The given problem is $\cot\left(\frac{8\pi}{3}\right)$. Let's find its exact value step by step:

Step 1: Simplify the angle

The given angle is $\frac{8\pi}{3}$. Since the cotangent function is periodic with a period of $\pi$, we can simplify the angle by reducing it within the range $[0, 2\pi)$.

To do this, subtract $2\pi$ from $\frac{8\pi}{3}$ until it falls within $[0, 2\pi)$: $\frac{8\pi}{3} - 2\pi = \frac{8\pi}{3} - \frac{6\pi}{3} = \frac{2\pi}{3}.$

So, $\frac{8\pi}{3} \equiv \frac{2\pi}{3}$ (mod $2\pi$).

Step 2: Evaluate $\cot\left(\frac{2\pi}{3}\right)$

The angle $\frac{2\pi}{3}$ is in the second quadrant. In the second quadrant:

• The cotangent function is negative because $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$, and $\cos(\theta)$ is negative while $\sin(\theta)$ is positive.

The reference angle for $\frac{2\pi}{3}$ is: $\pi - \frac{2\pi}{3} = \frac{\pi}{3}.$

Now, calculate $\cot\left(\frac{\pi}{3}\right)$: $\cot\left(\frac{\pi}{3}\right) = \frac{\cos\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}\right)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}.$

Thus: $\cot\left(\frac{2\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}.$

Step 3: Simplify the result

The simplified exact value is: $\boxed{-\frac{\sqrt{3}}{3}}$

Would you like further clarifications or a graph to visualize this result?

Related Questions:

1. What is the value of $\sin\left(\frac{8\pi}{3}\right)$ and $\cos\left(\frac{8\pi}{3}\right)$?
2. How does the periodicity of $\cot(x)$ affect this solution?
3. Can you verify this result using a unit circle diagram?
4. What is the relationship between $\cot(x)$ and $\tan(x)$?
5. How would the result differ if the problem was $\cot\left(\frac{7\pi}{3}\right)$?

Tip:

Always simplify trigonometric angles to their principal range (e.g., $[0, 2\pi)$ or $[0, \pi)$ for cotangent) to make evaluations straightforward!