# How do you evaluate arccos(sin(13pi/7))?

Jul 6, 2016

$\left(- \frac{19}{14}\right) \pi$

#### Explanation:

Use $\sin \left(\frac{\pi}{2} - a\right) = \cos a \mathmr{and} a r c \cos \left(\cos b\right) = b$.

Here,

$\sin \left(\left(\frac{13}{7}\right) \pi\right) = \cos \left(\frac{\pi}{2} - \left(\frac{13}{7}\right) \pi\right) = \cos \left(- \left(\frac{19}{14}\right) \pi\right)$.

The given expression

= arc cos (cos(-(19/14)pi)

$= - \left(\frac{19}{14}\right) \pi = - {244.3}^{o}$

Note that, for any arbitrary a, $\sin a = \cos \left(\frac{\pi}{2} - a\right)$.

Do not use cos (-b)=cos b, in any intermediate step..

Here, for the start angle $\left(\frac{13}{7}\right) \pi$ (not the principal value), the end

angle is $- \left(\frac{19}{14}\right) \pi$

${114.7}^{o} = \frac{9 \pi}{14} = \left(2 \pi - \left(\frac{19}{14}\right) \pi\right)$..