Unit Circle Symmetries

For any point (x, y) on the unit circle, x² + y² = 1, we know that -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1. If we know the x- or y-coordinate of a point on the unit circle, we can find the other coordinate. If x = 3/5, then

(3/5)2 + y2 = 1       y2 = 1 - 9/25 = 16/25       y = ± 4/5. Thus, (3/5 , 4/5) and (3/5 , -4/5) are points on the unit circle. There are two points with an x-coordinate of 3/5 .

The unit circle is symmetric with respect to the x-axis, the y-axis, and the origin, so we can use the coordinates of one point on the unit circle to find coordinates of its reflections.

Thus (3/5, 4/5) being on the unit circle not only implies that (3/5, -4/5) is on the unit circle as we already discussed, but it also implies that (-3/5, 4/5) and (-3/5, -4/5) are also on the unit circle.

There is also a symmetry with respect to the line y = x. Thus if the point (a, b) is on the circle then the point (b, a) is also on the circle. So in our example with the point (3/5, 4/5) on the circle, we know that (4/5, 3/5) would also be on the circle. Combine this idea with the symmetries mentioned above and one point on a circle would in general lead to the discovery of 7 more points on the circle. You don't need to worry about memorizing all of these points because if you plot them on a unit circle they just become obvious.

Note that you need to use your prior knowledge of point plotting to determine which coordinate is positive and which coordinate is negative at each point.

EXAMPLE 1
  a.) Find the point in the first quadrant of the unit circle whose x-coordinate is 7/12.
  b.) Find and plot all of the other points on the unit circle that we know exist as a result of knowing this point.

a.) Image to the left. Using the equation of the unit circle we know that (7/12)² + y² = 1. Solving for y and using only the positive answer since we want a first quadrant point, we get:
Thus we have the point: .

b.) All we need to do for this part is make sure we have all possible combinations of plus and minus of the coordinates for the point in part a and then plot them. Thus we have: Image above right.

Moving About the Unit Circle

You may want to print up a few circles to draw on while you are reading this material.

Recall that the Unit Circle is the circle of radius 1, centered at the origin. Also, recall that in the last lesson you learned that to get an angle in radians in general you just use the circumcribed arclength of the angle and divide by the radius. This leads to the standard way of measuring angles on the unit circle, which is to start at the point on the positive x-axis (the coordinates of this point are (1, 0)) and measure counterclockwise for positive angles and clockwise for negative angles. The angle's measure (in radians) then would equal the arclength of the circle from this starting point to the terminal point (the point where you stop). An angle measured in the way I just described is said to be in standard position.

1. Use this idea to mesure the angles in radians on the following circles. You may need to cut a string to measure with. These circles have a radius of one inch if the image when you print is the same as the image I made. Place one end of the string on the starting point and measure along the circumference.

Let P(t) be the point on the unit circle that lies on the terminal side of an angle in standard position.

2.) Plot the following points, P(t) on the given unit circle.

P(5π) P(-3π/4) P(5π/2) P(5) P(-3)

You need to get good at moving around the unit circle. I can't stress this enough. Get good at moving around the Unit Circle now and it will make your semester go much more smoothly.

3.) Let P(t) be as pictured on the following unit circle. Plot:

P(t + π/2) P(t - π/2) P(t + π) P(t - π) P(t + 2π) P(t + 3π/2) P(π/2 - t) P(π - t) P(-t) P(2π - t)

4.) P(t) in number 3 is located at the point (11/61, 60/61). Find the coordinates of each of the points listed in question 3.