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!
 A: 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
$$
\frac{1}{\sin18^\circ}=\frac{1}{\cos72^\circ}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1
$$
A: In the regular pentagon below we have $\displaystyle{1\over\sin 18°}={2BD\over AB}$. On the other hand: 
$$
{BD\over AB}={AE\over EF}={AE\over EC-FC}={AB\over BD-AB}.
$$
It follows that: 
$$
{BD\over AB}=\hbox{golden ratio}={1+\sqrt5\over2}.
$$

A: 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 $DA=DB=CB=x.$
$$\boxed{ \therefore 1=AC=AD+DC=x+x^2=2\sin \frac{\pi}{10} +4\sin^2\frac{\pi}{10} \ }$$
A: 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.
A: Note that $\forall x\in \mathbb{R}$
$$\sin x= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.
$$
which converges “very fast for small” x.
In this case $x=\frac{\pi}{10}$ so we have that
$$ \sin \left(\frac{\pi}{10}\right) \approx \frac{\pi}{10} - \frac{\pi^3}{6000} + \frac{\pi^5}{10^5\cdot 5!}\approx 0.30901705...$$
while the exact value is
$$\sin  \left( \frac{\pi}{10} \right)=0.30901699...$$
