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. 
 A: Using complex numbers we can derive $\sin(\frac{\pi}{5})$ and $\cos(\frac{\pi}{5})$
$$ z^5=-1 \implies z^4-z^3+z^2-z+1=0 \implies z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0 $$
By substitution $t=z+\frac{1}{z}$ we get:
$$ t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2} $$
And because $\cos(\frac{\pi}{5})$ is positive, we may consider just $t=\frac{1+\sqrt{5}}{2}$
$$ z^2-z\frac{1+\sqrt{5}}{2}+1=0 \implies z=\frac{1+\sqrt{5}}{4}\pm i\frac{\sqrt{10-2\sqrt{5}}}{4} $$
And we get that $\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4} $ and $\sin(\frac{\pi}{5})= \frac{\sqrt{10-2\sqrt{5}}}{4}$
A: There is a slightly shorter way:
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
You can also write this
$$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14$$
Then, using the formula $2\cos a\cos b=\cos(a+b)+\cos(a-b)$, you have:
$$\frac12=2\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}$$
Hence $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$ are the roots of $t^2-\frac12t-\frac14$. The rest is easy, and the positive root is $\cos\frac{\pi}5$.

To answer your question: in general, it's not possible to find values of $\cos\frac{\pi}{n}$ only by radicals, even though they are algebraic numbers. Even in the case you obtain an irreducible cubic equation (which can be solved by radicals), it's the "trigonometric case" here, which has no expression with real radicals (and a complex cubic root needs the trigonometric functions anyway). For instance, $\cos1^\circ$ can't be computed with real radicals. But $\cos3^\circ$ can.
Even if in general it's not possible, there are a few tricks you can apply, for instance
$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$$
Also, if you can compute $\cos x$ by radicals, then you can compute $\cos\dfrac{x}{2^n}$ too.

So, which angles of the form $\frac{\pi}{n}$ leads to trigonometric functions computable by radicals? The answer is given by the Gauss-Wantzel theorem. For instance, one can compute $\cos\dfrac{\pi}{17}$. However, the computation is not obvious, see http://mathworld.wolfram.com/TrigonometryAnglesPi17.html
A: Everything begins with finding $\sin\left(\frac{\pi}{10}\right)$
Okay... Lets say $x=\frac{\pi}{10}$
$$5x=\frac{\pi}{2}$$
$$2x=\frac{\pi}{2}-3x$$
$$\sin(2x)=\sin\left(\frac{\pi}{2}-3x\right)$$
$$\sin(2x)=\cos(3x)$$
$$2\sin(x)\cos(x)=4\cos^3(x)-3\cos(x)$$
$$2\sin(x)\cos(x)-4\cos^3(x)+3\cos(x)=0$$
$$\cos(x)\left(2\sin(x)-4\cos^2(x)+3\right)=0$$
$$2\sin(x)-4\cos^2(x)+3=0$$
$$2\sin(x)-4(1-sin^2(x))+3=0$$
$$2\sin(x)-4+4\sin^2(x)+3=0$$
$$4\sin^2(x)+2\sin(x)-1=0$$
Now use the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$$a=4,b=2,c=-1$$
$$=\frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$$
$$=\frac{-2\pm \sqrt{20}}{8}$$
$$\sin\left(\frac{\pi}{10}\right)=\frac{-1\pm \sqrt{5}}{4}$$
One another way of finding the value of $\cos\left(\frac{\pi}{5}\right)$
$$\cos\left(\frac{\pi}{5}\right)=\cos\left(2\left(\frac{\pi}{10}\right)\right)$$
$$=1-2\left(\sin\frac{\pi}{10}\right)^2$$
Since $\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$
$$=1-2\left(\frac{\sqrt{5}-1}{4}\right)^2$$
$$=1-\frac{(\sqrt{5}-1)^2}{8}$$
$$=\frac{8-5-1+2\sqrt{5}}{8}$$
$$=\frac{2+2\sqrt{5}}{8}$$
$$\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4}$$
Now lets find $\sin\left(\frac{\pi}{5}\right)$
$$\sin\left(\frac{\pi}{5}\right)=\sqrt{1-\cos^2\left(\frac{\pi}{5}\right)}$$
$$=\sqrt{1-\left(\frac{1+\sqrt{5}}{4}\right)^2}$$
$$=\sqrt{1-\frac{1+5+2\sqrt{5}}{16}}$$
$$=\sqrt{\frac{10-2\sqrt{5}}{16}}$$
$$\sin\left(\frac{\pi}{5}\right)=\frac{\sqrt{10-2\sqrt{5}}}{4}$$
A: 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, $$\cos 36^o=\frac{AE}{AB}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{1+\sqrt{5}}{4}.$$

