How do you find the value of #cot^-1 (-1/sqrt3)#?

1 Answer
Aug 12, 2015

#cot^(-1)(-1/sqrt(3)) = 150^o + n*180^o = (5pi)/6 + npi#
#color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXX")#for #AA n in ZZ#

Explanation:

If #theta = cot^(-1)(-1/sqrt(3))#
the
the ratio #("opposite side")/("adjacent side")# of the defining triangle
must be
#color(white)("XXXX")##(-1):(sqrt(3))# (or equivalently #(1):(-sqrt(3))#)

We therefore have the conditions below:enter image source here

These are standard reference triangles and based on their quadrants the corresponding angles (within the range #[0,360^o]#
are
#color(white)("XXXX")##180^o - 30^0 = 150^o = (5pi)/6# radians
and
#color(white)("XXXX")##360^o - 30^o = 330^o = (11pi)/6# radians

These can be combined and include angles outside the #[0,2pi]# range as
#color(white)("XXXX")##theta = 150^o +n*180^o#
or
#color(white)("XXXX")##theta = (5pi)/6 + n*pi#
#color(white)("XXXX")##color(white)("XXXX")##color(white)("XXXX")#for #AA n in ZZ#