Question

Prove that `sina+ sinb+ sinc- sin(a+b+c)= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2)` ?

Answer 1

Explanation:

`LHS=sina+ sinb+ sinc- sin(a+b+c)`

`=sina+ sinb+ sinc- sin((a+b+c)`

`=2sin((a+b)/2)cos((a-b)/2)- 2cos((a+b+2c)/2)sin((a+b)/2)]`

`=2sin((a+b)/2)[cos((a-b)/2)- cos((a+b+2c)/2)]`

`=2sin((a+b)/2)xx2sin((a+b+2c)/4-(a-b)/4)sin((a+b+2c)/4+(a-b)/4)`

`= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2)=RHS`