Question

cosA . sin(B-C)+cosB . sin(C-A)+ cosC . sin(A-B)=0

Answer 1

Answer:

To Prove: cos A . sin ( B - C ) + cos B . sin ( C - A ) + cos C . sin ( A - B ) = 0

We use the the following result,

sin ( x - y ) = sin x . cos y - cos x . sin y

LHS

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

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

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

= 0 + 0 + 0

= 0

RHS

= 0

LHS = RHS

Hence, Proved

Answer 2

use the basic trigonometry formula of sin(A-B)