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!

There are a couple of well-known approaches. The first parallels the answer of @egreg and goes back to Gauss. For a prime $p$ find a primitive root $g$ and then for and divisor $d$ of $p-1$ and integer $0\le j<d$ let
$$\sigma_{dj}=\sum_{k=0}^{\frac{p-1}d-1}\omega^{g^{j+kd}}$$
Where $\omega=e^{2\pi i/p}$. In our case $p=5$ and $g=2$ is a primitive root. Now, $\sigma_{10}=-1$ because the sum of the roots of $z^p-1=0$ is $\sigma_{10}+1=0$ because it's minus the coefficient of $z^{p-1}$. Then $\sigma_{20}+\sigma_{21}=\sigma_{10}=-1$ and we can work out the product $\sigma_{20}\sigma_{21}$ by constructing a table where each table entry is the product of its row and column headers:
$$\begin{array}{c|cc}
&\omega^1&\omega^4\\
\hline\omega^2&\omega^3&\omega^1\\
\omega^3&\omega^4&\omega^2\end{array}$$
And the product is the sum of the table entries, $\sigma_{20}\sigma_{21}=\sigma_{10}=-1$. Since we know the sum and product of $\sigma_{20}$ and $\sigma_{21}$ we can construct an equation they satisfy, $\sigma_{20}^2+\sigma_{20}-1=0$ with solution
$$2\cos\frac{2\pi}5=\omega+\omega^4=\sigma_{20}=\frac{-1\pm\sqrt5}2=\frac{-1+\sqrt5}2$$
because it's a first quadrant angle. So
$$\csc\frac{\pi}{10}=\frac1{\sin\frac{\pi}{10}}=\frac1{\cos\frac{2\pi}5}=\frac4{\sqrt5-1}=\sqrt5+1$$
It might seem like a lot of machinery to get this simple answer, but this approach is capable of getting to $\cos\frac{2\pi}{17}$ because just as in the above we can find an equation that $\sigma_{20}$ and $\sigma_{21}$ satisfy, then equations for $\sigma_{40}$ and $\sigma_{42}$ and for $\sigma_{41}$ and $\sigma_{43}$ and finally for $\sigma_{80}=2\cos\frac{2\pi}{17}$ and $\sigma_{84}$ and that's one construction they don't usually teach you in Euclidean geometry class.  
The other approach is similar to that proposed in the original question but is perhaps a bit more systematic. We start with the identity
$$\cos(n+1)\theta=\cos n\theta\cos\theta-\sin n\theta\sin\theta=2\cos n\theta\cos\theta-\cos(n-1)\theta$$
Starting with $\cos(0\theta)=1$, $\cos(1\theta)=\cos\theta$ we progress to
$$\begin{align}\cos2\theta&=2\cos^2\theta-1\\
\cos3\theta&=4\cos^3\theta-3\cos\theta\\
\cos4\theta&=8\cos^4\theta-8\cos^2\theta+1\\
\cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta\end{align}$$
Since our target is $\theta=\frac{2\pi}5$ and we know that $\cos5\theta=1$ and $\cos0=1$ is a root and also that there are two double roots because $\cos\theta=\cos4\theta$ and $\cos2\theta=\cos3\theta$ we can factor
$$16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=\left(\cos\theta-1\right)\left(4\cos^2\theta+2\cos\theta-1\right)^2=0$$
So we get
$$\cos\theta=\cos\frac{2\pi}5=\sin\frac{\pi}{10}=\frac{-2\pm\sqrt{20}}8=\frac{-1+\sqrt5}4$$
Because it's a first quadrant angle, as before.
