<s>[INST] 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!
 [/INST] 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
$$
\frac{1}{\sin18^\circ}=\frac{1}{\cos72^\circ}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1
$$
</s>