if sinA+sinB= root3(cos B-cos A) then sin3B+sin3A =

Hello Hemanth ....

sinA + sinB = root(3) cosB - root(3)cosA

Divide by 2 both side

sinA / 2  + sinB / 2  = root(3).cosB/2 - root(3).cosA/2

1/2 . sinA  + root(3)/2 .cosA  =  root(3)/2 . cos(B)  - 1/2 . sin(B)

sin(A + 60)  = cos(B + 30)  =>  sin(A+60) = sin ( 60 - B )

so finaly  A + 60 = 60 - B

so A = -B

Therefore   Sin3B + sin3A  =  0  

I hope i have answered you :)

Regards

Yagya

askiitians_expert