Question

How do you find the exact value of `tan^-1 (-1)`?

Answer 1

`tan^-1(-1) = -pi/4`

Explanation:

`sin(pi/4) = sqrt(2)/2` so `sin(-pi/4) = -sqrt(2)/2`

`cos(pi/4) = sqrt(2)/2` so `cos(-pi/4) = sqrt(2)/2`

`tan(-pi/4) = sin(-pi/4)/cos(-pi/4) = (-sqrt(2)/2)/(sqrt(2)/2) = -1`

Note that `tan(theta)` is periodic with period `pi`. So we find:

`tan(kpi-pi/4) = -1` for any integer `k`.

However, the principal value denoted `tan^(-1)` is chosen to lie in the range `(-pi/2, pi/2)`, which includes `-pi/4`. So that is the value of `tan^(-1)(-1)`