$ ext{Determine the exact values of trigonometric functions at specific angles on the unit circle.}$

To find the exact values of the trigonometric functions for the angle $\frac{7\pi}{6}$:

1. Find the reference angle by subtracting $\pi$: $\frac{7\pi}{6} – \pi = \frac{7\pi}{6} – \frac{6\pi}{6} = \frac{\pi}{6}$

2. Determine the coordinates on the unit circle for $\frac{\pi}{6}$, which are $(\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$. These are $\left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)$

3. Since $\frac{7\pi}{6}$ is in the third quadrant, both cosine and sine are negative: $\left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$

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

To determine the values of the trigonometric functions at the angle $frac{5pi}{4}$:

1. Calculate the reference angle: $frac{5pi}{4} – pi = frac{5pi}{4} – frac{4pi}{4} = frac{pi}{4}$

2. The coordinates corresponding to $frac{pi}{4}$ on the unit circle are $left( cos frac{pi}{4}, sin frac{pi}{4}
ight) = left( frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight)$

3. Identify the signs in the third quadrant, where both cosine and sine are negative: $left( -frac{sqrt{2}}{2}, -frac{sqrt{2}}{2}
ight)$

Hence, $cos left( frac{5pi}{4}
ight) = -frac{sqrt{2}}{2}$ and $sin left( frac{5pi}{4}
ight) = -frac{sqrt{2}}{2}$

To find the trigonometric values at $frac{11pi}{6}$:

1. Reference angle: $frac{11pi}{6} – 2pi = frac{11pi}{6} – frac{12pi}{6} = -frac{pi}{6}$

2. Coordinates for $frac{pi}{6}$: $left( cos frac{pi}{6}, sin frac{pi}{6}
ight) = left( frac{sqrt{3}}{2}, frac{1}{2}
ight)$

3. Fourth quadrant sign adjustment: $ left( frac{sqrt{3}}{2}, -frac{1}{2}
ight)$

Thus, $cos left( frac{11pi}{6}
ight) = frac{sqrt{3}}{2}$ and $sin left( frac{11pi}{6}
ight) = -frac{1}{2}$

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