Question

A+B+C=180 then sin2A+sin2B+sin2C

Answer 1

heya friends

here is ur answer ...

Sin2A+sin2B+sin2C

2×sin(2A+2B)/2×cos(2A-2B)/2+sin2C

=>2sin(180°-(A+B)×cos(A-B)+2sinC×cosC

=>2sinC×cos(A-B)+2sinC×cosC

=>2sinC[cos(A-B)+cosC]

=>2sinC[cos(A-B)×cos(180°-(A+B)]

=>2sinC[2sin(A-B+A+B)/2×sin(A+B-A+B)]

=>2sinC[2sinA×sinB]

=>4sinA×sinB×sinC...Ans ..

hope it helps you..

@Rajukumar

Answer 2

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