Question

How do you solve Arctan(-sqrt3)
?

Answer 1

It basically asks you at which angle is tangent equal to\[-\sqrt{3}\]So you can say\[\tan \alpha=-\sqrt{3}\]I hope that you know the angle where tangent equals minus square root of three.

Answer 2

tan: sin/cos

Answer 3

so 2pi/3 or 5pi/3

Answer 4

????

Answer 5

"How do you solve Arctan(-sqrt3)?"

First of all, how much familiarity have you with inverse trig functions?
\[y=\tan ^{-1}(-\sqrt{3})\]

Answer 6

can be verbalized as "the angle whose tangent (ratio) is \[\frac{ -\sqrt{3} }{ 1 }\]

Answer 7

yes. according to the unit circle, i got 2pi/3 or 5pi/3
,. I don't know which one to choose

Answer 8

Here we're using y to represent that angle.

It follows that \[\tan y = \frac{ -\sqrt{3} }{ 1 }=\frac{ opp~side }{ adj~side}\]

Answer 9

Are you able to draw this situation, namely, a triangle whose opposite side, being negative, is in Quadrant III or Quadrant IV, and whose adjacent side is +1?

Answer 10

wrong drawling lol :O