Prove that for $\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$ I have the following question in a mock exam:
$\beta = \frac{\pi}{2} - \alpha$, show that $\beta > \frac{1}{3}$
From the earlier part of the question we know that $\alpha$ is an angle between two vectors (so it's between $0$ and $\pi$) and that $\cos(\alpha) = \frac{1}{3}$ (so it's really between $0$ and $\frac{\pi}{2}$).
Using this information I simplified the inequality to:
$\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$
Now I believe the answer will have something to do with the monotonicity of cos between 0 and $\frac{\pi}{2}$, but I don't know how to show this. Thanks for your help.

You mean that: if $0<\alpha<\frac{\pi}{2}$ and $\cos\alpha=\frac{1}{3}$, then $\alpha<\frac{\pi}{2}-\frac{1}{3}$?
Let $F(x)=\cos x-\frac{1}{3}$, then $F(0)=\frac{2}{3}$ and $F(\frac{\pi}{2}-\frac{1}{2})=\sin\frac{1}{3}-\frac{1}{3}<0$ since $\sin x<x (0<x<\frac{\pi}{2})$.
This shows that $\alpha<\frac{\pi}{2}-\frac{1}{3}$.
