Show $\sin(a+b)+\sin(a+c)+\sin(b+c)-\sin a-\sin b-\sin c-\sin(a+b+c)=8\sin\frac{a}{2}\sin\frac{b}{2} \sin \frac{c}{2} \cos \frac{a+b+c}{2}$

Solution 1:

Consider that $$\sin(a+b)-\sin c=2\cos\left(\frac{a+b+c}{2}\right)\sin\left(\frac{a+b-c}{2}\right)$$

So if you pair off all the first six terms on the LHS in the same way you get a common factor of $$2\cos\left(\frac{a+b+c}{2}\right)$$

And the other factor is $$\sin\left(\frac{a+b-c}{2}\right)+\sin\left(\frac{b+c-a}{2}\right)+\sin\left(\frac{c+a-b}{2}\right)-\sin\left(\frac{a+b+c}{2}\right),$$ where the last term comes from splitting $\sin(a+b+c)$ into half angles.

This remaining factor can be collected pairwise as $$2\sin\left(\frac b2\right)\cos\left(\frac{a-c}{2}\right)+2\cos\left(\frac{a+c}{2}\right)\sin\left(-\frac b2\right)$$ $$=2\sin\left(\frac b2\right)\left[-2\sin\left(\frac a2\right)\sin\left(-\frac c2\right)\right]$$

And hence the result.