What are the exact sine and cosine values of \( \frac{11\pi}{6} \)?

\( \cos(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2} \) and \( \sin(\frac{11\pi}{6}) = -\frac{1}{2} \)

Find the exact value of \( \cos(-\frac{3\pi}{4}) \) without a calculator.

The angle \( -\frac{3\pi}{4} \) is coterminal with \( \frac{5\pi}{4} \) in Quadrant III. Therefore, \( \cos(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \).

Find the exact sine and cosine of \( \frac{16\pi}{3} \).

Subtracting \( 4\pi \) gives the coterminal angle \( \frac{4\pi}{3} \) in Quadrant III. \( \cos(\frac{16\pi}{3}) = -\frac{1}{2} \) and \( \sin(\frac{16\pi}{3}) = -\frac{\sqrt{3}}{2} \).

What are the special angles on the unit circle?

The primary special angles in the first quadrant are 0, \( \frac{\pi}{6} \) (30 degrees), \( \frac{\pi}{4} \) (45 degrees), \( \frac{\pi}{3} \) (60 degrees), and \( \frac{\pi}{2} \) (90 degrees). All other special angles around the circle are reflections of these base angles.

How does unit circle symmetry work?

Angles that reflect across the y-axis, x-axis, or origin share the same absolute values for sine and cosine as their first-quadrant reference angle. Only the positive or negative sign changes depending on which quadrant the angle lands in.

Which quadrant has a negative sine and a positive cosine?

Quadrant IV. Because cosine corresponds to the x-coordinate (which is positive to the right of the y-axis) and sine corresponds to the y-coordinate (which is negative below the x-axis).

Special Angles of the Unit Circle | Symmetry & Trigonometric Values

Understanding Unit Circle Symmetry

Any line through the center of a circle is a line of symmetry. We consider a unit circle with its center at the origin \( (0,0) \). We are interested in three specific types of symmetry:

Symmetry across the y-axis: Relates Quadrant I to Quadrant II.
Symmetry across the origin: Relates Quadrant I to Quadrant III.
Symmetry across the x-axis: Relates Quadrant I to Quadrant IV.

The unit circle below shows the cosine (x-coordinate) and sine (y-coordinate) for the special angles in radians and degrees: \( 0, \dfrac{\pi}{6} (30^\circ), \dfrac{\pi}{4} (45^\circ), \dfrac{\pi}{3} (60^\circ), \text{and } \dfrac{\pi}{2} (90^\circ) \).

Unit circle showing special angles in quadrant 1 — Figure 1. The Unit Circle and Quadrant 1 Special Angles

Angles Related to \( \dfrac{\pi}{3} \) (60°)

Consider the angles: \( \dfrac{2\pi}{3} \), \( \dfrac{4\pi}{3} \), and \( \dfrac{5\pi}{3} \). They all have a reference angle of \( \dfrac{\pi}{3} \).

Base Values: \( \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}, \quad \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} \)

1. Symmetry with respect to the y-axis (Quadrant II)

\[ \dfrac{2\pi}{3} = \pi - \dfrac{\pi}{3} \]

In Quadrant II, x is negative and y is positive.

\[ \cos\left(\dfrac{2\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = \mathbf{-\dfrac{1}{2}} \]

\[ \sin\left(\dfrac{2\pi}{3}\right) = \sin\left(\dfrac{\pi}{3}\right) = \mathbf{\dfrac{\sqrt{3}}{2}} \]

2. Symmetry with respect to the origin (Quadrant III)

\[ \dfrac{4\pi}{3} = \pi + \dfrac{\pi}{3} \]

In Quadrant III, both x and y are negative.

\[ \cos\left(\dfrac{4\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = \mathbf{-\dfrac{1}{2}} \]

\[ \sin\left(\dfrac{4\pi}{3}\right) = -\sin\left(\dfrac{\pi}{3}\right) = \mathbf{-\dfrac{\sqrt{3}}{2}} \]

3. Symmetry with respect to the x-axis (Quadrant IV)

\[ \dfrac{5\pi}{3} = 2\pi - \dfrac{\pi}{3} \]

In Quadrant IV, x is positive and y is negative.

\[ \cos\left(\dfrac{5\pi}{3}\right) = \cos\left(\dfrac{\pi}{3}\right) = \mathbf{\dfrac{1}{2}} \]

\[ \sin\left(\dfrac{5\pi}{3}\right) = -\sin\left(\dfrac{\pi}{3}\right) = \mathbf{-\dfrac{\sqrt{3}}{2}} \]

Angles Related to \( \dfrac{\pi}{6} \) (30°)

Consider the angles: \( \dfrac{5\pi}{6} \), \( \dfrac{7\pi}{6} \), and \( \dfrac{11\pi}{6} \). They all have a reference angle of \( \dfrac{\pi}{6} \).

Base Values: \( \cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}, \quad \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2} \)

1. Symmetry with respect to the y-axis (Quadrant II)

\[ \dfrac{5\pi}{6} = \pi - \dfrac{\pi}{6} \]

\[ \cos\left(\dfrac{5\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = \mathbf{-\dfrac{\sqrt{3}}{2}} \]

\[ \sin\left(\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \mathbf{\dfrac{1}{2}} \]

2. Symmetry with respect to the origin (Quadrant III)

\[ \dfrac{7\pi}{6} = \pi + \dfrac{\pi}{6} \]

\[ \cos\left(\dfrac{7\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = \mathbf{-\dfrac{\sqrt{3}}{2}} \]

\[ \sin\left(\dfrac{7\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = \mathbf{-\dfrac{1}{2}} \]

3. Symmetry with respect to the x-axis (Quadrant IV)

\[ \dfrac{11\pi}{6} = 2\pi - \dfrac{\pi}{6} \]

\[ \cos\left(\dfrac{11\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) = \mathbf{\dfrac{\sqrt{3}}{2}} \]

\[ \sin\left(\dfrac{11\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = \mathbf{-\dfrac{1}{2}} \]

Angles Related to \( \dfrac{\pi}{4} \) (45°)

Consider the angles: \( \dfrac{3\pi}{4} \), \( \dfrac{5\pi}{4} \), and \( \dfrac{7\pi}{4} \). They all have a reference angle of \( \dfrac{\pi}{4} \).

Base Values: \( \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}, \quad \sin\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \)

1. Symmetry with respect to the y-axis (Quadrant II)

\[ \dfrac{3\pi}{4} = \pi - \dfrac{\pi}{4} \]

\[ \cos\left(\dfrac{3\pi}{4}\right) = -\cos\left(\dfrac{\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \]

\[ \sin\left(\dfrac{3\pi}{4}\right) = \sin\left(\dfrac{\pi}{4}\right) = \mathbf{\dfrac{\sqrt{2}}{2}} \]

2. Symmetry with respect to the origin (Quadrant III)

\[ \dfrac{5\pi}{4} = \pi + \dfrac{\pi}{4} \]

\[ \cos\left(\dfrac{5\pi}{4}\right) = -\cos\left(\dfrac{\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \]

\[ \sin\left(\dfrac{5\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \]

3. Symmetry with respect to the x-axis (Quadrant IV)

\[ \dfrac{7\pi}{4} = 2\pi - \dfrac{\pi}{4} \]

\[ \cos\left(\dfrac{7\pi}{4}\right) = \cos\left(\dfrac{\pi}{4}\right) = \mathbf{\dfrac{\sqrt{2}}{2}} \]

\[ \sin\left(\dfrac{7\pi}{4}\right) = -\sin\left(\dfrac{\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \]

Challenge Questions

Apply your knowledge of the unit circle, reference angles, and coterminal angles to solve the following problems without a calculator.

Question 1 (Negative Angles): Find the exact value of \( \cos\left(-\dfrac{3\pi}{4}\right) \) and \( \sin\left(-\dfrac{3\pi}{4}\right) \).

Show Solution

Step 1: A negative angle means rotating clockwise. To find a positive coterminal angle, add \( 2\pi \).

\[ -\dfrac{3\pi}{4} + \dfrac{8\pi}{4} = \dfrac{5\pi}{4} \]

Step 2: The angle \( \dfrac{5\pi}{4} \) is in Quadrant III. Both sine and cosine are negative.

Answer: \( \cos\left(-\dfrac{3\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \) and \( \sin\left(-\dfrac{3\pi}{4}\right) = \mathbf{-\dfrac{\sqrt{2}}{2}} \)

Question 2 (Angles larger than \(2\pi\)): Find the exact sine and cosine of \( \dfrac{16\pi}{3} \).

Show Solution

Step 1: Find a coterminal angle between \( 0 \) and \( 2\pi \) by repeatedly subtracting full circles (\( 2\pi \) or \( \dfrac{6\pi}{3} \)).

\[ \dfrac{16\pi}{3} - \dfrac{6\pi}{3} = \dfrac{10\pi}{3} \]

\[ \dfrac{10\pi}{3} - \dfrac{6\pi}{3} = \dfrac{4\pi}{3} \]

Step 2: The angle \( \dfrac{4\pi}{3} \) is in Quadrant III with a reference angle of \( \dfrac{\pi}{3} \).

Answer: \( \cos\left(\dfrac{16\pi}{3}\right) = \mathbf{-\dfrac{1}{2}} \) and \( \sin\left(\dfrac{16\pi}{3}\right) = \mathbf{-\dfrac{\sqrt{3}}{2}} \)

Question 3 (Reciprocal Functions): Find the exact value of \( \sec\left(\dfrac{11\pi}{6}\right) \) and \( \csc\left(\dfrac{11\pi}{6}\right) \).

Show Solution

Step 1: Identify the basic sine and cosine values. The angle \( \dfrac{11\pi}{6} \) is in Quadrant IV.

\[ \cos\left(\dfrac{11\pi}{6}\right) = \dfrac{\sqrt{3}}{2} \quad \text{and} \quad \sin\left(\dfrac{11\pi}{6}\right) = -\dfrac{1}{2} \]

Step 2: Use the reciprocal identities: \( \sec(\theta) = \dfrac{1}{\cos(\theta)} \) and \( \csc(\theta) = \dfrac{1}{\sin(\theta)} \).

\[ \sec\left(\dfrac{11\pi}{6}\right) = \dfrac{2}{\sqrt{3}} = \mathbf{\dfrac{2\sqrt{3}}{3}} \]

\[ \csc\left(\dfrac{11\pi}{6}\right) = \dfrac{1}{-1/2} = \mathbf{-2} \]

Special Angles in the Unit Circle

Understanding Unit Circle Symmetry

Angles Related to \( \dfrac{\pi}{3} \) (60°)

Angles Related to \( \dfrac{\pi}{6} \) (30°)

Angles Related to \( \dfrac{\pi}{4} \) (45°)

Challenge Questions

Related Trigonometry Topics

More Math Resources