How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question: 
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.

As $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$
$S=\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ $
$=\sin(-138^\circ)+\sin(-66^\circ)+\sin6°+\sin78^\circ $
Observe that the angles are in Arithmetic Progression with common difference $=72^\circ$ 
Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? 
and Werner Formula: $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
and finally Prosthaphaeresis Formula, $\cos C+\cos D=2\cos\dfrac{C+D}2\cos\dfrac{C-D}2$
$2\sin36^\circ\cdot S=-2\cos54^\circ\cos60^\circ\implies S=-\dfrac12$
