# How do you simplify Cos[cos^-1 (3/5) - sin^-1 (-4/5)] ?

Jun 21, 2016

Let $a = {\cos}^{- 1} \left(\frac{3}{5}\right)$. Then, $\cos a = \frac{3}{5} > 0$.a is in either 1st

quadrant or in the 4th. Accordingly, $\sin a = \pm \frac{4}{5}$.

Let $b = {\sin}^{- 1} \left(- \frac{4}{5}\right)$. Then, $\sin b = - \frac{4}{5} > 0$.b is in either 3rd

quadrant or in the 4th. Accordingly, $\cos b = \pm \frac{3}{5}$.

Now, the given expression is

$\cos \left(a - b\right) = \cos a \cos b + \sin a \sin b$

=((3/5)(+-3/5)-(+-4/5)((-4/5))

$= \frac{\pm 9 \pm 16}{25}$

$= \pm 1 , \pm \frac{7}{25}$.