The terminal point corresponding to an angle $t = -\frac{\pi}{6}$ on the unit circle can be found as follows:

1. First, recognize that $-\frac{\pi}{6}$ radians is equivalent to rotating $\frac{\pi}{6}$ radians clockwise from the positive $x$-axis.
2. $\frac{\pi}{6}$ radians corresponds to 30 degrees, so we are rotating 30 degrees clockwise.
3. The reference angle is $30^\circ$, and the terminal point lies in the fourth quadrant.

Now, we calculate the coordinates of the terminal point using the unit circle:

• In the unit circle, for an angle $\frac{\pi}{6}$, the coordinates are $\left( \cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right) \right)$.
• $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Since the angle is in the fourth quadrant, the cosine remains positive and the sine becomes negative:

• Terminal point = $\left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$.

Thus, the terminal point for $t = -\frac{\pi}{6}$ is $\left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right)$.

Would you like further clarification or more details on this?

Here are 5 related questions to consider:

1. What is the terminal point for $t = \frac{\pi}{3}$?
2. How do you determine the reference angle for negative angles?
3. What is the terminal point for $t = -\frac{\pi}{4}$?
4. How are sine and cosine values affected by the quadrant in which the terminal point lies?
5. How would the terminal point change if $t = -\frac{2\pi}{3}$?

Tip: When dealing with angles in different quadrants, remember that sine and cosine take different signs depending on the quadrant.