Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle)
Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$
There is already a solution on Math-SE, however I want to avoid using the sum-to-product identity because technically the book I go by hasn't covered it yet.
So, is there a way to prove it with identities only as advanced as $\sin\frac{\alpha}{2}$?
Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).
Best Answer
You may go the other way around: $$ \cos\frac{\gamma}{2}=\cos\frac{\pi-\alpha-\beta}{2}= \sin\frac{\alpha+\beta}{2}= \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ so the right hand side becomes $$ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \sin\frac{\alpha}{2}\cos\frac{\beta}{2}+ 4\cos\frac{\alpha}{2}\cos\frac{\beta}{2} \cos\frac{\alpha}{2}\sin\frac{\beta}{2} $$ Recalling the duplication formula for the sine we get $$ 2\sin\alpha\cos^2\frac{\beta}{2}+2\sin\beta\cos^2\frac{\alpha}{2} $$ and we can recall $$ 2\cos^2\frac{\delta}{2}=1+\cos\delta $$ to get $$ \sin\alpha+\sin\alpha\cos\beta+\sin\beta+\sin\beta\cos\alpha = \sin\alpha+\sin\beta+\sin(\alpha+\beta)= \sin\alpha+\sin\beta+\sin\gamma $$