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: 
$\sin \alpha>0$ then $0<\alpha<\pi$ and $\cos \beta>0$ then $-\frac{\pi}2<\beta<\frac{\pi}2$ and $-\frac{\pi}2<-\beta<\frac{\pi}2$
Thus $$-\frac{\pi}2<\alpha-\beta<\frac{3\pi}2$$
and
$$-\frac{\pi}4<\frac{\alpha-\beta}2<\frac{3\pi}4$$
1) $\sin^2\alpha+\cos^2\alpha=1$, then $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$  and $\sin \beta=\pm\sqrt{1-\cos^2\beta}$
2) $\cos (\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$
3)$|\cos \frac{\alpha-\beta}2|=\sqrt{\frac{1+\cos(\alpha-\beta)}2}$
