To find the terminal point $P(x, y)$ on the unit circle for $t = -\frac{4\pi}{3}$, we follow these steps:

1. Identify the reference angle: $-\frac{4\pi}{3} \quad \text{corresponds to an angle of} \quad \frac{4\pi}{3} \quad \text{in the positive direction}$ since rotating by $-\frac{4\pi}{3}$ is equivalent to rotating in the opposite (clockwise) direction.

2. Locate the terminal angle: The angle $\frac{4\pi}{3}$ in standard position is in the third quadrant because it is greater than $\pi$ but less than $\frac{3\pi}{2}$.

3. Calculate coordinates on the unit circle:

   • For the unit circle, the coordinates $(x, y)$ for an angle of $\theta = \frac{4\pi}{3}$ are given by: $x = \cos\left(\frac{4\pi}{3}\right) \quad \text{and} \quad y = \sin\left(\frac{4\pi}{3}\right)$
   • Since $\frac{4\pi}{3}$ is in the third quadrant, both $\cos\left(\frac{4\pi}{3}\right)$ and $\sin\left(\frac{4\pi}{3}\right)$ are negative.

4. Compute the exact values: $\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \quad \text{and} \quad \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$

Thus, the terminal point $P(x, y)$ is: $P\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$

Would you like further details on any part of this solution, or do you have any questions?

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Here are five related questions:

1. How would you find the terminal point for $t = \frac{5\pi}{3}$?
2. What is the terminal point for $t = \pi$ on the unit circle?
3. How does one determine which quadrant a terminal point is in?
4. What are the coordinates of the terminal point for $t = -\frac{\pi}{6}$?
5. How do reference angles help in finding terminal points?

Tip: The coordinates $(\cos(t), \sin(t))$ always give the terminal point on the unit circle for any angle $t$.