A recent Quora question asks: How can you prove that [math]\cos{(\pi/15)} = (-1+\sqrt{5}+\sqrt{30+6\sqrt{5}})/4[/math]?
Well clearly you cannot prove that, because the RHS is bigger than sqrt(30)/4 which is bigger than sqrt(16)/4 which is 1.
But to evaluate $#\cos(\pi/15)#$ in terms of radicals a natural first step is to deal with the factor of 5 in the denominator by noting that $#\cos(5\theta)=\cos(4\theta +\theta)#$ and combine the cosine sum rule with two applications of the double angle rule to get the well known result that
$#\begin{align}\cos(5\theta)&=(((\cos^2(\theta)-\sin^2(\theta))^2-(2\sin(\theta)\cos(\theta))^2)\cos(\theta)\\&-2(2\sin(\theta)\cos(\theta))(\cos^2(\theta)-\sin^2(\theta))\sin(\theta)\\&=((2(2\cos^2(\theta)-1)^2–1)\cos(\theta)-2(2\sin(\theta)\cos(\theta))(2\cos^2(\theta)-1)\sin(\theta)\\&=((2(2\cos^2(\theta)-1)^2–1)\cos(\theta)-4(1-\cos^2(\theta)\cos(\theta))(2\cos^2(\theta)-1)\\&=16\cos^5(\theta)-20\cos^3(\theta)+5\cos(\theta)\end{align}#$
Now one approach would be to use this with $#5\theta=\pi/3#$ to get the equation $#16\cos^5(\pi/15)-20\cos^3(\pi/15)+5\cos(\pi/15)=cos(\pi/3)=1/2#$
But the 5th degree equation may be easier if we look first at $#5\theta=2\pi#$ to get
$#16\cos^5(2\pi/5)-20\cos^3(2\pi/5)+5\cos(2\pi/5)=\cos(2\pi)=1#$
For $#c=\cos(2\pi/5)#$ this gives $#16c^5–20c^3+5c-1=0#$ for which $#c=1#$ is clearly a solution so we can reduce the degree by long division to get $#16c^4+16c^3–4c^2–4c+1=0#$
Since this is of just 4th degree we know that there is an explicit solution in radicals which we can write down using Cardano’s formula; but I just noticed that the two highest degree terms match those of $#(4c^2+2c)^2=16c^4+16c^3+4c^2#$, and if we look at what’s left we see something interesting – namely $#16c^4+16c^3–4c^2–4c+1=(4c^2+2c)^2-8c^2-4c+1=(4c^2+2c)^2-2(4c^2+2c)+1=(4c^2+2c-1)^2#$.
So the only possibilities are for $#c#$ to be one of the roots of $#4c^2+2c-1#$ which are $#\frac{-1\pm \sqrt{5}}{4}#$ and only the $#+#$ sign gives a possible value for $#\cos#$ so we must have $#\cos(2\pi/5)=\frac{ \sqrt{5}-1}{4}#$.
And since $#\pi/15=2\pi/5-\pi/3#$ we can now easily use the cosine of a difference rule to get
$#\cos(\pi/15)=(-1+\sqrt{5}+\sqrt{30+6\sqrt{5}})/8#$.
So the suggested answer was just off by a factor of 2.