# How do you simplify cos[Arcsin(-2/5)-Arctan3]?

Jun 29, 2016

$\frac{\sqrt{21} - 6}{5 \sqrt{10}}$, against principal values of arc sin and arc tan.
$\frac{\pm \sqrt{2} \pm 6}{5 \sqrt{10}}$, against general values..

#### Explanation:

Let a = arc sin (-2/5)3. Then, $\sin a = \left(- \frac{2}{5}\right) < 0$. So,

$\cos a = \frac{\sqrt{21}}{5}$ for principal a and $\pm \frac{\sqrt{21}}{5}$ for general a.

Let b = arc tan 3. Then, $\tan b = 3 > 0$. So,

$\sin b = \frac{3}{\sqrt{10}}$, for principal b and $\pm \frac{3}{\sqrt{10}}$, for general b..

$\cos b = \frac{1}{\sqrt{10}}$, for principal b and $\pm \frac{1}{\sqrt{10}}$, for general b..

Note that sin b and cos b are of the same sign , when tan b >0..

Now, the given expression is

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

$\frac{\sqrt{21} - 6}{5 \sqrt{10}}$,

against principal values of arc sin and arc tan.

$\frac{\pm \sqrt{2} \pm 6}{5 \sqrt{10}}$, against general values..