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

Jul 20, 2015

${\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4}$

#### Explanation:

$\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ so $\sin \left(- \frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2}$

$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ so $\cos \left(- \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

$\tan \left(- \frac{\pi}{4}\right) = \sin \frac{- \frac{\pi}{4}}{\cos} \left(- \frac{\pi}{4}\right) = \frac{- \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = - 1$

Note that $\tan \left(\theta\right)$ is periodic with period $\pi$. So we find:

$\tan \left(k \pi - \frac{\pi}{4}\right) = - 1$ for any integer $k$.

However, the principal value denoted ${\tan}^{- 1}$ is chosen to lie in the range $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, which includes $- \frac{\pi}{4}$. So that is the value of ${\tan}^{- 1} \left(- 1\right)$