# How do you evaluate Sin(arcsin(3/5)+(arctan-2))?

May 21, 2017

$- \frac{1}{\sqrt{5.}}$

#### Explanation:

Let arc sin(3/5)=x, and, arc tan(-2)=y; x,y in (-pi/2,pi/2).

$\therefore \sin x = \frac{3}{5} , \mathmr{and} , \tan y = - 2.$

We note that, sinx > 0, x in (0,pi/2), &, tany <0, y in (-pi/2,0).

Since, the Reqd. Value

$= \sin \left(x + y\right) = \sin x \cos y + \cos x \sin y , \ldots . . \left(\ast\right)$

We have to find the values of $\cos y , \sin y \mathmr{and} \cos x .$

$\sin x = \frac{3}{5} \Rightarrow \cos x = \sqrt{1 - {\sin}^{2} x} = \sqrt{1 - \frac{9}{25}} = \pm \frac{4}{5} ,$

$\text{but, } \because , x \in \left(0 , \frac{\pi}{2}\right) , \cos x = + \frac{4}{5.} \ldots \ldots . . \left(1\right) .$

Again, $\tan y = - 2 \Rightarrow \sec y = \sqrt{1 + {\tan}^{2} y} = \pm \sqrt{5} ,$ but with

$y \in \left(- \frac{\pi}{2} , 0\right) , \sec y = + \sqrt{5} , \text{ giving, } \cos y = + \frac{1}{\sqrt{5.}} . . \left(2\right) .$

Finally, $\sin y = \tan y \cdot \cos y = - 2 \cdot \frac{1}{\sqrt{5}} = - \frac{2}{\sqrt{5.}} \ldots \ldots \ldots . \left(3\right) .$

Utilising $\left(1\right) , \left(2\right) \mathmr{and} \left(3\right) \in \left(\ast\right) ,$ we get,

$\text{The Reqd. Value=} \left(\frac{3}{5}\right) \left(\frac{1}{\sqrt{5}}\right) + \left(\frac{4}{5}\right) \left(- \frac{2}{\sqrt{5}}\right) = - \frac{1}{\sqrt{5.}}$

Enjoy Maths.!