Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show 
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have 
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$
Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something? 

The hint says to use the following:
$$\sin 3x = 3 \sin x - 4 \sin^3x$$
and
$$ \sin 2x = 2 \sin x \cos x$$
So if $x = \frac{\pi}{5}$, then we have, using $\sin 3x = \sin 2x$ that
$$3 \sin x - 4 \sin^3 x = 2 \sin x \cos x $$
Since $\sin x \neq 0$, cancel it, and use $\sin^2 x = 1 - \cos ^2x$.
For a geometric way to find $\cos \frac{\pi}{5}$ see: http://www.cut-the-knot.org/pythagoras/cos36.shtml
