## Finding the principle value of arctan

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
chiefboo
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### Finding the principle value of arctan

http://i.imgur.com/Hz4w1SI.png
$\mbox{Q. What is the principal value, in terms of radians, of }\,\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)?$

$\mbox{A. If }\,\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\, =\, \theta,\,\mbox{ then }\,\tan(\theta)\, =\, -\frac{\sqrt{3}}{3}.$

$\mbox{The range of the inverse tangent is }\,\left[-\frac{\pi}{2},\, \frac{\pi}{2}\right],\,$

. . . .$\mbox{ so we know that }\, -\frac{\pi}{2}\, \leq \,\theta\, \leq \, \frac{\pi}{2}.$

$\mbox{Also, we know that }\,\tan\left(-\frac{\pi}{6}\right)\, =\, -\frac{\sqrt{3}}{2}$

$\mbox{So }\,\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\, =\,-\frac{\pi}{6}$.
I don't understand how you get -(1/6)pi. It didn't really explain how that was calculated. I've been setting the neg square root of 3 over 3 equal to y/x, solving for y then pluging it in to the Pythagorean theorem, but it's not working out. Is this the correct way to solve this? Is anyone else about to get -1/6pi with this method?

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Joined: Sun Feb 22, 2009 11:12 pm

### Re: Finding the principle value of arctan

I don't understand how you get -(1/6)pi. It didn't really explain how that was calculated.
It probably wasn't. it's one of the angles youre supposed to memorize like they show here.
I've been setting the neg square root of 3 over 3 equal to y/x, solving for y then pluging it in to the Pythagorean theorem, but it's not working out.
What are you getting? plz show the steps. thnx.

chiefboo
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Joined: Fri Mar 22, 2013 3:46 pm
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### Re: Finding the principle value of arctan

Thanks for the reply shadow. I've moved on from this and have resigned myself to just memorizing this angle. I'm trying to prepare myself for the CLEP Calculus exam so I'm trying to move as quickly as possible to see if I even have a chance. I'll probably wind up having to take the CLEP Precal instead though.