I"ve been studying the unit circle and inverse trig functions on Khan Academy. One of the questions asked, is, what is the arctan of \$-fracsqrt33\$.

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
asked Dec 7 "11 at 23:55 drcdrc
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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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