7.1: The Unit Circle
The core concepts of trigonometry are developed from a circle with radius equal to \(1\) unit, drawn in the \(xy\)-coordinate plane, centered at the origin. This circle is given a name: the unit circle (Figure \(7.1.1\) below). Just like a \(12\)-hour clock with values of time from \(1\) to \(12\), trigonometric functions are periodic, meaning the same values are reproduced with each \(360˚\) revolution.
An angle is in standard position (see Figure \(7.1.2\) above) if its initial side is along the positive \(x\)-axis and its vertex is at the origin : point \((0,0)\). The following angles are in standard position. An angle that rotates in the counter-clockwise direction is a positive angle . An angle that rotates in the clockwise direction is a negative angle .
Coterminal Angles
Two or more standard angles that share common terminal sides are said to be coterminal angles . For example, \(30˚\) and \(390˚\) are coterminal angles.
More formally: Every angle \(B\) is coterminal with angle \(A\) where \(B = A + 360˚k\), \(k =\) any integer.
The expression \(315˚ + 360˚k\) gives the angles coterminal with \(315˚\). State the coterminal angles the equation generates using the following \(k\)-values: \(k = −2, −1, 1, 2\). Then sketch the angles.
Solution
We substitute the given \(k\)-values into the expression \(315˚ + 360˚k\).
| \(k = -2\) | \(k = -1\) | \(k = 1\) | \(k = 2\) |
| \(315˚ + 360˚(\textcolor{red}{-2}) = -405˚\) | \(315˚ + 360˚(\textcolor{red}{-1}) = -45˚\) | \(315˚ + 360˚(\textcolor{red}{1}) = -675˚\) | \(315˚ + 360˚(\textcolor{red}{2}) = 1035˚\) |
Below, the angles are sketched. Do you see that each angle shares the same terminal side? All four angles are coterminal with \(315˚\), and coterminal with each other.
Circle Centered at the Origin
Every ordered pair \((x, y)\) on a circle is associated with a right triangle. The right triangle has horizontal distance \(x\), vertical distance \(y\), and hypotenuse = radius = \(r\).
The equation of a circle of radius \(r\) centered at the origin:
\[x^2 + y^2 = r^2\]
Note: the \(x\)-coordinate and the \(y\)-coordinate can take on negative values, depending on the quadrant of the terminal side of the angle.
Find the \(y\)-coordinate of point A \(\left(−\dfrac{5}{9} , y \right)\) if point A lies in QIII on the unit circle.
Solution
The unit circle has radius \(r = 1\). Trigonometry weds algebra and geometry with visual sketches. Create a sketch before jumping into a solution. This helps you see the answer.
\(\begin{array} &\left(−\dfrac{5}{9}\right)^2 + y^2 &= 1^2 &\text{Substitute \(x = −\dfrac{5}{9}\) and \(r = 1\) into equation of a circle.} \\ \dfrac{25}{81} + y^2 &= 1 &\text{Simplify.} \\ y^2 &= 1 − \dfrac{25}{81} &\text{Subtract \(\dfrac{25}{81}\) to each side.} \\ y^2 &= \dfrac{81}{81} − \dfrac{25}{81} &\text{Find the LCD.} \\ y^2 &= \dfrac{56}{81} &\text{Simplify.} \\ \sqrt{y^2} &= ±\sqrt{\dfrac{56}{81}} &\text{Square root both sides.} \\ y &= − \dfrac{\sqrt{56}}{9} &\text{Choose the correct sign value. \(y < 0\) in QIII.} \\ y &= − \dfrac{2\sqrt{14}}{9} &\text{Simplify the radical.} \end{array}\)
Special Right Triangles
Using the Pythagorean Theorem, one can drive the following two templates for special right triangles: \(45˚\)-\(45˚\)-\(90˚\) triangles and \(30˚\)-\(60˚\)-\(90˚\) triangles.
Since right triangles and circles are inextricably tied to each other, the acute angles \(30˚\), \(45˚\), \(60˚\) are frequently useful for finding exact values in trigonometry; these solutions do not require the use of a calculator.
Examples of special right triangles and their solutions, can be viewed in these videos:
Try It! (Exercises)
For #1-5, draw the angle \(\theta\) in standard position.
- \(\theta = 210˚\)
- \(\theta = −300˚\)
- \(\theta = 150˚\)
- \(\theta = 270˚\)
- \(\theta = −135˚\)
For #6-10, state any two angles coterminal with the given angle \(\theta\). Give one positive coterminal angle and one negative coterminal angle.
- \(\theta = 30˚\)
- \(\theta = 60˚\)
- \(\theta = 90˚\)
- \(\theta = 180˚\)
- \(\theta = 240˚\)
For #11-15, state the equation of the circle with the given radius.
- \(r = 4\)
- \(r = \dfrac{1}{2}\)
- \(r = 3 \sqrt{2}\)
- \(r = \dfrac{\sqrt{6}}{2}\)
- \(r = \sqrt{\dfrac{101}{62}}\)
- Fill in the blanks: A unit circle is a circle with \(\underline{\;\;\;\;\;\;\;\;\;\;}\) equal to one unit. The circle is centered at \(\underline{\;\;\;\;\;\;\;\;\;\;}\). The circle has equation: \(\underline{\;\;\;\;\;\;\;\;\;\;}\).
For #17-24, The given point lies on a circle with given radius and terminal side in the given quadrant. Find the missing coordinate of the given ordered pair.
| Point on Circle | Radius | Quadrant in which \(\theta\) terminates. |
|---|---|---|
| 17. \((2, y)\) | \(r = 5\) | \(\theta ∈\) QI |
| 18. \((x, \sqrt{6})\) | \(r = 3\) | \(\theta ∈\) QII |
| 19. \((−3\sqrt{7}, y)\) | \(r = \sqrt{77}\) | \(\theta ∈\) QIII |
| 20. \((x, −4\sqrt{5})\) | \(r = 3\sqrt{15}\) | \(\theta ∈\) QIV |
| 21. \(\left( \dfrac{3}{4} , y \right)\) | \(r=1\) | \(\theta ∈\) QI |
| 22. \(\left(x, \dfrac{3}{16} \right)\) | \(r=1\) | \(\theta ∈\) QII |
| 23. \(\left(− \dfrac{\sqrt{5}}{4} , y \right)\) | \(r=1\) | \(\theta ∈\) QIII |
| 24. \(\left(x, − \dfrac{3\sqrt{2}}{8} \right)\) | \(r=1\) | \(\theta ∈\) QIV |
For #25-29, find the fraction of a full revolution. Then sketch the angle.
- \(\dfrac{1}{4} \cdot 360˚\)
- \(\dfrac{1}{2} \cdot 360˚\)
- \(\dfrac{3}{4} \cdot 360˚\)
- \(\dfrac{1}{8} \cdot 360˚\)
- \(\dfrac{1}{6} \cdot 360˚\)
For #30-35, solve for the \(2\) missing side lengths. Give exact answers. No calculators.
For #36-38, Find the coordinates of the ordered pair \((x, y)\) on the unit circle with the given standard angle. Use special right triangle relationships. Give exact values of \(x\) and \(y\).