$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ when $\alpha + \beta + \gamma = \pi$ 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).


You can easily solve the question.
What you need to do is to plug 
$$\beta=\pi-(\alpha +\gamma)$$
Now put it in your equation and you would get 
$$ sin(\alpha) + sin(\pi - (\alpha + \gamma)) + sin(\gamma)$$
$$ sin(\alpha) + sin(\alpha + \gamma) + sin(\gamma)$$
Now open $sin(\alpha + \gamma)$  and take $sin(\alpha)$ common from one side and $sin(\gamma)$ for another side and then you would get
$$ sin(\alpha)(1+cos(\gamma)) + sin(\gamma)(1+cos(\alpha))$$
Write $1+cos(\gamma) = 2cos^2(\gamma/2)$ and use the same for $\alpha$ 
Now write $sin(\alpha)=2sin(\alpha/2)cos(\alpha/2)$ and do the same thing for $sin(\gamma)$  also.
Now you just need to take suitable things common out of the terms and use the sine addition identity. Hope it helps.
I didn't do the whole proof as you only wanted some hint regarding that.
