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!

Not an ugly solution at all, but one can do better. Here's an analytical solution based on complex numbers.
Let $\alpha=72^\circ=2\pi/5$, so we can consider $z=e^{\alpha}=\cos\alpha+i\sin\alpha$ that satisfies $z^5-1=0$. Since $z\ne1$, we can deduce
$$
z^4+z^3+z^2+z+1=0
$$
and also, dividing by $z^2$,
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
On the other hand, $z^2+\frac{1}{z^2}=(z+\frac{1}{z})^2-2$ and we can observe that
$$
z+\frac{1}{z}=2\cos\alpha
$$
Therefore we see that $2\cos\alpha$ satisfies the equation $t^2+t-1=0$. Since it is positive, we conclude
$$
2\cos\alpha=\frac{-1+\sqrt{5}}{2}
$$
and therefore
$$
\cos\alpha=\frac{\sqrt{5}-1}{4}
$$
Thus
$$
\frac{1}{\sin18^\circ}=\frac{1}{\cos72^\circ}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1
$$
