## anonymous one year ago sin[tan^-1(-8)]

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1. anonymous

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

2. anonymous

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 $$\tan x=-8=\dfrac{-8}{1}$$: |dw:1439434906616:dw| If you can find the missing side, you can determine $$\sin x=\sin\left(\tan^{-1}(-8)\right)$$.