Question

sin2a+sin2b+sin2c=4cosacosbcosc​

Answer 1

Answer:

sin(2A) + sin(2B) - sin(2C) = 2sin(A + B)*cos(A - B) - 2*sin(C)*cos(C) 

wkt for a triangle  A + B + C = 180°, A + B = 180° - C 

==> sin(A + B) = sin(180° - C) = sin(C)  

Also cos(C) = cos{180° - (A + B)} = -cos(A + B)  

Thus , sin(2A) + sin(2B) - sin(2C) = 2sin(C)*cos(A - B) - 2*sin(C)*cos(C)  

= 2sin(C){cos(A - B) - cos(C)}  

= 2sin(C){cos(A - B) + cos(A + B)}  

= 2sin(C)*2cos(A)*cos(B) [Identity]  

= 4*cos(A)*cos(B)*sin(C)  

Thus sin(2A) + sin(2B) - sin(2C) = 4*cos(A)*cos(B)*sin(C)