The Purplemath Forums

by **chiefboo** on Sun Apr 07, 2013 1:24 pm

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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by **theshadow** on Sun Apr 07, 2013 8:14 pm

chiefboo wrote: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.

chiefboo wrote: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.

by **chiefboo** on Mon Apr 08, 2013 12:28 am

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.

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