cos C )

2 sin C ( cos (A-B ) + cos (π- ( A+B ))

2 sin C ( cos ( A-B ) - cos ( A+B ))

2 sin C ( 2 sin A sin B)

4 sin A sin B sin C