26.AB+C=t, prove that sin2B + sin20= 4 cOSA sinB.sin C.

1.	26.AB+C=t, prove that sin2B + sin20= 4 cOSA sinB.sin C.
Answer» Given:A,B,CA,B,Care angles of a triangle, we have thatA+B+C=π Wehave that sin(2A)+sin(2B) =2sin(A+B)cos(A−B) =2sin(π−C)cos(A−B) =2sin(C)cos(A−B)sin⁡(2A)+sin⁡(2B)=2sin⁡(A+B)cos⁡(A−B)=2sin⁡(π−C)cos⁡(A−B)=2sin⁡(C)cos⁡(A−B) Hence, sin(2A)+sin(2B)+sin(2C)=2sin(C)cos(A−B)+2sin(C)cos(C) =2sin(C)(cos(A−B)+cos(C)) =2sin(C)(cos(A−B)+cos(π−(A+B)))=2sin(C)(cos(A−B)−cos(A+B))=2sin(C)×2sin(A)sin(B)=4sin(A)sin(B)sin(C)Hence proved

Answer»

Given:A,B,CA,B,Care angles of a triangle, we have thatA+B+C=π

Wehave that

sin(2A)+sin(2B)

=2sin(A+B)cos(A−B)

=2sin(π−C)cos(A−B)

=2sin(C)cos(A−B)sin⁡(2A)+sin⁡(2B)=2sin⁡(A+B)cos⁡(A−B)=2sin⁡(π−C)cos⁡(A−B)=2sin⁡(C)cos⁡(A−B)

Hence,

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

=2sin(C)(cos(A−B)+cos(C))

=2sin(C)(cos(A−B)+cos(π−(A+B)))=2sin(C)(cos(A−B)−cos(A+B))=2sin(C)×2sin(A)sin(B)=4sin(A)sin(B)sin(C)Hence proved

Discussion