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

We need to find the exact values of sine, cosine, and tangent for the angle $\frac{7\pi}{6}$ radians.

1. Find the reference angle:

The reference angle for $\frac{7\pi}{6}$ is $\pi – \frac{7\pi}{6} = \frac{\pi}{6}$.

2. Determine the signs in the third quadrant:

In the third quadrant, sine and cosine are negative, and tangent is positive.

3. Use the reference angle to find the values:

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

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

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

To determine the trigonometric values for $frac{7pi}{6}$:

1. Identify the reference angle:

The reference angle is $frac{7pi}{6} – pi = frac{pi}{6}$.

2. Confirm the quadrant:

This angle is in the third quadrant where sine and cosine are negative, and tangent is positive.

3. Calculate using the reference angle:

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

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

$ anleft(frac{7pi}{6}
ight) = frac{1}{sqrt{3}} = frac{sqrt{3}}{3}$

Solving for $frac{7pi}{6}$ in radians:

In the third quadrant:

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

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

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

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