7.1: The Unit Circle

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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.

Figure 7.1.2 : Angles in Standard Position

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.

Figure 7.1.3 : Positive angle in standard position.

Figure 7.1.4 : : Negative angle in standard position.

Figure 7.1.5 : Positive angle in standard position.

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.

$clipboard_edae1eab80f059ab6160333bcf21dad75.png$

More formally: Every angle $B$ is coterminal with angle $A$ where $B = A + 360˚k$, $k =$ any integer.

Example 7.1.1

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.

$clipboard_e817c28fbb916447571993b1ba5b92c18.png$

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 = $r$.

$clipboard_ea4f396a40a55f4929eb6dd6b1ad1cbda.png$

Equation of a Circle

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.

Example 7.1.2

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.

$clipboard_ed6aa5e0ff457f2cdadcf335269d53625.png$

$\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.

$clipboard_e2026a8b9c765a82686d89e448e76dfa7.png$

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.

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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 $y$.

$clipboard_ea02e11ad76d1b60bcb971065f6a9a1eb.png$
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\(k = -2\)	\(k = -1\)	\(k = 1\)	\(k = 2\)
\(315˚ + 360˚(\textcolor{red}{-2}) = -405˚\)	\(315˚ + 360˚(\textcolor{red}{-1}) = -45˚\)	\(315˚ + 360˚(\textcolor{red}{1}) = -675˚\)	\(315˚ + 360˚(\textcolor{red}{2}) = 1035˚\)

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

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Example 7.1.1

Equation of a Circle

Example 7.1.2