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

[trigonometry]

Prove that in an ABC triangle occurs sin A - sin B + sin C = 4 sin (A/2) cos (B/2) sin (C/2).
I know how to do LHS, but I don't know how to resolve RHS.
I know that there is sin (A/2) = sqrt((1-cos A)/2) and cos (B/2) = sqrt((1+cos B)/2).
Please help...

 

[Deleted member 4993]

Prove that in an ABC triangle occurs sin A - sin B + sin C = 4 sin (A/2) cos (B/2) sin (C/2).
I know how to do LHS, but I don't know how to resolve RHS.
I know that there is sin (A/2) = sqrt((1-cos A)/2) and cos (B/2) = sqrt((1+cos B)/2).
Please help...

I'll do a similar but different problem.
If a+b+c=π = 180°, prove that the identity is true.
sin a + sin b + sin c=4cos(a/2)cos(b/2)cos(c/2)

LHS = sina+sinb +sinc
= 2sin(a+b)/2 * cos(a-b)/2 + sinc
= 2sin(90-c/2)*cos(a-b)/2+2sinc/2*cosc/2
=2cosc/2*cos(a-b)/2 + 2sinc/2*cosc/2
=2cosc/2* { cos(a-b)/2 + sin c/2}
=2cosc/2 {cos(a-b)/2 + cos (a+b)/2)}
=2cosc/2{ 2 cosa/2 * cos b/2}
=4cos a/2 * cosb/2 * cos c/2

Follow a similar logic!