Unit Circle Properties, Reference Numbers, and Terminal Points

Fundamentals of the Unit Circle

  • A circle of radius 11 (r=1r = 1) centered at the origin (0,0)(0, 0) is defined as the unit circle.

  • The unit circle represents the set of all points located at a distance of 11 unit from the origin.

  • Properties of the unit circle are foundational when studying and discussing trigonometric functions.

  • The standard algebraic equation of the unit circle is:   x2+y2=1x^2 + y^2 = 1

  • Key axis-intercept points directly visible on the unit circle include:

    • Positive xx-axis intercept: (1,0)(1, 0)

    • Positive yy-axis intercept: (0,1)(0, 1)

    • Negative xx-axis intercept: (1,0)(-1, 0)

    • Negative yy-axis intercept: (0,1)(0, -1)

  • To find coordinates of points on the unit circle that do not lie on the xx-axis or yy-axis, the equation x2+y2=1x^2 + y^2 = 1 must be utilized directly along with given quadrant constraints.

Reference Numbers and Quadrant Rules

  • Definition of Reference Number:

    • Let tt be a real number.

    • The reference number associated with tt is denoted as tˉ\bar{t}.

    • The reference number tˉ\bar{t} represents the shortest distance along the perimeter of the unit circle between the terminal point determined by tt and the xx-axis.

  • Relationship Between Reference Numbers and Terminal Points:

    • Finding a terminal point in any quadrant only requires knowing the corresponding terminal point in the first quadrant.

    • Determining the quadrant in which the terminal point determined by tt lies is necessary to calculate its reference number tˉ\bar{t}.

  • Quadrant-Specific Rules for Finding Reference Numbers tˉ\bar{t}:

    • First Quadrant (Quadrant I) and Fourth Quadrant (Quadrant IV):

    • In Quadrants I and IV, the xx-coordinate of the terminal point is positive (x > 0).

    • The reference number tˉ\bar{t} is found by measuring the shortest distance along the circle to the positive xx-axis.

    • Second Quadrant (Quadrant II) and Third Quadrant (Quadrant III):

    • In Quadrants II and III, the xx-coordinate of the terminal point is negative (x < 0).

    • The reference number tˉ\bar{t} is found by measuring the shortest distance along the circle to the negative xx-axis.

Step-by-Step Examples: Finding Coordinates and Reference Numbers

  • Finding an Unknown Coordinate on the Unit Circle:

    • Problem Statement: A point lies on the unit circle in the third quadrant with an xx-coordinate equal to 32-\frac{\sqrt{3}}{2}. Find its yy-coordinate.

    • Step 1: Substitute the known xx-coordinate x=32x = -\frac{\sqrt{3}}{2} into the unit circle equation x2+y2=1x^2 + y^2 = 1:     (32)2+y2=1\left(-\frac{\sqrt{3}}{2}\right)^2 + y^2 = 1

    • Step 2: Simplify the squared term:     34+y2=1\frac{3}{4} + y^2 = 1

    • Step 3: Subtract 34\frac{3}{4} from both sides:     y2=134y^2 = 1 - \frac{3}{4}     y2=14y^2 = \frac{1}{4}

    • Step 4: Take the square root of both sides:     y=±12y = \pm \frac{1}{2}

    • Step 5: Apply quadrant rules to select the sign:

    • Since the point is located in the third quadrant, its yy-coordinate must be negative.

    • Therefore, y=12y = -\frac{1}{2}.

    • Complete Coordinates: (32,12)\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

  • Finding Reference Numbers for Specific Values of tt:

    • Case 1: t=5π6t = \frac{5\pi}{6}

    • The terminal point for t=5π6t = \frac{5\pi}{6} lies in the second quadrant.

    • The shortest distance along the circle to the xx-axis is to the negative xx-axis.

    • Calculation:       tˉ=π5π6=π6\bar{t} = \pi - \frac{5\pi}{6} = \frac{\pi}{6}

    • Case 2: t=7π4t = \frac{7\pi}{4}

    • The terminal point for t=7π4t = \frac{7\pi}{4} lies in the fourth quadrant.

    • The shortest distance along the circle to the xx-axis is to the positive xx-axis.

    • Calculation:       tˉ=2π7π4=π4\bar{t} = 2\pi - \frac{7\pi}{4} = \frac{\pi}{4}

  • Determining Terminal Points Using Reference Numbers:

    • Finding Terminal Point for t=5π6t = \frac{5\pi}{6}:

    • Reference number is tˉ=π6\bar{t} = \frac{\pi}{6}.

    • In Quadrant I, tˉ=π6\bar{t} = \frac{\pi}{6} determines the terminal point (32,12)\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right).

    • Since t=5π6t = \frac{5\pi}{6} lies in Quadrant II, its xx-coordinate is negative (x < 0) and its yy-coordinate is positive (y > 0).

    • Final desired terminal point: (32,12)\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

    • Finding Terminal Point for t=7π4t = \frac{7\pi}{4}:

    • Reference number is tˉ=π4\bar{t} = \frac{\pi}{4}.

    • In Quadrant I, tˉ=π4\bar{t} = \frac{\pi}{4} determines the terminal point (22,22)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).

    • Since t=7π4t = \frac{7\pi}{4} lies in Quadrant IV, its xx-coordinate is positive (x > 0) and its yy-coordinate is negative (y < 0).

    • Final desired terminal point: (22,22)\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

Terminal Points and Circular Motion

  • Definition of Circumference and Terminal Point Motion:

    • A terminal point on the unit circle is the point reached after traveling a given distance tt along the circle, starting at (1,0)(1, 0).

    • Circumference formula for a circle of radius rr:     C=2πrC = 2\pi r

    • Substituting r=1r = 1 gives total circumference:     C=2πC = 2\pi

  • Counterclockwise Motion and Terminal Points:

    • Complete trip around the unit circle once corresponds to a distance of 2π2\pi.

    • Starting Position (t=0t = 0):

    • Terminal point: (1,0)(1, 0)

    • Quarter trip around the circle (t=14×2π=π2t = \frac{1}{4} \times 2\pi = \frac{\pi}{2}):

    • Terminal point: (0,1)(0, 1)

    • Halfway trip around the circle (t=12×2π=πt = \frac{1}{2} \times 2\pi = \pi):

    • Terminal point: (1,0)(-1, 0)

    • Three-quarters trip around the circle (t=34×2π=3π2t = \frac{3}{4} \times 2\pi = \frac{3\pi}{2}):

    • Terminal point: (0,1)(0, -1)

    • Full trip around the circle (t=2πt = 2\pi):

    • Returns to terminal point (1,0)(1, 0)

  • Clockwise Motion (Negative Values of tt):

    • Traveling clockwise (opposite direction) is indicated by a negative sign.

    • Quarter trip clockwise (t=π2t = -\frac{\pi}{2}):

    • Reaches terminal point (0,1)(0, -1)

    • Half trip clockwise (t=πt = -\pi):

    • Reaches terminal point (1,0)(-1, 0)

  • Equivalence of Multiple Terminal Points:

    • Different values and multiples of π\pi can lead to the exact same terminal points (for example, t=3π2t = \frac{3\pi}{2} and t=π2t = -\frac{\pi}{2} both share the terminal point (0,1)(0, -1); t=πt = \pi and t=πt = -\pi both share (1,0)(-1, 0)).


A unit circle is a circle with a radius of $1$ centered at the origin $(0, 0)$.

  • It shows all points that are $1$ unit away from the center.

  • The equation for the unit circle is:
    x2+y2=1x^2 + y^2 = 1

  • Important points on the unit circle are:

    • On the right: $(1, 0)$

    • On the top: $(0, 1)$

    • On the left: $(-1, 0)$

    • On the bottom: $(0, -1)$

  • To find other points on the unit circle, you can use the equation and the quadrant where the point is located (four sections of the circle).

Reference Numbers and Quadrants

  • A reference number helps you find where a point is on the circle based on its angle.

  • Every angle can be simplified by finding its position in the first quadrant.

  • Here’s how to use reference numbers:

    • 1st Quadrant: Both x and y are positive.

    • 2nd Quadrant: x is negative, y is positive.

    • 3rd Quadrant: Both x and y are negative.

    • 4th Quadrant: x is positive, y is negative.

Examples

  • If a point in the third quadrant has x=32x = -\frac{\sqrt{3}}{2}, here’s how to find $y$:

    1. Plug it into the circle’s equation:
      (32)2+y2=1\left(-\frac{\sqrt{3}}{2}\right)^2 + y^2 = 1.

    2. Solve: 34+y2=1\frac{3}{4} + y^2 = 1.

    3. This gives: y2=14y^2 = \frac{1}{4}, so y=12y = -\frac{1}{2} (because it’s in the third quadrant).

    4. Complete point: (32,12)\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right).

  • If t=5π6t = \frac{5\pi}{6}, it is in the second quadrant. Reference number: tˉ=π5π6=π6\bar{t} = \pi - \frac{5\pi}{6} = \frac{\pi}{6}.