# What is the sin^(-1)(sqrt(3)/2)?

Sep 7, 2015

${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ (radians)

#### Explanation:

${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = \theta$
$\textcolor{w h i t e}{\text{XX}}$means
$\sin \left(\theta\right) = \frac{\sqrt{3}}{2}$
$\textcolor{w h i t e}{\text{XXXX}}$with the restriction that $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

The possible triangles with this ratio of $y$ to $\text{hypotenuse}$ (i.e. $\sin$) are pictured below:

...but note that only $\frac{\pi}{3}$ falls within the restricted range for ${\sin}^{- 1}$

May 10, 2016

$\theta$ = 60° or 120°

#### Explanation:

It helps if one recognises the lengths 3 and $\sqrt{2}$ as
being two of the sides in one of the special triangle,
namely 30°, 60° 90°.

Sketch this by drawing an equilateral triangle of sides 2 units, and drawing in a line of symmetry to form two right-angled triangles.

The hypotenuse is of length 2 units and the opposite side
is of length $\sqrt{3}$. The angle is therefore 60°.

However, the sign is positive. Sin is positive in the first and second quadrants.

Therefore $\theta$ = 60° or 120°