If $ \sin \alpha = \frac 45 $ and $ \cos \beta = \frac{5}{13} $, prove that $ \cos \frac{\alpha-\beta}{2} = \frac{8}{\sqrt {65}} $ I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$
But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.
Is the question wrong ?

HINT:
\begin{align}
\cos(\alpha/2-\beta/2)&=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}=\sqrt{(1+\cos\alpha\cos\beta+\sin\alpha\sin\beta)/2}\\
\end{align}
$$ \cos\alpha =\frac35, \cos\beta = \frac{5}{13},\, \sin\alpha=\frac45,\, \sin\beta =\frac{12}{13}, $$
plug in to get $ 8/\sqrt{65}$
For other quadrants calculate other three values also which are possible with inverse trig functions.
$$\sqrt{(1\pm\cos\alpha\cos\beta\pm\sin\alpha\sin\beta)/2}$$
