<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] From $\sin(\alpha+\beta)=\frac1{\sqrt2}$, we should expect $|\cos(\alpha+\beta)|=\frac1{\sqrt2}$ from the formula $\sin^2 \theta + \cos^2 \theta = 1$.
\begin{align}
\cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)\color{blue}-\sin(\alpha)\sin(\beta)\\
&=\frac{2}{\sqrt5}\cdot \frac{3}{\sqrt{10}}-\frac1{\sqrt5}\cdot \frac{1}{\sqrt{10}}\\
&=\frac{6}{5\sqrt{2}}-\frac{1}{5\sqrt2}\\
&=\frac{5}{5\sqrt2}\\
&=\frac{1}{\sqrt2}
\end{align}
</s>