# How do you find the value for sin^-1 (-1/sqrt2)?

$- \setminus \frac{\pi}{4}$

#### Explanation:

Notice, $- \setminus \frac{\pi}{2} \setminus \le \setminus \sin x \setminus \le \setminus \frac{\pi}{2} \setminus \setminus \setminus \forall \setminus \setminus x \setminus \in R$

$\setminus \therefore \setminus {\sin}^{- 1} \left(- \frac{1}{\setminus} \sqrt{2}\right)$

$= - \setminus {\sin}^{- 1} \left(\frac{1}{\setminus} \sqrt{2}\right)$

$= - \setminus \frac{\pi}{4}$

Jul 27, 2018

As below

#### Explanation:

$\theta = {\sin}^{-} 1 \left(- \frac{1}{\sqrt{2}}\right)$

$\sin \theta = - \frac{1}{\sqrt{2}}$

$\theta = {135}^{\circ}$ or $- {45}^{\circ}$ $= {\left(\frac{5 \pi}{4}\right)}^{c}$ or ${\left(\frac{7 \pi}{4}\right)}^{c}$

Generalizing, $\theta = {\left(n \pi + \left(\frac{\pi}{4}\right)\right)}^{c}$, where n is odd integer & $\theta = {\left(n \pi - \left(\frac{\pi}{4}\right)\right)}^{c}$ where n is even integer.