# How do you simplify sin[cos^-1( - sqrt5 / 5 ) + tan^-1 ( - 1 / 3) ]?

May 26, 2016

$= \pm 0.7 \sqrt{2} \mathmr{and} \pm \frac{1}{\sqrt{2}} = \pm 0.98995 \mathmr{and} \pm 0.70711$, nearly.

#### Explanation:

Let $a = {\cos}^{- 1} \left(- \frac{\sqrt{5}}{5}\right)$. Then $\cos a = - \frac{\sqrt{5}}{5} = - \frac{1}{\sqrt{5}} < 0$. so, a is in the 2nd quadrant or in the 4th. Accordingly, $\sin a = \pm \frac{2}{\sqrt{5}}$.

Let $b = {\tan}^{- 1} \left(- \frac{1}{3}\right)$. Then $\tan b = - \frac{1}{3} < 0$. So, b is in the 2nd quadrant or in the 4th. Accordingly, $\sin b = \pm \frac{1}{\sqrt{10}} \mathmr{and} \cos b = \frac{3}{\sqrt{10}}$. Also, sin b and cos b have opposite signs.

Now, the given expression is

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

$= \left(\pm \frac{2}{\sqrt{5}}\right) \left(\pm \frac{3}{\sqrt{10}}\right) - \left(- \frac{1}{\sqrt{5}}\right) \left(\pm \frac{1}{\sqrt{10}}\right)$

$= \pm \frac{6}{\sqrt{50}}$ $-$ or + $\frac{1}{\sqrt{50}}$

$= \left(\frac{\sqrt{2}}{10}\right) \left(\pm 6 \pm 1\right)$

$= \pm 0.7 \sqrt{2} \mathmr{and} \pm \frac{1}{\sqrt{2}}$

In each case, the angles a and b can be obtained separately. .