Math, asked by mohinipasumarti, 1 year ago


Sin2A+sin2B+sin2C=4sinAsinBsinC

Answers

Answered by 1RADHIKAA1
12
sin2A + sin 2B +sin2C 

= 2 sin(A+B)cos(A-B) + 2sinC cosC 

=2sinC cos(A-B)+2sinC cosC 

=2sinC (cos(A-b) + cos C) 

=2sin C(cos(A-B) - cos(A+B) ) 

= 2sinC . 2sin A sin B 

=4 sinA sin B sin C
Answered by karanamsaiganesh
4
sin2a+sin2b+sin2c
2sin(a+b)cos(a-b)+sin2c
2sinccos(a-b)+2sinccosc
2sinc(cos(a-b)+cosc)
2sinc(cos(a-b)+sin(a+b))
2sinc(2sinasinb)
4sinasinbsinc
Similar questions