# How do you prove cos^-1(-sin((2/3) pi)?

Jun 22, 2016

Note there is nothing here to prove.
If the intended question was to evaluate
color(white)("XXX")cos^(-1)(-sin((2pi)/3))=color(green)(-pi/6

#### Explanation:

$\frac{2 \pi}{3}$ is equivalent to a reference angle of $\frac{\pi}{3}$ in Quadrant II.
In Quadrant II the $\sin$ of the reference angle is equal to the $\sin$ of the actual angle.

$\frac{\pi}{3}$ is a standard angle with $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

So
$\textcolor{w h i t e}{\text{XXX}} {\cos}^{1} \left(- \sin \left(\frac{2 \pi}{3}\right)\right)$

$\textcolor{w h i t e}{\text{XXXXXX}} = {\cos}^{- 1} \left(- \frac{\sqrt{3}}{2}\right)$

${\cos}^{- 1}$ or ($\arccos$) using the standard function definitions is restricted to the range $\left(- \frac{\pi}{2} , + \frac{\pi}{2}\right]$

Within this interval only
$\textcolor{w h i t e}{\text{XXX}} \cos \left(- \frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$
(again using standard trigonometric triangles)

So
$\textcolor{w h i t e}{\text{XXX}} {\cos}^{- 1} \left(- \frac{\sqrt{3}}{2}\right) = - \frac{\pi}{6}$

Jun 23, 2016

$\frac{5 \pi}{6}$ in $\left[0. \pi\right]$.

#### Explanation:

Let us use ${\cos}^{- 1} \left(\cos x\right) = x$

Here, $- \sin \left(\frac{2 \pi}{3}\right) = \sin \left(- \frac{2 \pi}{3}\right) = \cos \left(\left(\frac{\pi}{2}\right) - \left(- \frac{2 \pi}{3}\right)\right) = \cos \left(\frac{5 \pi}{6}\right)$.

Now, the given expression is ${\cos}^{- 1} \left(\cos \left(\frac{5 \pi}{6}\right)\right) = \frac{5 \pi}{6}$.