TITLE: A closed-form of product the gamma functions containing $\pi$ and $\phi$
QUESTION [6 upvotes]: Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result

\begin{equation}
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi}
\end{equation}

where $\phi$ is golden ratio.
Could anyone here please help me to prove it by hand? I mean without using table for the specific values of $\Gamma(x)$ except for $\Gamma\left(\frac{1}{2}\right)$. As usual, preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

REPLY [2 votes]: In this answer, we derive Gauss's Multiplication Formula:
$$
\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)
=\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{1}
$$
In this answer, we derive Euler's Reflection Formula:
$$
\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{2}
$$
Multiply $(2)$ by $\frac{\Gamma\left(\frac8{10}\right)}{\Gamma\left(\frac8{10}\right)}$ to get
$$
\begin{align}
\frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)}
&=\frac{\color{#C00000}{\Gamma\left(\frac3{10}\right)}\Gamma\left(\frac4{10}\right)\color{#C00000}{\Gamma\left(\frac8{10}\right)}}{\color{#00A000}{\Gamma\left(\frac2{10}\right)\Gamma\left(\frac8{10}\right)}}\tag{3a}\\
&=\frac{\color{#C00000}{2\sqrt{\pi}\frac{\Gamma\left(\frac35\right)}{2^{3/5}}}\Gamma\left(\frac25\right)}{\color{#00A000}{\pi\csc\left(\frac\pi5\right)}}\tag{3b}\\
&=\frac{2^{2/5}\sqrt\pi\,\pi\csc\left(\frac{2\pi}5\right)}{\pi\csc\left(\frac\pi5\right)}\tag{3c}\\
&=\frac{2^{2/5}\sqrt\pi}{2\cos\left(\frac\pi5\right)}\tag{3d}
\end{align}
$$
Explanation:
$\mathrm{(3a)}$: multiply numerator and denominator by $\Gamma\left(\frac8{10}\right)$
$\mathrm{(3b)}$: in red, apply $(1)$ with $x=\frac3{10}$ and $n=2$; in green, apply $(2)$ with $x=\frac15$
$\mathrm{(3c)}$: apply $(2)$ with $x=\frac25$
$\mathrm{(3d)}$: $\sin(2x)=2\sin(x)\cos(x)$
We can use the identity $\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)$ to get
$$
\begin{align}
-1&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)\\
0&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)+1\\
&=\left[4\cos^2\left(\frac\pi5\right)-2\cos\left(\frac\pi5\right)-1\right]^2\left[\cos\left(\frac\pi5\right)+1\right]\tag{4}
\end{align}
$$
which implies, since $\cos\left(\frac\pi5\right)\gt0$, that
$$
2\cos\left(\frac\pi5\right)=\phi\tag{5}
$$
Combining $(3)$ and $(5)$ yields
$$
\frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)}
=\frac{2^{2/5}\sqrt\pi}{\phi}\tag{6}
$$