Question

How do you evaluate `cot (- pi/6)`?

Answer 1

`cot(-pi/6) = -sqrt(3)`

Explanation:

We will use the following:
(1) `cot(x) = cos(x)/sin(x)`
(2) `sin(-x) = -sin(x)`
(3) `cos(-x) = cos(x)`
(4) `cos(pi/6) = sqrt(3)/2`
(5) `sin(pi/6) = 1/2`
Verification of these facts is a good exercise, and may be done using basic definitions together with the unit circle for 1-3.
For 4 and 5, as a hint, bisect an equilateral triangle with side length `1` into two equal right triangles. What are the angles of the right triangles?

With this, we have
`cot(-pi/6) =cos(-pi/6)/sin(-pi/6)` (by 1)
`=>cot(-pi/6) = cos(pi/6)/(-sin(pi/6))` (by 2 and 3)
`=>cot(-pi/6) = -(sqrt(3)/2)/(1/2) = -sqrt(3)` (by 4 and 5)