# Find arcsin(-sqrt3/2)?

Apr 12, 2016

$\arcsin \left(- \frac{\sqrt{3}}{2}\right)$ is equal to $\left(2 n + 1\right) \pi \pm \frac{\pi}{3}$, where $n$ is an integer

#### Explanation:

$\arcsin \left(- \frac{\sqrt{3}}{2}\right)$ is the angle whose sine is $- \frac{\sqrt{3}}{2}$.

As sine is negative in second and third quadrant and $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$,

we have $\sin \left(\pi - \frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$ and

$\sin \left(p + \frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$

Hence $\arcsin \left(- \frac{\sqrt{3}}{2}\right)$ is equal to $\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$.

As adding or subtracting $2 \pi$ does not affect trigonometric ratios of angles, we can have infinite solutions given by

$\left(2 n + 1\right) \pi \pm \frac{\pi}{3}$, where $n$ is an integer.

Apr 12, 2016

$n \pi + {\left(- 1\right)}^{n} \left(- \frac{\pi}{3}\right)$, n = 0, 1, 2, 3,...

#### Explanation:

If sin x = a and $\alpha$ is the principal value of $x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, then the general value of x = $n \pi + {\left(- 1\right)}^{n} \alpha$, n = 0, 1, 2, 3,...

Here $x = a r c \sin \left(- \frac{\sqrt{3}}{2}\right)$
$\sin x = - \frac{\sqrt{3}}{2}$.
So, $\alpha = - \frac{\pi}{3}$..

May 20, 2018

$x = \frac{4 \pi}{3} + 2 k \pi$
$x = \frac{5 \pi}{3} + 2 k \pi$

#### Explanation:

$\sin x = \left(- \frac{\sqrt{3}}{2}\right)$. Find arc x (or angle x)
Unit circle gives 2 solutions for arc x (or angle x) -->
$x = - \frac{\pi}{3} + 2 k \pi$, and
$x = \pi - \left(- \frac{\pi}{3}\right) = \pi + \frac{\pi}{3} = \frac{4 \pi}{3} + 2 k \pi$
Note that $x = - \frac{\pi}{3}$ is co-terminal to $x = \frac{5 \pi}{3.}$
$x = \frac{4 \pi}{3} + 2 k \pi$
$x = \frac{5 \pi}{3} + 2 k \pi$