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.

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.
