<s>[INST] Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$ Evaluate: 
$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$
We let 
$$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad  \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$
Then we have:
$$\begin{align}
\cos(\alpha) = \frac{2}{\sqrt5} &\qquad \cos(\beta) = \frac{3}{\sqrt{10}} \\[4pt]
\sin(\alpha) = \frac{1}{\sqrt5} &\qquad \sin(\beta) = \frac{1}{\sqrt{10}}
\end{align}$$
In order to evaluate, we are told, we first determine $\sin(\alpha + \beta)$; we wind up with $1/\sqrt2$, thus we have $\pi/4$. 
What I am confused about is why we have to use sin($\alpha + \beta$). For example, if I were to use $\cos(\alpha + \beta)$, I would get the answer $7/(\sqrt{10}\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\sin(\alpha + \beta)$ since he/she checked cos and saw nothing comes out of this? 
Any help is much appreciated, thank you
 [/INST] You can directly use the formula $$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$
$$x=\frac{2}{\sqrt{5}},y=\frac{3}{\sqrt{10}}$$
$$=\cos^{-1}\left(\left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)-\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{1-\frac45}\sqrt{1-\frac{9}{10}}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac15}\sqrt{\frac{1}{10}}\right)$$
$$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$$
$$=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$
$$\cos^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}\frac{3}{\sqrt{10}}=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$
</s>