# How do you evaluate  sin(arctan(-3))?

Apr 8, 2016

I want to add some nuances relating to $\theta$ in the fine answer given by Eric Sea. See the explanation.

#### Explanation:

The angle for $\sin \theta = \frac{3}{\sqrt{10}}$, in the 1st quadrant, is nearly ${71.57}^{o}$.

Here,tan $\theta$ is negative.

So, the angle for positive $\sin \theta = \frac{3}{\sqrt{10}}$ is in the 2nd quadrant.

The angle for the opposite sign is in the 4th quadrant,

$\sin \left(\pi - \theta\right) = \sin \theta \mathmr{and} \sin \left(- \theta\right) = \sin \left(2 \pi - \theta\right) = - \sin \theta$.
So, $\theta = 180 - 71.57 = {108.43}^{o} \mathmr{and} - {71.57}^{o} \mathmr{and} {288.43}^{o}$. nearly.-

I think that I have elucidated on the fact that $\theta$ is not ${71.57}^{o}$.

The nuances like this are important when direction is important, particularly in space mechanics, to avoid blunders.

Library functions in the calculator/computer might give ${\sin}^{- 1} \left(\frac{3}{\sqrt{10}}\right) = 71.565 {\ldots}^{o}$, only...