# How do you find the exact value of Arctan (-1) ?

Apr 25, 2016

$a r c \tan \left(- 1\right) = n \pi + \frac{3 \pi}{4} , n = 0 , \pm 1 , \pm 2 , \pm 3. . .$ n=0, gives its value as $\frac{3 \pi}{4} \in \left[0 , \pi\right]$. Including n=1, we get two values $\frac{3 \pi}{4} \mathmr{and} \frac{7 \pi}{4} \in \left[0 , 2 \pi\right]$.

#### Explanation:

Use $\tan \left(\pi - x\right) = \tan x$.
$\tan \left(\pi - \frac{\pi}{4}\right) = - \tan \left(\frac{\pi}{4}\right) = - 1$

So, a principal value of $a r c \tan \left(- 1\right) \in \left[0 , \pi\right] i s \frac{3 \pi}{4}$.
There is another in [0, 2pi]. tan(2pi-pi/4)=-tan (pi/4)=-1

So, there are two, $\frac{3 \pi}{4} \mathmr{and} \frac{7 \pi}{4} , \in \left[0 , 2 \pi\right]$.

Either can be taken as principal value, for the general value

$a r c \tan \left(- 1\right) = n \pi + \frac{3 \pi}{4} , n = 0 , \pm 1 , \pm 2 , \pm 3. . .$ . .