trigonometry - $sinalpha + sinbeta + singamma = 4cos{frac{alpha}{2}}cos{frac{beta}{2}}cos{frac{gamma}{2}}$ when $alpha + beta + gamma = pi$

Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle)

Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$

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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 $\sin\frac{\alpha}{2}$?

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Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).

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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
$$
\sin\alpha+\sin\alpha\cos\beta+\sin\beta+\sin\beta\cos\alpha
=
\sin\alpha+\sin\beta+\sin(\alpha+\beta)=
\sin\alpha+\sin\beta+\sin\gamma
$$