Trigamma identity $4\,\psi_1\!\left(\frac15\right)+\psi_1\!\left(\frac25\right)-\psi_1\!\left(\frac1{10}\right)=\frac{4\pi^2}{\phi\,\sqrt5}.$ I heuristically discovered the following identity for the trigamma function, that I could not find in any tables or papers or infer from existing formulae (e.g. [1], [2], [3], [4], [5], [6]):
$$4\,\psi_1\!\left(\frac15\right)+\psi_1\!\left(\frac25\right)-\psi_1\!\left(\frac1{10}\right)=\frac{4\pi^2}{\phi\,\sqrt5}.\tag1$$
It also seems to be unknown to Mathematica, but numerically checks with at least $20000$ decimal digits. It might be provable through some application of reflection and multiplication theorems, but I couldn't do this.
Please suggest how to prove it.

Update:
Another identity is
$$3\,\psi_1\!\left(\frac1{12}\right)-30\,\psi_1\!\left(\frac13\right)=120\,G+\left(6\sqrt3-8\right)\pi^2.\tag2$$
 A: Consider the multiplication formula for the polygamma function
$$
\psi_n(mz)=\frac{1}{m^{n+1}}\sum_{k=0}^{m-1}\psi_n\left(z+\frac km\right)\quad;\quad\text{for}\ n\ge1\tag1
$$
and
its reflection formula
$$
\psi_n(1-z)+(-1)^{n+1}\psi_n(1-z)=(-1)^n\pi\frac{d^n}{dz^n}\cot\pi z.\tag2
$$
Using $(1)$ by setting $n=1,\ m=2,$ and $z=\dfrac1{10}$, we obtain
\begin{align}
\psi_1\left(\frac15\right)&=\frac{1}{4}\left[\psi_1\left(\frac1{10}\right)+\psi_1\left(\frac35\right)\right]\\
\psi_1\left(\frac35\right)&=4\psi_1\left(\frac15\right)-\psi_1\left(\frac1{10}\right)\tag3
\end{align}
then using $(3)$ and with helping $(2)$, we obtain
\begin{align}
4\psi_1\left(\frac15\right)+\psi_1\left(\frac25\right)-\psi_1\left(\frac1{10}\right)&=\psi_1\left(\frac35\right)+\psi_1\left(\frac25\right)\\
&=-\pi\left.\frac{d}{dz}\cot\pi z\right|_{z=\frac25}\\
&=\frac{8\pi^2}{5+\sqrt{5}}\\
&=\frac{4\pi^2}{\phi\sqrt{5}}.\tag{Q.E.D.}
\end{align}
A: I will use only the standard reflection and duplication identities:
\begin{align}
&\psi_1(z)+\psi_1(1-z)=\frac{\pi^2}{\sin^2\pi z},\\
&\psi_1(z)+\psi_1\bigl(z+\text{$\frac12$}\bigr)=4\psi_1(2z).
\end{align}
Use the first of them to replace $\psi_1\bigl(\frac1{10}\bigr)$ by $ -\psi_1\bigl(\frac9{10}\bigr)+\displaystyle\frac{\pi^2}{\sin^2\frac{\pi}{10}}$. Then use the second to replace
$\psi_1\bigl(\frac25\bigr)+\psi_1\bigl(\frac{9}{10}\bigr)$ by $4\psi_1\bigl(\frac45\bigr)$. Finally use the first the second time to replace $4\psi_1\bigl(\frac15\bigr)+4\psi_1\bigl(\frac45\bigr)$ by $\displaystyle\frac{4\pi^2}{\sin^2\frac{\pi}{5}}$. In this way we obtain
$$4\psi_1\left(\frac15\right)+\psi_1\left(\frac25\right)-\psi_1\left(\frac1{10}\right)=\frac{4\pi^2}{\sin^2\frac{\pi}{5}}-\frac{\pi^2}{\sin^2\frac{\pi}{10}},$$
and the rest is straightforward.

As for the second identity, we will use in addition that
$$\sum_{k=0}^2\psi_1\bigl(z+\text{$\frac{k}3$}\bigr)=9\psi_1(3z),\qquad \sum_{k=0}^3\psi_1\bigl(z+\text{$\frac{k}4$}\bigr)=16\psi_1(4z).$$
This allows to write
$$\psi_1\left(\frac1{12}\right)+\psi_1\left(\frac5{12}\right)+\psi_1\left(\frac3{4}\right)=9\psi_1\left(\frac14\right),$$
$$\psi_1\left(\frac1{12}\right)+\psi_1\left(\frac13\right)+\psi_1\left(\frac7{12}\right)+\psi_1\left(\frac5{6}\right)=16\psi_1\left(\frac13\right).$$
Adding the two identities, we find
$$2\psi_1\left(\frac1{12}\right)+\frac{\pi^2}{\sin^2\frac{5\pi}{12}}+4\psi_1\left(\frac23\right)+\psi_1\left(\frac34\right)=9\psi_1\left(\frac14\right)+16\psi_1\left(\frac13\right).$$
Now using two times the reflection formula, we can rewrite the last identity as
$$2\psi_1\left(\frac1{12}\right)+\frac{\pi^2}{\sin^2\frac{5\pi}{12}}+\frac{4\pi^2}{\sin^2\frac{2\pi}{3}}+\frac{\pi^2}{\sin^2\frac{3\pi}{4}}=10\psi_1\left(\frac14\right)+20\psi_1\left(\frac13\right),$$
or, equivalently,
$$2\psi_1\left(\frac1{12}\right)-20\psi_1\left(\frac13\right)=10\psi_1\left(\frac14\right)-\frac{46-12\sqrt3}{3}\pi^2.$$
Now it suffices to use the formula (4) from here.
