Question

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

Answer 1

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 = 3/sqrt10`, in the 1st quadrant, is nearly `71.57^o`.

Here,tan `theta` is negative.

So, the angle for positive `sin theta = 3/sqrt10` is in the 2nd quadrant.

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

`sin(pi-theta)=sin theta and sin (-theta) = sin (2pi-theta)= -sin theta`.
So, `theta = 180-71.57=108.43^o and -71.57^o or 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)(3/sqrt10) = 71.565...^o`, only...