Let's go through each part of the problem in the image.

Given: $t = \frac{2\pi}{3}$

Part (a): Find the reference number $t'$ for the value of $t$.

The reference number $t'$ is the positive acute angle between the terminal side of $t$ and the x-axis. For angles in the second quadrant, the reference angle $t'$ is calculated as: $t' = \pi - t$

Since $t = \frac{2\pi}{3}$, we can substitute to find $t'$: $t' = \pi - \frac{2\pi}{3} = \frac{3\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{3}$

Thus, $t' = \frac{\pi}{3}$

Part (b): Find the terminal point determined by $t$.

To find the terminal point on the unit circle for $t = \frac{2\pi}{3}$, we use the cosine and sine of $t$: $x = \cos\left(\frac{2\pi}{3}\right) \quad \text{and} \quad y = \sin\left(\frac{2\pi}{3}\right)$

In the second quadrant:

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

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

Summary of Answers:

• (a) $t' = \frac{\pi}{3}$
• (b) $(x, y) = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

Would you like further details or have any questions?

Here are some related questions to deepen understanding:

1. What is the reference angle for other angles in the second quadrant, like $\frac{3\pi}{4}$?
2. How does the reference angle change for angles in the third quadrant?
3. What are the cosine and sine values of other common angles in the unit circle?
4. How do terminal points on the unit circle relate to trigonometric identities?
5. How can you use the unit circle to find values for angles greater than $2\pi$?

Tip: Knowing the reference angle and the quadrant can quickly help determine the sine and cosine values for various angles.