Given $\cos2a$ and $\cos b$, find $a+b$. High School Level Mathematics:

If $\cos(2\alpha) = -\frac{63}{65}$ for $\alpha \in (0,\frac{\pi}{2})$ and $\cos(\beta) = \frac{7}{\sqrt{130}}$ for $\beta \in (0,\frac{\pi}{2})$ then, without using a calculator, what is $\alpha + \beta$?

The answer is ${3\pi\over 4}$ but I need steps.
 A: HINT:
$$2\cos^2\alpha-1=-\dfrac{63}{65}\implies\cos\alpha=+\dfrac1{\sqrt{65}}$$ as $0<\alpha<\dfrac\pi2$
For the same reason, $\sin\alpha=+\sqrt{1-\cos^2\alpha}=\cdots$
and $\sin\beta=+\sqrt{1-\cos^2\beta}=\cdots$
Now $\cos(\alpha+\beta)=\cdots$
Finally use $0<\alpha+\beta<\pi$
A: Notice $$\cos \alpha=\sqrt{\frac{1+\cos 2\alpha}{2}} \quad (\forall \space 0<\alpha<\frac{\pi}{2})$$ $$=\sqrt{\frac{1-\frac{63}{65}}{2}}=\sqrt{\frac{2}{2\times65}}=\color{blue}{\frac{1}{\sqrt{65}}}$$ $$\implies \sin\alpha=\sqrt{1-\cos^2\alpha} \quad (\forall \space 0<\alpha<\frac{\pi}{2})$$ $$=\sqrt{1-\left(\frac{1}{\sqrt{65}}\right)^2}=\color{blue}{\frac{8}{\sqrt{65}}}$$
$$\implies \sin\beta=\sqrt{1-\cos^2\beta} \quad (\forall \space 0<\beta<\frac{\pi}{2})$$ $$=\sqrt{1-\left(\frac{7}{\sqrt{130}}\right)^2}=\color{blue}{\frac{9}{\sqrt{130}}}$$ Now, we have $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ Substitute all the values in the above expression, we get $$\sin(\alpha+\beta)=\left(\frac{8}{\sqrt{65}}\right)\left(\frac{7}{\sqrt{130}}\right)+\left(\frac{1}{\sqrt{65}}\right)\left(\frac{9}{\sqrt{130}}\right)$$$$=\frac{65}{\sqrt{65\times130}}=\sqrt{\frac{65}{130}}=\frac{1}{\sqrt{2}}$$ $$\implies \alpha+\beta=\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \quad \text{or} \quad \pi-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) $$
By the given values of $\alpha$ & $\beta$ we have $\color{blue}{0<(\alpha+\beta)<\pi}$ 
$$\color{blue}{\alpha+\beta}=\pi-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) =\pi-\frac{\pi}{4}$$$$\color{blue}{=\frac{3\pi}{4}}$$
A: Starting from 
$\cos 2\alpha=\frac{-63}{65}$
Gives us
$\cos \alpha =\frac{1}{\sqrt {65}}$  (using $\cos 2a=2\cos^2a-1$)
and
$\sin \alpha=\frac{8}{\sqrt {65}}$
Similarly,
From $\cos \beta=\frac{7}{\sqrt {130}}$
we get
$\sin \beta=\frac{9}{\sqrt {130}}$
Plugin these values in $$\cos (\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$
A: With all angles in the first quadrant,
$$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\
=\sqrt{\frac{1+\cos(2\alpha)}2}\cos(\beta)-\sqrt{\frac{1-\cos(2\alpha)}2}\sqrt{1-\cos^2(\beta)}\\
=\sqrt{\frac{2}{130}}\frac7{\sqrt{130}}-\sqrt{\frac{128}{130}}\frac{9}{\sqrt{130}}=-\frac{\sqrt2}2.$$
$\alpha+\beta$ is $\dfrac{3\pi}4.$
