Verifying $\sqrt{\frac{2}{3}}\cos\left(\frac{1}{3}\arccos\sqrt{\frac{3^3}{2^5}}\,\right)=\cos\frac\pi5$ I can verify the equation using Excel. But is there another derivation?
$$\sqrt{\frac{2}{3}}\cos\left[\frac{1}{3}\arccos\left(\!\!\sqrt{\frac{3^3}{2^5}}\right)\right]=\cos\left(\frac{\pi}{5}\right)$$
Thank you for help.
 A: Let $u=\sqrt{\frac{2}{3}}\cos\left(\theta\right)$, where $\theta=\frac{1}{3}\arccos\left(\sqrt{\frac{3^3}{2^5}}\right)$.
By the triple angle formula, we have
$$4\cos^3(\theta) -3\cos(\theta)-\frac{3}{4}\sqrt{\frac{3}{2}} = 0.$$
i.e.  $u$ is a root of the cubic $$q(x) = 6\sqrt{\frac{3}{2}}x^3-3\sqrt\frac{3}{2}x-\frac{3}{4}\sqrt{\frac{3}{2}}.$$
or, after clearing coefficients,
$$ p(x)=8x^3-4x-1.$$
On the other hand, $\cos\left(\frac{\pi}{5}\right)$ is also a root of $p(x)$, as
\begin{align*}
p\left(\cos\left(\frac{\pi}{5}\right)\right)&=8\cos^3\left(\frac{\pi}{5}\right)-4\cos\left(\frac{\pi}{5}\right)-1\\
&= 2\left(3\cos\left(\frac{\pi}{5}\right)+\cos\left(\frac{3\pi}{5}\right)\right)-4\cos\left(\frac{\pi}{5}\right)-1\\
&= 2\cos\left(\frac{\pi}{5}\right)+2\cos\left(\frac{3\pi}{5}\right)-1\\
&= 0.
\end{align*}
Factoring $p(x)$, we have
$$p(x)=(2x+1)(4x^2-2x-1).$$
We see that $p(x)$ has a unique positive root. Since $u$ and $\cos\left(\frac{\pi}{5}\right)$ are both positive, we must have $u=\cos\left(\frac{\pi}{5}\right)$.
