<s>[INST] Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$ Let $\varphi=\frac{1+\sqrt5}2$ (the golden ratio).
How can I simplify the following expression?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
 [/INST] We will use a well-known$^{[1]}$ formula for sum of arctangents:
$$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$
The exact equality holds for $uv<1$, for other values there is additional term — an integer multiple of $\pi$ (it is easy to determine in each case).
Using this formula, we can establish the following identities:
$$\begin{align}
\arctan\varphi\phantom{^1}&=\frac\pi2-\frac12\,\arctan2,\tag2\\
\arctan\varphi^3&=\frac\pi4+\frac12\,\arctan2,\tag3\\
\arctan\varphi^5&=\pi-\frac32\,\arctan2.\tag4
\end{align}$$
Plugging these into the original expression in the question and expanding parentheses, we can see that all $\arctan2$ terms cancel, and we get the result
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5=\frac{7\pi^2}8.\tag5$$
</s>