Question

if a+b+c=π then sin2a+sin2b-sin2c= 4cosAcosBcosC
prove this using only LHS

Answer 1

Step-by-step explanation:

=> A + B + C = π

=> sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C

From the double angle formula:

sin 2Θ = 2 sin Θ cos Θ

=> sin 2A + sin 2B - sin 2C = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

Since A + B + C = π ;

> A is a supplement angle of ( B + C )

> B is a supplement angle of ( A + C )

> C is a supplement angle of ( A + B )

sin 2A + sin 2B - sin 2C

... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

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

From the Sum of Angle Identity:

sin ( α + ß ) = sin α cos ß + cos α sin ß

sin 2A + sin 2B - sin 2C

... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

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

... = 2 ( sin B cos C + cos B sin C ) cos A

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

..... ..... – 2 ( sin A cos B + cos A sin B ) cos C

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

..... ..... + 2 sin A cos B cos C + 2 cos A cos B sin C

..... ..... – 2 sin A cos B cos C – 2 cos A sin B cos C

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

... = 4 cos A cos B sin C

HOPE IT HELPS YOU,

THANK YOU. ☺️

Answer 2

Answer:

Step-by-step explanation:

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