Determine the exact values of trigonometric functions at specific angles using the unit circle

To find the exact values of trigonometric functions for $ \theta = \frac{5\pi}{6} $, we first recognize that this angle corresponds to a reference angle of $ \frac{\pi}{6} $ in the second quadrant.

The coordinates of the point on the unit circle at $ \frac{\pi}{6} $ are $ (\frac{\sqrt{3}}{2}, \frac{1}{2}) $. Since $ \frac{5\pi}{6} $ lies in the second quadrant, the x-coordinate becomes negative:

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

$ \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} $

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

To determine the trigonometric values for an angle of $ frac{7pi}{4} $, note that this angle is equivalent to $ -frac{pi}{4} $, putting it in the fourth quadrant.

The coordinates on the unit circle for $ frac{pi}{4} $ are $ (frac{sqrt{2}}{2}, frac{sqrt{2}}{2}) $. For $ -frac{pi}{4} $, the y-coordinate is negative:

$ cos left( frac{7pi}{4}
ight) = frac{sqrt{2}}{2} $

$ sin left( frac{7pi}{4}
ight) = -frac{sqrt{2}}{2} $

$ an left( frac{7pi}{4}
ight) = frac{sin left( frac{7pi}{4}
ight)}{cos left( frac{7pi}{4}
ight)} = frac{-frac{sqrt{2}}{2}}{frac{sqrt{2}}{2}} = -1 $

For $ heta = frac{3pi}{2} $, which lies on the negative y-axis:

$ cos left( frac{3pi}{2}
ight) = 0 $

$ sin left( frac{3pi}{2}
ight) = -1 $

$ an left( frac{3pi}{2}
ight) = frac{sin left( frac{3pi}{2}
ight)}{cos left( frac{3pi}{2}
ight)} = frac{-1}{0} ext{ (undefined)} $

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