How do you evaluate arctan(−1√3)?

1 Answer
Ratnaker Mehta
Jul 11, 2016

arctan(−1√3)=−π6.

Explanation:

Let arctan(−1√3)=θ,θ∈(−π2,π2)

Then, by defn. of arctan fun, tanθ=−1√3<0, so that θ∉(0,π2)

Now, tan(−π6)=−tan(π6)=−1√3, where, (−π6)∈(−π2,0)

Thus, tan(−π6)=−1√3=tanθ, and, tan fun. is injective i.e., 1−1 in (−π2,0), we conclude that θ=arctan(−1√3)=−π6.

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
32323 views around the world
You can reuse this answer
Creative Commons License