Question

sin[tan^-1(-8)]

Answer 1

Is tan^-1(-8) the arctan of -8 ?

Answer 2

Suppose you let \(x=\tan^{-1}(-8)\). This would mean that \(\tan x=-8\). Keep in mind that \(\tan x\) is defined for \(\left(\dfrac{(2n+1)\pi}{2},\dfrac{(2n+3)\pi}{2}\right)\) for integers \(n\). Its inverse is typically defined by restricting the domain of \(\tan x\) to \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\) (i.e. the first and fourth quadrants of the unit circle).

In terms of the quadrants, \(\tan x\) is negative for values of \(x\) within the second and fourth quadrants. Draw up a reference triangle with angle \(x\) such that \(\tan x=-8=\dfrac{-8}{1}\):

If you can find the missing side, you can determine \(\sin x=\sin\left(\tan^{-1}(-8)\right)\).