DWG NO. 4.2 — Lesson 2 of 6

The Unit Circle and Special Angles

Unit 4: Trigonometric Functions and the Unit Circle · ~25 min

Objective Find the coordinates of any special angle on the unit circle and use them to evaluate sine and cosine exactly.

The unit circle is the circle of radius 1 centered at the origin. For any angle θ in standard position, the terminal side crosses the circle at exactly one point — and that point's coordinates are the definitions of cosine and sine:

(cos θ, sin θ)

This is the bridge from Unit 1's function language to trigonometry: cosine and sine are functions of θ, each one just reading off a coordinate of a point that moves around a circle as θ changes.

Reference angles

Every angle, no matter how large or which quadrant it lands in, has a reference angle — the acute angle between its terminal side and the x-axis. The reference angle tells you the size of cos θ and sin θ; the quadrant tells you the sign.

Quadrant I (0° to 90°): both cos θ and sin θ are positive.
Quadrant II (90° to 180°): sin θ positive, cos θ negative.
Quadrant III (180° to 270°): both cos θ and sin θ negative.
Quadrant IV (270° to 360°): cos θ positive, sin θ negative.

The point at θ = 45° on the unit circle: (cos 45°, sin 45°) = (\(\frac{\sqrt{2}}{2,}\) \(\frac{\sqrt{2}}{2}\))

Special angles in Quadrant I

θ (deg)	θ (rad)	cos θ	sin θ
0°	0	1	0
30°	\(\frac{\pi}{6}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)
45°	\(\frac{\pi}{4}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{2}}{2}\)
60°	\(\frac{\pi}{3}\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)
90°	\(\frac{\pi}{2}\)	0	1

Every other special angle on the circle uses one of these four values — you just attach the correct sign for the quadrant.

Worked Example 1 · Quadrant II

ProblemFind cos 150° and sin 150°.

1150° is in Quadrant II. Its reference angle is 180° − 150° = 30°.

2From the table, cos 30° = \(\frac{\sqrt{3}}{2}\) and sin 30° = \(\frac{1}{2.}\) In Quadrant II, cosine is negative and sine is positive.

cos 150° = −\(\frac{\sqrt{3}}{2}\) · sin 150° = \(\frac{1}{2}\)

Worked Example 2 · Radians, Quadrant III

ProblemFind cos \(\frac{4\pi}{3}\) and sin \(\frac{4\pi}{3}\).

1\(\frac{4\pi}{3}\) is between π and \(\frac{3\pi}{2}\), so it's in Quadrant III. Reference angle: \(\frac{4\pi}{3}\) − π = \(\frac{\pi}{3}\).

2From the table, cos \(\frac{\pi}{3}\) = \(\frac{1}{2}\) and sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2.}\) In Quadrant III, both are negative.

cos \(\frac{4\pi}{3}\) = \(\frac{-1}{2}\) · sin \(\frac{4\pi}{3}\) = −\(\frac{\sqrt{3}}{2}\)

Worked Example 3 · Quadrant IV

ProblemFind cos 315° and sin 315°.

1315° is in Quadrant IV. Reference angle: 360° − 315° = 45°.

2cos 45° = sin 45° = \(\frac{\sqrt{2}}{2.}\) In Quadrant IV, cosine is positive and sine is negative.

cos 315° = \(\frac{\sqrt{2}}{2}\) · sin 315° = −\(\frac{\sqrt{2}}{2}\)

Guided practice

1.Find cos 120° and sin 120°.
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Quadrant II, reference angle 60°. cos 120° = \(\frac{-1}{2,}\) sin 120° = \(\frac{\sqrt{3}}{2}\).
2.Find cos \(\frac{5\pi}{4}\) and sin \(\frac{5\pi}{4}\).
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Quadrant III, reference angle \(\frac{\pi}{4}\). cos \(\frac{5\pi}{4}\) = −\(\frac{\sqrt{2}}{2,}\) sin \(\frac{5\pi}{4}\) = −\(\frac{\sqrt{2}}{2}\).
3.Find cos 330° and sin 330°.
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Quadrant IV, reference angle 30°. cos 330° = \(\frac{\sqrt{3}}{2,}\) sin 330° = \(\frac{-1}{2}\).
4.State the coordinates of the unit-circle point at θ = 180° and at θ = 270° without a table.
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These land exactly on the axes: θ = 180° gives (−1, 0), and θ = 270° gives (0, −1).
5.Explain why cos θ and sin θ are always between −1 and 1, using the definition of the unit circle.
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Every point (cos θ, sin θ) lies on a circle of radius 1, so both coordinates are bounded by the circle's extent — neither can go farther than 1 unit from the origin in either the x- or y-direction.

← 4.1 Angles, Degree and Radian Measure Unit Overview 4.3 Right-Triangle Trigonometry Revisited →