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After trying to solve a geometry problem, represented by the following image

enter image description here

I've arrived at this expression:

$\alpha=\arccos\left(\frac{d+r \cos \left(\varphi+\frac{\vartheta}2\right)}{\sqrt{d^2+r^2+2 dr\cos \left(\varphi+\frac{\vartheta}2\right)}}\right)+\arccos\left(\frac{d+r\cos \left(\varphi-\frac{\vartheta}2\right)}{\sqrt{d^2+r^2+2dr\cos\left(\varphi-\frac{\vartheta}2\right)}}\right)$

Is there a way to simplify such expression?

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$$\arccos \beta +\arccos \gamma=\arccos \left(\beta\cdot\gamma-\sqrt{1-\beta^2} \cdot \sqrt{1-\gamma^2}\right)$$

On Wikipedia there are also these identities:

$$\arccos x_1+\arccos x_2= \begin{cases} \arccos\left(x_1x_2-\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)& x_1+x_2\ge0\\ 2\pi-\arccos\left(x_1x_2-\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)& x_1+x_2<0 \end{cases} $$

$$\arccos x_1-\arccos x_2= \begin{cases} -\arccos\left(x_1x_2+\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)& x_1\ge x_2\\ \arccos\left(x_1x_2+\sqrt{1-x_1^2}\sqrt{1-x_2^2}\right)& x_1<x_2 \end{cases} $$

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    $\begingroup$ Nice. This comes from the angle addition formula (Let $\beta = \cos B$ and $\gamma = \cos C$. $\endgroup$
    – Benjamin Wang
    Commented Jun 28, 2020 at 15:55

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