A terminal point is the (x,y) coordinate that is associate with a certain degree angle on the unit circle. For example, for 0º the terminal point would be (1,0) because that is where it would hit on a traditional graph (x,y coordinate plane). However this can feel trickier with angles in-between 90º intervals like the ones you are asking.
Each point on the unit circle is made from cosine and sine where cos is x and sin is y. For example, cosine 30º (pi/6) is √3/2 and sine 30º (pi/6) is 1/2 therefore the terminal point on the unit circle is simply (√3/2 , 1/2). Because of this and the radius of 1, the pattern will repeat making it easy to figure out once you know the first quadrant.
For instance, to solve your first problem it is negative pi/6. The negative just means that you go in the opposite direction; instead of going up to the left, you'd go down pi/6 to the right making you land at 11pi/6 or 330º, which is in the 4th quadrant. You can also think of just subtracting pi/6 from 2pi. Because the reference angle (fastest way to get to the x axis) is still pi/6, the coordinate values will still be the same just the signs will differ due to the new quadrant. In the 4th quadrant the x values are always positive and the y values are always negative therefore the answer is (√3/2 , -1/2).
Hopefully that explanation helped you to understand how to get the answers for the following. Feel free to respond if you have a question!
Question = Terminal Point
- -pi/6 (same as 11pi/3) = (√3/2 , -1/2)
- -7pi/4 (same as positive pi/4) = (√2/2, √2/2)
- -2pi/3 (same as 4pi/3 meaning the reference angle is pi/3) = (-1/2 , -√3/2)
- 10pi/3 (bigger than 2pi, 6pi/3, by 4pi/3, so it is the same as 4pi/3. This means the reference angle is pi/3) = (-1/2 , -√3/2)
- 13pi/6 (one more than 2pi, so it is the same as pi/6 which is also its reference angle) = (√3/2, 1/2)