Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,

Evaluate: $$\frac {1}{\sin 18°}$$

I tried to do something myself.

It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× 2\sin(x) \cos (x)×(1-2\sin^2(x)), \cos(x)>0$$
$$8\sin^3(x)-4\sin(x)+1=0$$
$$(2\sin(x)-1)(4\sin^2(x)+2\sin (x)-1)=0$$
$$4\sin^2(x)+2\sin (x)-1=0, \sin(x)≠\frac 12$$
$$4t^2+2t-1=0$$
$$t_{1,2}=\frac {-1±\sqrt 5}{4}$$
$$t=\frac {\sqrt5-1}{4} ,t>0$$
$$\sin 18°=\frac {\sqrt5-1}{4} .$$
Finally, $$\frac {1}{\sin 18°}=\frac {4}{\sqrt5-1}=\sqrt5+1$$

Is this way correct and is there a better/elegant way to do it? As always, it was an ugly solution.
Thank you!

Take $\Delta ABC$ with $AB=AC=1$ and $BC=x,$ and with $\angle ABC=\angle ACB=2\pi/5.$ Then $\angle BAC=\pi/5$ and $x=2AB\sin \frac {1}{2}\angle BAC=2\sin \pi/10.$ 
Take $D$ on side $AC$ with $\angle  DBC=\pi/5.$ Then $ABC$ and $BDC$ are similar triangles so $CD/x= CD / CB=  CB/CA=x/1$  . So $CD=x^2.$ 
And $\Delta BDA$ is isosceles because $\angle DCA=\angle DAC=\pi/5 .$ So $DA=DB=CB=x.$
$$\boxed{ \therefore 1=AC=AD+DC=x+x^2=2\sin \frac{\pi}{10} +4\sin^2\frac{\pi}{10} \ }$$
