Find the value of $ an( heta)$ on the unit circle where $ heta$ is a special angle
Answer 1
Lucas Brown
To find the value of $\tan(\theta)$ on the unit circle, we need to use the relationship $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.
Let’s consider $\theta = \frac{\pi}{4}$. On the unit circle, $\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.
Thus,
$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1.$
So, $\tan\left(\frac{\pi}{4}\right) = 1$.
Answer 2
Mia Harris
To find $ an( heta)$ on the unit circle, we use $ an( heta) = frac{sin( heta)}{cos( heta)}$.
Consider $ heta = frac{2pi}{3}$. On the unit circle, $sinleft(frac{2pi}{3}
ight) = frac{sqrt{3}}{2}$ and $cosleft(frac{2pi}{3}
ight) = -frac{1}{2}$.
Therefore,
$ anleft(frac{2pi}{3}
ight) = frac{sinleft(frac{2pi}{3}
ight)}{cosleft(frac{2pi}{3}
ight)} = frac{frac{sqrt{3}}{2}}{-frac{1}{2}} = -sqrt{3}.$
So, $ anleft(frac{2pi}{3}
ight) = -sqrt{3}$.
Answer 3
Henry Green
Find $ an( heta)$ on the unit circle using $ an( heta) = frac{sin( heta)}{cos( heta)}$ for $ heta = frac{pi}{6}$.
On the unit circle, $sinleft(frac{pi}{6}
ight) = frac{1}{2}$ and $cosleft(frac{pi}{6}
ight) = frac{sqrt{3}}{2}$.
Therefore,
$ anleft(frac{pi}{6}
ight) = frac{sinleft(frac{pi}{6}
ight)}{cosleft(frac{pi}{6}
ight)} = frac{frac{1}{2}}{frac{sqrt{3}}{2}} = frac{1}{sqrt{3}} = frac{sqrt{3}}{3}.$
So, $ anleft(frac{pi}{6}
ight) = frac{sqrt{3}}{3}$.
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