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

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 $$ \frac{1}{\sin18^\circ}=\frac{1}{\cos72^\circ}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1 $$

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: $$ {BD\over AB}=\hbox{golden ratio}={1+\sqrt5\over2}. $$

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 $DA=DB=CB=x.$

$$\boxed{ \therefore 1=AC=AD+DC=x+x^2=2\sin \frac{\pi}{10} +4\sin^2\frac{\pi}{10} \ }$$