Exact value for $\cos 36°$

\begin{align} \sin 108^\circ&=\sin (72^\circ+36^\circ)\\ \sin(180^\circ-72^\circ)&=\sin 72^\circ\cos36^\circ+\cos 72^\circ\sin36^\circ\\ \sin72^\circ&=\sin (2\cdot36^\circ)\cos36^\circ+\cos (2\cdot36^\circ)\sin36^\circ\\ \sin (2\cdot36^\circ)&=2\sin36^\circ\cos36^\circ\cos36^\circ+(2\cos^236^\circ-1)\sin36^\circ\\ 2\sin36^\circ\cos36^\circ&=\sin36^\circ(2\cos^236^\circ+2\cos^236^\circ-1)\\ 2\cos36^\circ&=4\cos^236^\circ-1\\ 4\cos^236^\circ-2\cos36^\circ-1&=0\tag1 \end{align} Let $y=\cos36^\circ$, then $(1)$ becomes $$ 4y^2-2y-1=0\tag2 $$ Use complete square method or quadratic formula to obtain the roots of $(2)$ and take the positive value only because $\cos36^\circ>0$ since it lies on the first quadrant.

Hint: We have $$\sin(108^\circ)=\sin(72^\circ)=2\sin(36^\circ)\cos(36^\circ). \tag{1}$$ Also, $$\sin(108^\circ)=\sin(36^\circ +72^\circ)=\sin(36^\circ)\cos(72^\circ)+\cos(36^\circ)\sin(72^\circ). \tag{2}$$

As suggested in the OP, use double-angle identities on the right-hand side of (2).

Awesome proof :

This is a regular pentagon. As every angle is $108^\circ$, $\angle CAD=36^\circ$ from symmetry.

Triangles $ABP$ and $AEP$ are similar.

$\frac{BE}{AB}=\frac{AE}{EP}$

$BE×EP=AB^2$

$(BP+EP)×EP=AB^2$

$(AB+EP)×EP=AB^2$

For the ratio $x=\frac{AB}{EP}$ we have the equation

$x+1=x^2$

with one positive solution $x=\Phi$, the golden ratio.

In $△AEP$, $AE=AB$ and $EP$ is one of the sides such that $\frac{AE}{EP}=\Phi$. Drop a perpendicular from $P$ to $AE$ to obtain two right triangles. Then say,

$cos(∠AEP)=(AE/2)/EP=(AE/EP)/2=ϕ/2$

But $∠AEP=36^\circ$ and we get the desired result.

Hence the answer is $$\frac{1+\sqrt 5}{4}$$

"Good Evening"

Use the following trig identities s(x) = s(180 - x)

s(x + y) = s(x)c(y) + s(y)c(x) and in particular s(2x) = 2s(x)c(x)

c(2x) = $c^2(x) - s^2(x)$

when you get near the end note that $nc^2 - s^2 = (n+1)c^2 - 1$

It works out quite easily.

For the final part note the allowed range of the cosine function.