Question

Simplify
${\cos }^{-1}$ (cos $\frac{8\pi }{7}$).

Answer 1

we know that the range of ${\cos }^{-1}x$ is $[0,\pi ]$

Now, ${\cos }^{-1}\left(\cos \right(8\pi /7\left)\right)$

$={\cos }^{-1}\left(\cos \right(\pi +\pi /7\left)\right)$

$={\cos }^{-1}(-\cos (\pi /7\left)\right)$

$={\cos }^{-1}\left(\cos \right(\pi -\pi /7\left)\right)$

$={\cos }^{-1}\left(\cos \right(6\pi /7\left)\right)$

$=\frac{6\pi }{7}$