Find the value of $ an(-frac{pi}{6})$ using the unit circle.
Answer 1
Benjamin Clark
We start by recognizing that the angle $-\frac{\pi}{6}$ is equivalent to rotating $\frac{\pi}{6}$ radians in the clockwise direction.
On the unit circle, the point corresponding to $\frac{\pi}{6}$ radians is $(\frac{\sqrt{3}}{2}, \frac{1}{2})$. When we rotate in the clockwise direction to $-\frac{\pi}{6}$, the coordinates of the point become $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$.
The tangent of an angle is given by the ratio of the y-coordinate to the x-coordinate:
$\tan(-\frac{\pi}{6}) = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$
So, the value of $\tan(-\frac{\pi}{6})$ is $-\frac{\sqrt{3}}{3}$.
Answer 2
Christopher Garcia
To find the value of $ an(-frac{pi}{6})$ using the unit circle, we note that $-frac{pi}{6}$ radians corresponds to a 30-degree rotation clockwise.
On the unit circle, the coordinates for $frac{pi}{6}$ are $(frac{sqrt{3}}{2}, frac{1}{2})$. Rotating clockwise to $-frac{pi}{6}$, the coordinates change to $(frac{sqrt{3}}{2}, -frac{1}{2})$. The tangent function is given as:
$ an(-frac{pi}{6}) = frac{-frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}}.$
Rationalizing the denominator, we get:
$ an(-frac{pi}{6}) = -frac{1}{sqrt{3}} imes frac{sqrt{3}}{sqrt{3}} = -frac{sqrt{3}}{3}.$
Thus, $ an(-frac{pi}{6})$ is $-frac{sqrt{3}}{3}$.
Answer 3
Thomas Walker
To find $ an(-frac{pi}{6})$ using the unit circle, recognize that it corresponds to $(frac{sqrt{3}}{2}, -frac{1}{2})$.
Then:
$ an(-frac{pi}{6}) = frac{-frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3}.$
So, $ an(-frac{pi}{6}) = -frac{sqrt{3}}{3}$.
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