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

Question

3163 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T06:04:56

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

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