<s>[INST] 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.
 [/INST] 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.
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