The text says $|\cos x| \le 1/\sqrt3$, and hence, $$\arccos(1/\sqrt3) \le x \le \pi - \arccos(1\sqrt3)$$ How is this? Could someone explain how we can say this?
Relationship between $|\cos|$ and $\arccos$ in inequality?
1
$\begingroup$
trigonometry
inequality
1 Answers
1
$|\cos x| \le 1/\sqrt3 \Rightarrow -1/\sqrt3\leq\cos(x)\leq 1/\sqrt3\Rightarrow\ \arccos(1/\sqrt3) \le x \le \arccos(-1\sqrt3)$
and since $\arccos(-x)=\pi-\arccos(x)$ you get what you need.