1. ## Help

arcsin (sin(9Pi/5) = ?
arctg (tg (7Pi/5) = ?

2. Originally Posted by Netochka
arcsin (sin(9Pi/5) = ?
arctg (tg (7Pi/5) = ?
Hi

If you can find k integer so that $-\frac{\pi}{2} \leq \theta + 2k \pi \leq \frac{\pi}{2}$ then $Arcsin(\sin(\theta)) = \theta + 2k \pi$

If not then exist n integer so that $-\frac{\pi}{2} \leq -\theta + (2n+1) \pi \leq \frac{\pi}{2}$ and then $Arcsin(\sin(\theta)) = -\theta + (2n+1) \pi$

For instance
For $\theta = \frac{9\pi}{5}$
$\theta -2 \pi = -\frac{\pi}{5}$ and $-\frac{\pi}{2} \leq -\frac{\pi}{5} \leq \frac{\pi}{2}$ then $Arcsin(\sin(\frac{9\pi}{5})) = -\frac{\pi}{5}$

For $\theta = \frac{7\pi}{5}$
$\theta -2 \pi = -\frac{3\pi}{5}$ and $-\frac{3\pi}{5} < -\frac{\pi}{2}$
$-\theta + \pi = -\frac{2\pi}{5}$
$-\frac{\pi}{2} \leq -\frac{2\pi}{5} \leq \frac{\pi}{2}$ then
$Arcsin(\sin(\frac{7\pi}{5})) = -\frac{2\pi}{5}$

For Arctan it is more simple
$Arctan(\tan(\theta)) = \theta + k \pi$ so that $-\frac{\pi}{2} < \theta + k \pi < \frac{\pi}{2}$
$Arctan(\tan(\frac{7\pi}{5})) = \frac{2\pi}{5}$

3. Thank you very much