How do you evaluate #arctan(-1/sqrt3)#?

1 Answer
Ratnaker Mehta
Jul 11, 2016

#arctan(-1/sqrt3)=-pi/6.#

Explanation:

Let #arctan(-1/sqrt3)=theta, theta in (-pi/2,pi/2)#

Then, by defn. of #arctan# fun, #tantheta=-1/sqrt3 <0,# so that #theta !in (0,pi/2)#

Now, #tan(-pi/6)=-tan(pi/6)=-1/sqrt3,# where, #(-pi/6) in (-pi/2,0)#

Thus, #tan(-pi/6)=-1/sqrt3=tantheta,# and, #tan# fun. is injective i.e., #1-1# in #(-pi/2,0)#, we conclude that #theta=arctan(-1/sqrt3)=-pi/6.#

Related questions
See all questions in Basic Inverse Trigonometric Functions
Impact of this question
30638 views around the world
You can reuse this answer
Creative Commons License