Ask Me Help Desk - Verifying trigonometric identities solver

Ok...

$sinA + sinB + sinC \equiv sin(A+B+C) + 4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2})$

I think it's better if we start on the right part.

$sin(A+B+C) + 4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2})$

Breaking sin(A+B+C);

$sin(A+B+C) = sin(A+B)cosC + cos(A+B)sinC$

$= (sinAcosB + cosAsinB)cosC + (cosAcosB - sinAsinB)sinC$

$= sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC$

Breaking the other part, using the identity $sinA = 2sin(\frac{A}{2})cos(\frac{A}{2})$

$4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2}) = 4(\frac{sin(A+B)}{2cos(A+B)})(\frac{sin(B+C)}{2cos (B+C)})(\frac{sin(A+C)}{2cos(A+C)})$

$= \frac12 (\frac{sinAcosB+ cosAsinB}{cosAcosB - sinAsinB})(\frac{sinBcosC+cosBsinC}{cosBcosC-sinBsinC})(\frac{sinAcosC+cosAsinC}{cosAcosC-sinAsinC})$

Phew! I give up here. That's terrifically long. I expanded giving 8 dissimilar terms in the numerator and denominator (actually, there are two sinAcosAsinBcosBsinCcosC which becomes 2sinAcosAsinBcosBsinCcosC in the numerator) . Anyone wants to continue? :o