A closed-form of product the gamma functions containing $\pi$ and $\phi$ 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.
 A: We will show that

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

where $\phi$ is the golden ratio.
Because of the Gauss's multiplication formula we know that
$$\Gamma(2z)=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$
Because $\Gamma$ is nowhere zero, we can divide the formula by $\Gamma(z)$, and we get
$$\frac{\Gamma(2z)}{\Gamma(z)}=\frac{1}{\sqrt{\pi}}2^{2z-1}\Gamma\left(z+\frac{1}{2}\right).$$
We put $z:=2/10$ into the formula and get
$$\color{red}{\frac{\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\color{blue}{\Gamma\left(\frac{7}{10}\right)}.$$
Now take a look at the Euler's reflection formula.
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(z\pi)}.$$
With $z:=3/10$ we get
$$\color{green}{\Gamma\left(\frac{3}{10}\right)}\color{blue}{\Gamma\left(\frac{7}{10}\right)}=\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$
Putting this all together, we get
$$
\color{red}{\frac{\color{green}{\Gamma\left(\frac{3}{10}\right)}\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}}=\frac{1}{\sqrt{\pi}}2^{\frac{4}{10}-1}\frac{\pi}{\sin\left(\frac{3}{10}\pi\right)}.$$
Now we need that
$$\sin\left(\frac{3}{10}\pi\right)=\frac{1+\sqrt{5}}{4},$$
and of course we also know that $\pi / \sqrt{\pi} = \sqrt{\pi}$.
Using this we get
$$
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=2^{-\frac{3}{5}} \cdot \sqrt{\pi} \cdot\frac{4}{1+\sqrt{5}} = \frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi},$$
and this completes the proof.

One last fun fact, that we can generalize the problem like this

$$\frac{\Gamma\left(\frac{1}{2}-z\right)\Gamma(2z)}{\Gamma(z)}=\frac{2^{2z-1}
 \cdot \sqrt{\pi}}{\cos(z\pi)},$$

for all $z \notin -\mathbb{N}$ and $z \neq n-1/2, \ n \in \mathbb{Z}$. You can get your result by $z:=2/10.$
And really at last an other related formula is the following

$$\frac{\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)}{\Gamma\left(\frac{1}{10}\right)}=\frac{\sqrt[10]{3} \cdot \sqrt[5]{2} \cdot \sqrt{\pi}}{\phi}.$$

This is even more interesting, because with the same idea you have to use the multiplication formula twice for $3z$ and also for $2z$ and after that the reflection formula. Because the equation $6z=z+2/3$ only has one solution, and it is $2/15$, that's why $2/15$ has a very important role in the formula, and this problem does not have a generalization like I gave above. I could imagine other ways to generalize, but it would not be nice.
A: You need two standard results about the Gamma function: the Euler reflection formula $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$ and the Legendre duplication formula $\Gamma(s)\Gamma(s+\frac{1}{2})=2^{1-2s}\sqrt{\pi}\Gamma(2s)$. If you start with the duplication formula with $s=0.2$ and then substitute for $\Gamma(0.7)$ from the Euler formula you get your result (except you also need to know that $\sin 0.7\pi=\frac{1+\sqrt{5}}{4}$.
A: 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
$$
\frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)}
=\frac{2^{2/5}\sqrt\pi}{\phi}\tag{6}
$$
