Question

Prove that sin^2 A+ sin^2B - sin^2C = 2 sinA× sinB× sinC​

Answer 1

Answer:

The given conditional identity cannot be proved. It should be "If A + B + C = 180⁰, then prove that sin²A + sin²B - sin²C = 2*sin(A)*sin(B)*cos(C)" and not as equal to 2*sin(A)*sin(B)*sin(C). Kindly verify your statement.

ii) Anyway let me give the solution for sin²A + sin²B - sin²C = 2*sin(A)*sin(B)*cos(C)

iii) By multiple angle identity, sin²A + sin²B = {1 - cos(2A)}/2 + {1 - cos(2B)}/2 &

sin²C = 1 - cos²C

==> sin²A + sin²B - sin²C = (1/2) + (1/2) - (1/2){cos(2A) + cos(2B)} - 1 + cos²C

= -(1/2){2cos(A+B)*cos(A-B)} + cos²C

= -(1/2){-2cos(C)*cos(A-B)} + cos²C

[Since A + B + C = 180 deg, A + B = 180 - C; ==> cos(A+B) = -cos(C)]

= cos(C)[cos(A-B) + cos(C)] = cos(C)[cos(A - B) - cos(A + B)]

= cos(C)[2sin(A)*sin(B)] = 2*sin(A)*sin(B)*cos(C)