Unit Circle Worksheet Answer Key – Precalculus

Problem 1 Degrees to Radians

Multiply by π/180.

(a) 45π/180 = π/4. (b) 150π/180 = 5π/6. (c) 225π/180 = 5π/4. (d) 330π/180 = 11π/6.

Problem 2 Radians to Degrees

Multiply by 180/π.

(a) 120°. (b) 315°. (c) 3 · 18° = 54°. (d) 5 · 15° = 75°.

Problem 3 Arc Length Basics

Use s = rθ.

(a) 12 · 5π/6 = 10π ≈ 31.42 inches.

(b) 9 · π/3 = 3π ≈ 9.42 feet.

Problem 4 Special Angle Values

(a) √3/2. (b) √3/2. (c) sin ÷ cos = (√2/2) ÷ (√2/2) = 1. (d) The point is (0, 1), so 0.

Problem 5 Coordinates from the Unit Circle

Quadrant II. Reference angle: π − 5π/6 = π/6.

At π/6 the point is (√3/2, 1/2); in Quadrant II the x-coordinate is negative: (−√3/2, 1/2).

Problem 6 Quadrants and Reference Angles

(a) π < 4π/3 < 3π/2: Quadrant III, reference 4π/3 − π = π/3.

(b) Quadrant IV, reference 2π − 7π/4 = π/4.

(c) Quadrant II, reference π − 5π/6 = π/6.

Problem 7 Exact Values in Every Quadrant

(a) Quadrant III, reference π/4: −√2/2.

(b) Quadrant II, reference π/6: −√3/2.

(c) Quadrant III, tan = (−1/2) ÷ (−√3/2) = 1/√3 = √3/3.

(d) Quadrant IV, reference π/6: √3/2.

Problem 8 Negative and Large Angles

(a) Cosine is even: cos(π/3) = 1/2.

(b) Sine is odd: −sin(3π/4) = −√2/2.

(c) 13π/6 − 2π = π/6, so 1/2.

(d) 17π/4 − 4π = π/4, so √2/2.

Problem 9 Even and Odd Relationships

(a) Odd: −a. (b) Even: b.

(c) 9π/8 = π + π/8 is in Quadrant III, so sine is negative: −a.

(d) 7π/8 = π − π/8 is in Quadrant II, so cosine is negative: −b.

Problem 10 Find the Angle from a Point

(a) Quadrant III with reference π/6: 7π/6.

(b) The bottom of the circle: 3π/2.

(c) Quadrant II with reference π/4: 3π/4. All angles: 3π/4 + 2πk for any integer k.

Problem 11 Find Cosine and Tangent from Sine

On the unit circle x² + y² = 1, so cos² θ = 1 − 9/25 = 16/25. In Quadrant II cosine is negative: cos θ = −4/5.

tan θ = (3/5) ÷ (−4/5) = −3/4.

Problem 12 Angle from Arc Length

θ = s/r = 15/6 = 2.5 radians.

In degrees: 2.5 · 180/π ≈ 143.2°.

Problem 13 Wheel Speeds

One revolution is 2π radians: ω = 240 · 2π = 480π ≈ 1,508.0 rad/min.

v = rω = 14 · 480π = 6,720π ≈ 21,112 in/min. Then 21,112 × 60 ÷ 63,360 ≈ 20.0 mph.

Problem 14 Coterminal Angles

Add or subtract 2π: 5π/6 + 2π = 17π/6; 5π/6 + 4π = 29π/6; 5π/6 − 2π = −7π/6.

Problem 15 Tangent Values

(a) (√3/2) ÷ (−1/2) = −√3. (b) (−√2/2) ÷ (−√2/2) = 1.

(c) Tangent is odd: −√3/3. (d) cos(3π/2) = 0, so undefined.

Problem 16 Ferris Wheel Travel

One turn is 2π radians in 240 seconds: ω = 2π/240 = π/120 rad/s.

In 50 seconds, θ = 50π/120 = 5π/12, so s = 25 · 5π/12 = 125π/12 ≈ 32.72 feet.

v = rω = 25π/120 = 5π/24 ≈ 0.654 ft/s.

Problem 17 Minute Hand Tip

The hand turns 2π in 60 minutes, so in 20 minutes θ = 2π/3 and s = 9 · 2π/3 = 6π ≈ 18.85 inches.

In 50 minutes θ = 5π/3 and s = 9 · 5π/3 = 15π ≈ 47.12 inches.

Problem 18 Bicycle Wheel Rotation

Speed: 12 mi/h = 12 · 63,360 ÷ 60 = 12,672 in/min. The radius is 13.5 in.

ω = v/r = 12,672 ÷ 13.5 ≈ 938.7 rad/min.

Divide by 2π: ≈ 149.4 rpm.

Problem 19 Lighthouse Beam

ω = 2 · 2π = 4π rad/min, so v = rω = 3 · 4π = 12π ≈ 37.7 mi/min.

30° = π/6 rad, so s = 3 · π/6 = π/2 ≈ 1.57 miles.

Problem 20 A Particle on the Unit Circle

After 2 s: θ = 3π/2, position (0, −1).

After 5 s: θ = 15π/4 − 2π = 7π/4, position (√2/2, −√2/2).

First return when (3π/4)t = 2π, so t = 8/3 seconds.