The solution is $-\frac16\pi$, ns don"t understand why?

If i pull up $-\frac16\pi$ top top a unit circle device (in this case on cannes Academy). The $y$ (the sine) value on the exact same angle $-\frac16\pi$ is $-\frac12$. I check out that the $x$ worth (the cosine) top top unit one of $-\frac16\pi$ is $\frac\sqrt32$.

Tan is opposite side / adjacent side, or in the unit circle"s instance sine/cosine. Which says to me the the tan of $-\frac16\pi$ is $-(1/2)/(\sqrt3/2)$ which equals $-1/\sqrt3$ no $-\frac16\pi$.

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

Any insight would be great!

trigonometry

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edited Dec 8 "11 in ~ 0:19

André Nicolas

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drcdrc

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## 2 answers 2

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Maybe you"re just lacking the reality that $1/\sqrt3$ is the very same as $\sqrt3/3$. You acquire that by rationalizing the denominator:

$$\frac1\sqrt3 = \frac1\cdot\sqrt3\sqrt3\sqrt3 = \frac\sqrt33.$$

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reply Dec 8 "11 in ~ 0:13

Michael HardyMichael durable

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Maybe you"re simply messing with the concept of

*inverse function*... Remember:

$\arctan x=y$ if and only if $\tan y =x$ and $y\in >-\pi/2,\pi/2<$.

Now, you have uncovered $\tan (-\pi/6) =-1/\sqrt3=-\sqrt3/3$ and you additionally have $-\pi/6 \in >-\pi/2, \pi/2<$, for this reason previous declare applies and also it returns $\arctan (-\sqrt3/3)=-\pi/6$ together claimed.

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edited Mar 5 "12 at 19:37

answered Dec 8 "11 at 0:28

PacciuPacciu

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