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.
 A: Angles that are multple of $18$ degrees occur in the regular pentagon with its diagonals and their trigonometric ratios are related to the golden ratio.
To find them, take an isosceles triangle with angles $A=36$ and $B=C=72$ degrees. Draw the bisector of $B$ that intersects $AC$ in a point $D$. Note that the triangles $ABC$ and $BCD$ are similar. Now let $1=AB$, $x=AD$. Note now that $ABD$ is also isosceles, and $AD=BD=BC=x$. The mentioned similarity yields
$$\frac1x=\frac x{1-x}$$
which is precisely the equation that defines the golden ratio.
To find $\cos 36^\circ$ draw the altitude of the triangle $ABD$ from $D$, that bisects the side $AB$. Then
$$\cos 36^\circ=\frac{1/2}{x}$$
Now, with the classical trigonometric identities, you can easily find closed expressions for the sine and the cosine of $18$, $36$, $54$ and $72$ degrees. For the angle of $18$ degrees, don't use half-angle formulae, but
$$\sin 18^\circ=\sin(90^\circ-72^\circ)=\cos 72^\circ=\cos^2 36^\circ-\sin^2 36^\circ$$
Remark: Perhaps you are wandering what that pentagon had to do with all of this. If you choose a vertex $A$ of a regular pentagon and draw the two diagonals $AB$ and $AC$ from it, the triangle $ABC$ is similar to the one we have just used.
A: There is not fast but strightforward way. $18^\circ = \pi/10$; let $s=\sin{(\pi/10)}$ so, we should evaluate
$$
a=\sin\frac\pi{10}-\cos\frac{\pi}{5}=2s^2 + s - 1.
$$
We may express $\sin5x$ in terms of $\sin x$:
$$
\sin5x=16\sin^5 x - 20\sin^3 x + 5\sin x
$$
(i.e. from $\sin5x=\sin(4x + x), \sin4x=2\sin2x\cos2x$, or by de Moivre). For $x=\frac{\pi}{10}$ we have $\sin5x=1$:
$$
16s^5 - 20s^3 + 5s -1 =0,
$$
or
$$
(s-1)(-1+2s+4s^2)^2 = 0;
$$
So,
$$
4s^2 + 2s - 1= 0=2a+1 \Longrightarrow a = -\frac12
$$
is answer.
A: You have done your question already. Note that  $sin18^\circ=\frac{\sqrt{5}-1}{4}$ and  $cos36^\circ=\frac{\sqrt{5}+1}{4}$.If you had done mulitiple angles in trigonometry you can easily calculate these.
For $sin18^\circ$ let, $A=18^\circ$. Then $5A=90^\circ$
$2A=90^\circ-3A$
$Sin2A=cos3A$. Use $Sin2A=2sinAcosA$ and $Cos3A=4cos^3A-3CosA.$
A: $\sin(5\cdot78^\circ)=\sin(360^\circ+30^\circ)=\sin30^\circ$
$\sin\{5(-66^\circ)\}=\sin(-360^\circ+30^\circ)=\sin30^\circ$
If $\sin5x=\sin30^\circ\implies5x=n180^\circ+(-1)^n30^\circ$ where $n$ is any integer
$\implies x=n72^\circ+6^\circ$ where $n\equiv-2,-1,0,1,2\pmod5$
Again, $\sin5x=16\sin^5x-20\sin^3x+5\sin x$
So, the roots of $\displaystyle16\sin^5x-20\sin^3x+5\sin x=\dfrac12$ are $\sin\left(n72^\circ+6^\circ\right)$ where $n\equiv-2,-1,0,1,2\pmod5$
Using Vieta's formula, $\displaystyle\sum_{n=-2}^2\sin\left(n72^\circ+6^\circ\right)=0$
$n=-2\implies$ $-2\cdot72^\circ+6^\circ=-138^\circ\implies\sin(-138^\circ)=-\sin(138^\circ)=-\sin(180^\circ-42^\circ)=-\sin42^\circ$
$n=-1\implies$ $-1\cdot72^\circ+6^\circ=-66^\circ\implies\sin(-66^\circ)=-\sin66^\circ$
$n=0\implies ?$
$n=1\implies ?$
and $n=2\implies\sin\left(2\cdot72^\circ+6^\circ\right)=\sin150^\circ=\dfrac12$
Do you see the destination?
A: 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, $\cos C+\cos D=2\cos\dfrac{C+D}2\cos\dfrac{C-D}2$
$2\sin36^\circ\cdot S=-2\cos54^\circ\cos60^\circ\implies S=-\dfrac12$
