trigonometry
Without using the surd value for $\sin 36^\circ$ or $\cos 18^\circ$, prove that
$$ 4 \sin 36^\circ \cos 18^\circ = \sqrt{5} $$
My attempt:
$ 4\sin36^\circ \cos18^\circ \!\cdot\!\dfrac{\sin18^\circ}{\sin18^\circ}=$
$=\dfrac{2\sin^2 36^\circ}{\sin18^\circ}=$
$=\dfrac{1-\cos72^\circ}{\sin18^\circ}=$
$=\dfrac{1-\sin18^\circ}{\sin18^\circ}=$
$=\dfrac{1}{\sin18^\circ}-1\,.$
This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.
As pointed out in a comment, of course you can solve for $x$ in the equation $$ \frac{(\sqrt{2}+\sqrt{6})\cdot(-1+\sqrt{5})-(\sqrt{6}-\sqrt{2})\cdot\sqrt{10+2\sqrt{5}}}{16}=3x-4x^3. $$ It will just be an even much bigger and uglier expression. And it will involve square roots of negative numbers at some point while you are evaluating it; this is unavoidable. (See the answers to Can we express the value of $\sin 1^\circ$ without using the imaginary unit? and Is it possible get an algebraic expression for $\sin 1^\circ$ that does not contain roots of negative values?.)
But if you start with a larger angle than $3^\circ$ there are expressions that are not so ugly. Here is one that is very neat (albeit possibly a little cryptic at first) from The Sine of a Single Degree by Travis Kowalski: $$ \sin 1^\circ = \frac{\sqrt[90]{\sqrt{-1}} - \sqrt[90]{-\sqrt{-1}}}{2 \sqrt{-1}} $$ where all the roots are principal roots.
Best Answer
First of all, it results that
$\begin{align}4\cos^2\!18\unicode{176}&=\dfrac{4\cos^3\!18\unicode{176}-3\cos18\unicode{176}}{\cos18\unicode{176}}+3=\dfrac{\cos54\unicode{176}}{\cos18\unicode{176}}+3=\\[3pt]&=\dfrac{\sin36\unicode{176}}{\cos18\unicode{176}}+3=\dfrac{2\sin18\unicode{176}\cos18\unicode{176}}{\cos18\unicode{176}}+3=\\[3pt]&=2\sin18\unicode{176}+3\;\;,\quad\color{blue}{(1)}\end{align}$
$4\sin^2\!18\unicode{176}=1-2\sin18\unicode{176}\,.\;\;\quad\color{blue}{(2)}$
Moreover ,
$\begin{align}4\sin36\unicode{176}\cos18\unicode{176}&=8\sin18\unicode{176}\cos^2\!18\unicode{176}=\\[3pt]&\overset{\color{brown}{(1)}}{=}2\sin18\unicode{176}\big(2\sin18\unicode{176}\!+3\big)=\\[3pt]&=4\sin^2\!18\unicode{176}\!+6\sin18\unicode{176}\overset{\color{brown}{(2)}}{=}4\sin18\unicode{176}+1=\\[3pt]&=\sqrt{\big(4\sin18\unicode{176}\!+1\big)^{\!2}}=\\[3pt]&=\sqrt{16\sin^2\!18\unicode{176}+8\sin18\unicode{176}+1}=\\[3pt]&\overset{\color{brown}{(2)}}{=}\sqrt{4-8\sin18\unicode{176}+8\sin18\unicode{176}+1}=\\[3pt]&=\sqrt5\;.\end{align}$