There is nothing amazing about that, I am sure you can derive it yourself.
$\displaystyle \cos 3x=\cos 2x\cos x-\sin 2x\sin x$
$\displaystyle \cos 3x=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x$
$\displaystyle \cos 3x=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x$
$\displaystyle \cos 3x=2\cos ^3x-\cos x-2\cos x-2\cos^3x$
$\displaystyle \cos 3x=4\cos^3 x-3\cos x$
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So let,
$\displaystyle x=\cos 15^o$
Thus,
$\displaystyle \frac{\sqrt{2}}{2}=4x^3-3x$
We need to find all the values of $\displaystyle x$ that satisfy this equation.
Thus,
$\displaystyle 4x^3-3x=\frac{\sqrt{2}}{2}$
Divide through by 4,
$\displaystyle x^3-\frac{3}{4}x=\frac{\sqrt{2}}{8}$
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Similarly, we can do the 1/5 angle formula.
But that leads to a quintic which is not solvable by the premutation group of radicals. Which answers the question to why there is no $\displaystyle \sin 1^o$ through radicals in this
thread.
Basically, not being able to express the unit angle through radicals is the same problem as not being able to construct some things through "Euclidean toys" (compass and straightedge).