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

1 Answer
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 = 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...