To solve $\cot\left(\frac{8\pi}{3}\right)$ using the unit circle:

1. Step 1: Simplify the angle $\frac{8\pi}{3}$ to a coterminal angle between $0$ and $2\pi$. \frac{8\pi}{3} \quad \text{is greater than 2\pi$. Subtract$2\pi:} $\frac{8\pi}{3} - 2\pi = \frac{8\pi}{3} - \frac{6\pi}{3} = \frac{2\pi}{3}.$ So, $\frac{8\pi}{3}$ is coterminal with $\frac{2\pi}{3}$.

2. Step 2: Identify the reference angle. The angle $\frac{2\pi}{3}$ is in the second quadrant (because $\pi < \frac{2\pi}{3} < \pi/2$). The reference angle is: $\pi - \frac{2\pi}{3} = \frac{\pi}{3}.$

3. Step 3: Use the unit circle to find $\cot\left(\frac{2\pi}{3}\right)$.

   • In the second quadrant, $\cot \theta = -\cot(\text{reference angle})$ (cotangent is negative in quadrant II).
   • The reference angle is $\frac{\pi}{3}$, where: $\cot\left(\frac{\pi}{3}\right) = \frac{\cos\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}\right)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.$
   • So: $\cot\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{3}.$

4. Step 4: Conclude the solution. Since $\cot\left(\frac{8\pi}{3}\right)$ is coterminal with $\frac{2\pi}{3}$: $\cot\left(\frac{8\pi}{3}\right) = -\frac{\sqrt{3}}{3}.$

If you have any further questions or would like a deeper breakdown, let me know!

Related Questions:

1. How do you find the reference angle for angles in different quadrants?
2. Why is the cotangent negative in the second quadrant?
3. What are the steps to simplify angles larger than $2\pi$?
4. How can you quickly determine $\sin$ and $\cos$ values for common angles like $\frac{\pi}{3}$?
5. How can you use the reciprocal property of cotangent to check your result?

Tip: Memorize the unit circle values for $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3},$ and $\pi/2$. This makes solving trigonometric problems easier and faster!