# How do you calculate Cos^-1 [cos ((7pi)/6) ]?

May 24, 2018

The answer will be $\frac{5 \pi}{6}$

#### Explanation:

You can write $\frac{7 \pi}{6}$ as $\left(\pi + \frac{\pi}{6}\right)$
Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative.

Hence, $\cos \left(\pi + \frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right)$

coming back to the question
${\cos}^{-} 1 \left[\cos \left(\frac{7 \pi}{6}\right)\right] = {\cos}^{-} 1 \left[- \cos \left(\frac{\pi}{6}\right)\right]$
=$\pi - {\cos}^{-} 1 \left[\cos \left(\frac{\pi}{6}\right)\right]$
=$\pi - \frac{\pi}{6}$
=$\frac{5 \pi}{6}$

May 24, 2018

$\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \pm \frac{7 \pi}{6} + 2 \pi k ,$ integer $k$

$\textrm{A r c} \textrm{\cos} \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \frac{5 \pi}{6}$

#### Explanation:

I treat $\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right)$ as a multivalued expression, all the angles whose cosine equals $\cos \left(\frac{7 \pi}{6}\right) .$

In general $\cos x = \cos a$ has solution $x = \pm a + 2 \pi k ,$ integer $k$

$x = \arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right)$

$\cos x = \cos \left(\frac{7 \pi}{6}\right)$

$x = \pm \frac{7 \pi}{6} + 2 \pi k$

$\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \pm \frac{7 \pi}{6} + 2 \pi k ,$ integer $k$

The principal value is in the second quadrant, given here by the minus sign and $k = 1.$

$\textrm{A r c} \textrm{\cos} \left(\cos \left(\frac{7 \pi}{6}\right)\right) = - \frac{7 \pi}{6} + 2 \pi = \frac{5 \pi}{6}$