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


Hint:
i) Let $\alpha + \beta + \gamma = \pi \Rightarrow $ 
$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ 
ii) Let $\alpha + \beta + \gamma > \pi \Rightarrow $ 
$\sin\alpha + \sin\beta + \sin\gamma > 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ 
iii) Let $\alpha + \beta + \gamma < \pi \Rightarrow $ 
$\sin\alpha + \sin\beta + \sin\gamma < 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ 
For example:
If $\alpha, \beta,\gamma \in \left(0,\frac{\pi}{2}\right] $ then
$$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)= 4\sin(\alpha)\sin(\beta)\sin(\gamma) \Leftrightarrow \alpha+\beta+\gamma=\pi$$
Proof:
1.) $\alpha+\beta+\gamma = \pi$ - equality holds.
Let $\alpha \ge \frac{\pi}{3}$ , then with increasing $\alpha$ decreases the left side and the right side increases $\Rightarrow $ 
if $\alpha+\beta+\gamma > \pi$ equality is not achieved.
2.) $\alpha+\beta+\gamma \le \pi$

$R -$ the radius of the circle circumscribed around $\triangle ABC$
$|AD|=|BD|=|CD|=L \ge R$
$$S_{ACD}+S_{CBD}+S_{BAD} \ge S_{ABC}  $$
$$S_{ACD}+S_{CBD}+S_{BAD}=\frac{1}{2}L^2(\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma))$$
$$S_{ABC}=\frac{|AB|\cdot|BC|\cdot|CA|}{4R}=\frac{2L^3}{R} \cdot\sin(\alpha)\cdot\sin(\beta)\cdot\sin(\gamma) \ge 2L^2\cdot\sin(\alpha)\cdot\sin(\beta)\cdot\sin(\gamma)$$
Then : $\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)= 4\sin(\alpha)\sin(\beta)\sin(\gamma)$ at $ \alpha, \beta,\gamma \in \left(0,\frac{\pi}{2}\right] $ 
is performed only when  : $\alpha+\beta+\gamma=\pi$
