Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$
Using the difference of cosines identity, we have
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
So,
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$.
$$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$
Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have
$$y-y^2+1=2y(2y^2-1)$$
$$4y^3+2y^2-3y-1=0$$
which has the correct solution
$$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$
One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too. 

Not so cool. The following may be cooler.
Let $\Delta ABC$ be a triangle with $\angle BAC=36^o$, $\angle ABC=\angle ACB=72^o.$ $BD$ bisects $\angle ABC$. $BE\perp AC$. You would find that $AD=BD=BC$ and $\Delta ABC \sim \Delta BDC.$ Thus, as the figure shows, we may obtain $$\frac{x}{2y}=\frac{x+2y}{x}.$$ Thus $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$
Solve it. We have $$\frac{x}{y}=1+\sqrt{5}.$$
Another negative root is not what we want.Thus, $$\cos 36^o=\frac{AE}{AB}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{1+\sqrt{5}}{4}.$$

