Determine the tangent values for points on the unit circle at angles $ heta = frac{pi}{4}$, $ heta = frac{2pi}{3}$, and $ heta = frac{5pi}{6}$

To find the tangent values for the given angles on the unit circle, we use the definition of the tangent function, which is $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.

For $ \theta = \frac{\pi}{4} $:

$ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \text{ and } \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $

Therefore,

$ \tan\left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $

For $ \theta = \frac{2\pi}{3} $:

$ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \text{ and } \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} $

Therefore,

$ \tan\left(\frac{2\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3} $

For $ \theta = \frac{5\pi}{6} $:

$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \text{ and } \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} $

Therefore,

$ \tan\left(\frac{5\pi}{6}\right) = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $

To find the tangent values at the given angles:

For $ heta = frac{pi}{4} $:

$ anleft(frac{pi}{4}
ight) = 1 $

For $ heta = frac{2pi}{3} $:

$ anleft(frac{2pi}{3}
ight) = -sqrt{3} $

For $ heta = frac{5pi}{6} $:

$ anleft(frac{5pi}{6}
ight) = -frac{sqrt{3}}{3} $

For $ heta = frac{pi}{4} $, $ an( heta) = 1 $.

For $ heta = frac{2pi}{3} $, $ an( heta) = -sqrt{3} $.

For $ heta = frac{5pi}{6} $, $ an( heta) = -frac{sqrt{3}}{3} $.

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