Alternate proof that $\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin(\frac{3A}2)\sin(\frac{3B}2)\sin(\frac{3C}2)$ for $A$, $B$, $C$ the angles of a triangle 
$A$, $B$, $C$ being the angles of a triangle, we need to prove that:
$$\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin\left(\frac{3A}2\right)\sin\left(\frac{3B}2\right)\sin\left(\frac{3C}2\right)\tag*{}$$

The proof which I got from my prof:
$\begin{align*}& \cos 3A + \cos 3B -\cos 3(A+B) \tag{01}\\ & = 1+ \sin 3A \sin 3B -(1-\cos 3A)(1-\cos 3B)\tag{02}\\ &= 1+4\sin\frac{3A}2\sin\frac{3B}2\left(\cos \frac{3A}2\cos\frac{3B}2-\sin \frac{3A}2\sin\frac{3B}2\right)\tag{03}\end{align*}$
(01) We use the fact that $C=\pi-(A+B) $
(02) $\cos3(A+B)$ is expanded using compound angle formula. 1 is added and subtracted. Following that some factorisation is done.
(03) Half angle formula is applied on $\sin 3A$ and $\sin 3B$. The identity $1-\cos x = 2\sin^2\frac x2$ is aplied on likes of $1-\cos 3A$
Now I believe the rest of the procedure is obvious.

I would like to know if it is possible to get the proof done without a lot of algebra. Does complex definition of sine and cosine help us to get it done nicely?


It's easier to work backwards.  Still very much "a lot of algebra", though.
$$1 - 4 \sin(\frac{3A}{2}) \sin(\frac{3B}{2}) \sin(\frac{3C}{2})$$
$$= 1 - 4(\frac{e^{i(3/2)A} - e^{-i(3/2)A}}{2i})(\frac{e^{i(3/2)B} - e^{-i(3/2)B}}{2i})(\frac{e^{i(3/2)C} - e^{-i(3/2)C}}{2i})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)A} - e^{-i(3/2)A})(e^{i(3/2)B} - e^{-i(3/2)B})(e^{i(3/2)C} - e^{-i(3/2)C})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)(A+B)} - e^{i(3/2)(A-B)} - e^{i(3/2)(-A+B)} + e^{i(3/2)(-A-B)})(e^{i(3/2)C} - e^{-i(3/2)C})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)(A+B+C)} - e^{i(3/2)(A+B-C)} - e^{i(3/2)(A-B+C)} + e^{i(3/2)(A-B-C)} - e^{i(3/2)(-A+B+C)} + e^{i(3/2)(-A+B-C)} + e^{i(3/2)(-A-B+C)} - e^{i(3/2)(-A-B-C)})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)\pi} - e^{i(3/2)(\pi-2C)} - e^{i(3/2)(\pi-2B)} + e^{i(3/2)(-\pi+2A)} - e^{i(3/2)(\pi - 2A)} + e^{i(3/2)(-\pi + 2B)} + e^{i(3/2)(-\pi+2C)} - e^{i(3/2)(-\pi)})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)\pi} - e^{i(3/2)\pi}e^{-i3C} - e^{i(3/2)\pi}e^{-i3B} + e^{-i(3/2)\pi}e^{i3A} - e^{i(3/2)\pi}e^{-i3A} + e^{-i(3/2)\pi}e^{i3B} + e^{-i(3/2)\pi}e^{i3C} - e^{-i(3/2)\pi})$$
$$= 1 - \frac{i}{2} ((-i) - (-i)e^{-i3C} - (-i)e^{-i3B} + ie^{i3A} - (-i)e^{-i3A} + ie^{i3B} + ie^{i3C} - i)$$
$$= 1 - \frac{i}{2} (-i + ie^{-i3C} + ie^{-i3B} + ie^{i3A} + ie^{-i3A} + ie^{i3B} + ie^{i3C} - i)$$
$$= 1 + \frac{1}{2} (-2 + e^{i3A} + e^{-i3A} + e^{-i3B} + e^{i3B} + e^{-i3C} + e^{i3C})$$
$$= 1 + \frac{1}{2} (-2 + 2 \cos(3A) + 2 \cos(3B) + 2\cos(3C))$$
$$= \cos(3A) + \cos(3B) + \cos(3C)$$
