The solution is $-frac16pi$, I don"t understand why?

If I pull up $-frac16pi$ on a unit circle tool (in this case on Khan Academy). The $y$ (the sine) value on the same angle $-frac16pi$ is $-frac12$. I see that the $x$ value (the cosine) on unit circle of $-frac16pi$ is $fracsqrt32$.

Tan is opposite side / adjacent side, or in the unit circle"s case sine/cosine. Which suggests to me that the tan of $-frac16pi$ is $-(1/2)/(sqrt3/2)$ which equals $-1/sqrt3$ not $-frac16pi$.

You are watching: Arctan(-sqrt(3))

Any insight would be great!

trigonometry

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edited Dec 8 "11 at 0:19

André Nicolas

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asked Dec 7 "11 at 23:55

drcdrc

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## 2 Answers 2

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Maybe you"re just missing the fact that $1/sqrt3$ is the same as $sqrt3/3$. You get that by rationalizing the denominator:

$$frac1sqrt3 = frac1cdotsqrt3sqrt3sqrt3 = fracsqrt33.$$

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answered Dec 8 "11 at 0:13

Michael HardyMichael Hardy

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Maybe you"re just messing with the concept of

*inverse function*... Remember:

$arctan x=y$ if and only if $ an y =x$ and $yin >-pi/2,pi/2<$.

Now, you have found $ an (-pi/6) =-1/sqrt3=-sqrt3/3$ and you also have $-pi/6 in >-pi/2, pi/2<$, hence previous statement applies and it yields $arctan (-sqrt3/3)=-pi/6$ as claimed.

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Folshort

edited Mar 5 "12 at 19:37

answered Dec 8 "11 at 0:28

PacciuPacciu

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