Question

If A+B+C=270 THEN sin2A+sin2B-sin2C=-4sinAsinBsinC

Answer 1

sin2A + sin 2B +sin2C = 2 sin(A+B)cos(A-B) + 2sinC cosC =2sinC cos(A-B)+2sinC cosC =2sinC (cos(A-b) + cos C) =2sin C(cos(A-B) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C

Answer 2

Given : A+B+C=270

So ,

sin2A+sin2B-sin2C

= 2 cos (A + C) sin (A – C) + 2 sinB cosB

= 2 cos (270 – B) sin (A – C) + 2 sinB cosB

= 2 sin B ( – sin (A – C) + cos (270 – ( A+C))

= 2 sin B (  – sin ( A – C) – sin (A + C))

= – 4 sinA sinB sinC

Used formulas :

$sin(x) + sin(y) = 2[cos(\frac{x+y}{2} ) cos(\frac{x-y}{2} )]$

sin2x = 2sin(x)cos(x)

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