<s>[INST] 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? 
 [/INST] You don't have to use the hint. Let $x = {\pi \over 5}$. Then $\cos(3x) = -\cos(2x)$ since $3x + 2x = \pi$. Since $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ and $\cos(2x) = 2\cos^2(x) - 1$, you get
$$4\cos^3(x) - 3\cos(x) = - ( 2\cos^2(x) - 1)$$
This simplifies to 
$$4\cos^3(x) + 2\cos^2(x) - 3\cos(x) - 1 = 0$$
This factors into 
$$(4\cos^2(x) - 2\cos(x) - 1)(\cos(x) + 1) = 0$$
Since $\cos(x)$ is not $-1$, you get the desired equation $4\cos^2(x) - 2\cos(x) - 1 = 0$.
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