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

##### 1 Answer
May 15, 2016

= 1/(4 sqrt 10)(( +-1 +- 3 sqrt 15 ).

#### Explanation:

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

So, a is in 1st or 2nd quadrant.

Accordingly, $\cos a = \pm \frac{\sqrt{15}}{4}$

Let $b = {\tan}^{- 1} \left(- 3\right)$. Then, $\tan b = - 3 < 0$.

So, b is in 2nd or 4th quadrant.

Accordingly, $\sin b = \pm \frac{3}{\sqrt{10}} \mathmr{and} \cos b = \pm \frac{1}{\sqrt{10}}$.

Now, the given expression = sin ( a + b ) = sin a cos b + cos a sin b

$= \left(\frac{1}{4}\right) \left(\pm \frac{1}{\sqrt{10}}\right) + \left(\pm \frac{\sqrt{15}}{4}\right) \left(\pm \frac{3}{\sqrt{10}}\right)$

= 1/(4 sqrt 10)(( +-1 +- 3 sqrt 15 ).