In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$

To prove $$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$

My approach :

$$ \begin{align} \text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt] & = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt] & = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt] & =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\} \end{align} $$

Unable to move further please help. thanks.

Note that $$\sin\frac{C}{2}=\cos\frac{A+B}{2}$$ and now use following formulas $$\sin p+\sin q=2\sin\frac{p+q}2\cos\frac{p-q}2$$ $$\cos p-\cos q=-2\sin\frac{p+q}2\sin\frac{p-q}2$$ to obtain $$\begin{align}\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}&=\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}\\&=2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+1-2\sin^2\frac{A+B}{4}\\ &=2\sin\frac{A+B}{4}\left(\cos\frac{A-B}{4}-\sin\frac{A+B}{4}\right)+1\\ &=2\sin\frac{A+B}{4}\left(\cos\frac{A-B}{4}-\cos\left(\frac{\pi}2-\frac{A+B}{4}\right)\right)+1\\ &=4\sin\frac{\pi-A}4\sin\frac{\pi-B}4\sin\frac{\pi-C}4+1\end{align}$$

I will prove,

$$\sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} - \sin \frac{\pi}{2}= 4\sin \frac{\pi -A}{4}\sin\frac{\pi -B}{4} \sin\frac{\pi-C}{4}$$

Let $z_1=e^{iA/2}$, $z_2=e^{iB/2}$, $z_3 = e^{iC/2}$, $z_4=e^{i\pi/2}=i$

If $z_0=e^{i\theta}$, $\sin(\theta)= \frac1{2i}\left(z_0 - \frac1{z_0}\right)$

Therefore, LHS,

$$\frac{1}{2i}\left(z_1-\frac{1}{z_1}+z_2-\frac{1}{z_2}+z_3-\frac{1}{z_3} -z_4+\frac{1}{z_4}\right)$$

Substituting, $z_4=i$,

$$\frac{1}{2i}\left(z_1-\frac{1}{z_1}+z_2-\frac{1}{z_2}+z_3-\frac{1}{z_3} -2i\right)$$

Now for the RHS,

$\sin\left(\frac{\pi-A}{4}\right) = \sin((\pi/2-A/2)/2) = \frac{1}{2i}(t-\frac{1}{t})$ where $t=\sqrt{\frac{z_4}{z_1}}$

Similarly for the other two terms. Multiply and we are done.