# How do you evaluate sin (tan^-1(1/4) + cos^-1(-2/3))?

May 14, 2016

$= \frac{2}{3 \sqrt{17}} \left(\pm 1 \pm 2 \sqrt{5}\right)$

#### Explanation:

Let $a = {\tan}^{- 1} \left(\frac{1}{4}\right) . \tan a = \frac{1}{4} > 0.$.

So, the angle a is either in the 1st quadrant or in the 3rd.

Accordingly,

sin a = +-1/sqrt 17 and cos a = +-4/sqrt 17#.

Let $b = {\cos}^{- 1} \left(- \frac{2}{3}\right) . \cos b = - \frac{2}{3} < 0.$.

So, the angle b is either in the 2nd quadrant or in the 3rd.

Accordingly, $\sin b = \pm \frac{\sqrt{5}}{3}$..

Now, the given expression is

sin (a + b ) = sin a cos b + cos a sin b

$= \pm \left(\frac{1}{\sqrt{17}}\right) \left(- \frac{2}{3}\right) \pm \left(\frac{4}{\sqrt{17}}\right) \left(\frac{\sqrt{5}}{3}\right)$

$= \frac{2}{3 \sqrt{17}} \left(\pm 1 \pm 2 \sqrt{5}\right)$