Inverse Trig Functions: Evaluating Expressions in Radians

[shiri]

Evaluate the following expressions. Your answer must be in radians.

a) arctan(-(sqrt3)/3)

b) arctan((sqrt3)/3)

c) arctan(-sqrt3)


What I got are:

a) 5pi/6

b) pi/6

c) 2pi/3


Do I got these answers right?

 

[tiny-tim]

Hi shiri!

(have a pi: π and a square-root: √ )

Yes and no.

The tangents of those angles are the numbers given.

But "arctan" mean the principal value, which is usually taken to be between -π/2 and π/2 …

see http://en.wikipedia.org/wiki/Arctan" .

(The reason why that range is chosen (and not the [0,2π) for arccos and arcsin) is because tan goes smoothly from -∞ to ∞ in that range.)

You've chosen the range from 0 to 2π (of course, if your professor told you to do so, ignore wikipedia and me ).