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!
edited Dec 8 "11 in ~ 0:19
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asked Dec 7 "11 at 23:55
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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:
$$\frac1\sqrt3 = \frac1\cdot\sqrt3\sqrt3\sqrt3 = \frac\sqrt33.$$
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.
edited Mar 5 "12 at 19:37
answered Dec 8 "11 at 0:28
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