# How do you evaluate arcsin(cos(10pi/9))?

Nov 24, 2015

$\frac{- 7 \pi}{18}$

#### Explanation:

$\left[1\right] \text{ } \arcsin \left[\cos \left(\frac{10 \pi}{9}\right)\right]$

Convert $\cos \left(\frac{10 \pi}{9}\right)$ to $\sin$ using the co-function identity.

$\left[2\right] \text{ } = \arcsin \left[\sin \left(\frac{\pi}{2} - \frac{10 \pi}{9}\right)\right]$

Simplify.

$\left[3\right] \text{ } = \arcsin \left[\sin \left(\frac{9 \pi}{18} - \frac{20 \pi}{18}\right)\right]$

$\left[4\right] \text{ } = \arcsin \left[\sin \left(\frac{- 11 \pi}{18}\right)\right]$

The restricted domain for sine is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ since this involves an $\arcsin$. We must look for an angle in this restricted domain whose sine has an equal value to $\sin \left(\frac{- 11 \pi}{18}\right)$. That angle is $\frac{- 7 \pi}{18}$ because it has the same reference angle and sign as $\frac{- 11 \pi}{18}$.

$\left[5\right] \text{ } = \arcsin \left[\sin \left(\frac{- 7 \pi}{18}\right)\right]$

Arcsin and sin will cancel because of the definition of inverse functions.

$\left[6\right] \text{ } \textcolor{b l u e}{= \frac{- 7 \pi}{18}}$