Find the exact values of the trigonometric functions at an angle of $frac{7pi}{6}$ on the unit circle.

To solve this problem, let’s first locate the angle $\frac{7\pi}{6}$ on the unit circle. This angle is in the third quadrant because $\frac{7\pi}{6}$ is greater than $\pi$ but less than $\frac{3\pi}{2}$.

The reference angle is calculated by subtracting $\pi$ from $\frac{7\pi}{6}$:

$\frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6}$

In the third quadrant, both sine and cosine are negative. The reference angle $\frac{\pi}{6}$ has sine and cosine values of $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ respectively.

Therefore, the exact values of the trigonometric functions at $\frac{7\pi}{6}$ are:

$ \sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2} $

$ \cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2} $

$ \tan \left( \frac{7\pi}{6} \right) = \frac{\sin \left( \frac{7\pi}{6} \right)}{\cos \left( \frac{7\pi}{6} \right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $

To determine the trigonometric values at $frac{7pi}{6}$, identify the reference angle:

$ pi – frac{7pi}{6} = frac{7pi}{6} – frac{6pi}{6} = frac{pi}{6} $

Sine and cosine are negative in the third quadrant:

$ sin left( frac{7pi}{6}
ight) = -sin left( frac{pi}{6}
ight) = -frac{1}{2} $

$ cos left( frac{7pi}{6}
ight) = -cos left( frac{pi}{6}
ight) = -frac{sqrt{3}}{2} $

$ an left( frac{7pi}{6}
ight) = frac{-frac{1}{2}}{-frac{sqrt{3}}{2}} = frac{1}{sqrt{3}} = frac{sqrt{3}}{3} $

In the third quadrant:

$ frac{7pi}{6} = pi + frac{pi}{6} $

Sine and cosine:

$ sin left( frac{7pi}{6}
ight) = -frac{1}{2} $

$ cos left( frac{7pi}{6}
ight) = -frac{sqrt{3}}{2} $

$ an left( frac{7pi}{6}
ight) = frac{sqrt{3}}{3} $

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