Unit Circle and Radian Measure

Understanding the unit circle, radian measure, and reference angles

Unit Circle and Radian Measure

The Unit Circle

The unit circle is a circle with:

  • Center at the origin (0,0)(0, 0)
  • Radius of 1

Its equation is: x2+y2=1x^2 + y^2 = 1

Why the Unit Circle Matters

The unit circle allows us to define trigonometric functions for all angles, not just acute angles in right triangles.

For any angle θ\theta in standard position:

  • cos⁡(θ)=x\cos(\theta) = x-coordinate of the point on the unit circle
  • sin⁡(θ)=y\sin(\theta) = y-coordinate of the point on the unit circle
  • tan⁡(θ)=yx=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{y}{x} = \frac{\sin(\theta)}{\cos(\theta)}

Radian Measure

A radian is the angle formed when the arc length equals the radius.

Key Conversions

180°=π radians180° = \pi \text{ radians}

To convert:

  • Degrees to radians: multiply by π180\frac{\pi}{180}
  • Radians to degrees: multiply by 180π\frac{180}{\pi}

Common Angle Conversions

| Degrees | Radians | |---------|---------| | 0°0° | 00 | | 30°30° | π6\frac{\pi}{6} | | 45°45° | π4\frac{\pi}{4} | | 60°60° | π3\frac{\pi}{3} | | 90°90° | π2\frac{\pi}{2} | | 180°180° | π\pi | | 270°270° | 3π2\frac{3\pi}{2} | | 360°360° | 2π2\pi |

Special Angles on the Unit Circle

First Quadrant (0° to 90°0° \text{ to } 90° or 0 to π20 \text{ to } \frac{\pi}{2})

| Angle | Degrees | Radians | cos⁡\cos | sin⁡\sin | tan⁡\tan | |-------|---------|---------|---------|---------|---------| | 00 | 0°0° | 00 | 11 | 00 | 00 | | 30°30° | 30°30° | π6\frac{\pi}{6} | 32\frac{\sqrt{3}}{2} | 12\frac{1}{2} | 33\frac{\sqrt{3}}{3} | | 45°45° | 45°45° | π4\frac{\pi}{4} | 22\frac{\sqrt{2}}{2} | 22\frac{\sqrt{2}}{2} | 11 | | 60°60° | 60°60° | π3\frac{\pi}{3} | 12\frac{1}{2} | 32\frac{\sqrt{3}}{2} | 3\sqrt{3} | | 90°90° | 90°90° | π2\frac{\pi}{2} | 00 | 11 | undefined |

Reference Angles

A reference angle is the acute angle formed between the terminal side and the x-axis.

Finding Reference Angles

  • Quadrant I: reference angle = θ\theta
  • Quadrant II: reference angle = 180°−θ180° - \theta or π−θ\pi - \theta
  • Quadrant III: reference angle = θ−180°\theta - 180° or θ−π\theta - \pi
  • Quadrant IV: reference angle = 360°−θ360° - \theta or 2π−θ2\pi - \theta

CAST Rule (Signs in Each Quadrant)

Remembering which trig functions are positive in each quadrant:

  • Quadrant I: All positive (sine, cosine, tangent)
  • Quadrant II: Sine positive only
  • Quadrant III: Tangent positive only
  • Quadrant IV: Cosine positive only

Memory trick: "All Students Take Calculus"

Arc Length and Sector Area

For a circle with radius rr and central angle θ\theta (in radians):

Arc length: s=rθs = r\theta

Sector area: A=12r2θA = \frac{1}{2}r^2\theta

📚 Practice Problems

1Problem 1easy

❓ Question:

Convert the following angles: (a) 135°135° to radians, (b) 5π6\frac{5\pi}{6} radians to degrees

💡 Show Solution

Solution:

Part a) Convert 135°135° to radians

Multiply by π180\frac{\pi}{180}:

135°×π180=135π180=3π4 radians135° \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4} \text{ radians}

Part b) Convert 5π6\frac{5\pi}{6} radians to degrees

Multiply by 180π\frac{180}{\pi}:

5π6×180π=5×1806=9006=150°\frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = \frac{900}{6} = 150°

Answers:

  • a) 3π4\frac{3\pi}{4} radians
  • b) 150°150°

2Problem 2easy

❓ Question:

a) Convert 225°225° to radians. b) Convert 5π6\frac{5\pi}{6} radians to degrees. c) Find the exact values of sin⁡5π6\sin\frac{5\pi}{6}, cos⁡5π6\cos\frac{5\pi}{6}, and tan⁡5π6\tan\frac{5\pi}{6}.

💡 Show Solution

Solution:

Part (a): To convert degrees to radians, multiply by π180\frac{\pi}{180}:

225°⋅π180=225π180=5π4225° \cdot \frac{\pi}{180} = \frac{225\pi}{180} = \frac{5\pi}{4} radians

Part (b): To convert radians to degrees, multiply by 180π\frac{180}{\pi}:

5π6⋅180π=5⋅1806=9006=150°\frac{5\pi}{6} \cdot \frac{180}{\pi} = \frac{5 \cdot 180}{6} = \frac{900}{6} = 150°

Part (c): 5π6\frac{5\pi}{6} is in Quadrant II (between π2\frac{\pi}{2} and π\pi).

Reference angle: π−5π6=6π−5π6=π6\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}

In Quadrant II: sine is positive, cosine and tangent are negative.

Using reference angle π6\frac{\pi}{6}:

sin⁡5π6=+sin⁡π6=12\sin\frac{5\pi}{6} = +\sin\frac{\pi}{6} = \frac{1}{2}

cos⁡5π6=−cos⁡π6=−32\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}

tan⁡5π6=sin⁡5π6cos⁡5π6=1/2−3/2=−13=−33\tan\frac{5\pi}{6} = \frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}

3Problem 3medium

❓ Question:

Find the exact values: (a) sin⁡(7π6)\sin\left(\frac{7\pi}{6}\right), (b) cos⁡(5π4)\cos\left(\frac{5\pi}{4}\right), (c) tan⁡(5π3)\tan\left(\frac{5\pi}{3}\right)

💡 Show Solution

Solution:

Part a) sin⁡(7π6)\sin\left(\frac{7\pi}{6}\right)

Step 1: Determine the quadrant. 7π6\frac{7\pi}{6} is between π\pi and 3π2\frac{3\pi}{2}, so it's in Quadrant III.

Step 2: Find the reference angle. 7π6−π=7π6−6π6=π6\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}

Step 3: Determine the sign. In Quadrant III, sine is negative.

Step 4: Evaluate. sin⁡(7π6)=−sin⁡(π6)=−12\sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}

Part b) cos⁡(5π4)\cos\left(\frac{5\pi}{4}\right)

Quadrant III, reference angle: 5π4−π=π4\frac{5\pi}{4} - \pi = \frac{\pi}{4}

Cosine is negative in Quadrant III.

cos⁡(5π4)=−cos⁡(π4)=−22\cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}

Part c) tan⁡(5π3)\tan\left(\frac{5\pi}{3}\right)

Quadrant IV, reference angle: 2π−5π3=π32\pi - \frac{5\pi}{3} = \frac{\pi}{3}

Tangent is negative in Quadrant IV.

tan⁡(5π3)=−tan⁡(π3)=−3\tan\left(\frac{5\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}

Answers:

  • a) −12-\frac{1}{2}
  • b) −22-\frac{\sqrt{2}}{2}
  • c) −3-\sqrt{3}

4Problem 4medium

❓ Question:

A point PP on the unit circle has coordinates (−35,45)\left(-\frac{3}{5}, \frac{4}{5}\right).

a) In which quadrant is point PP? b) If PP corresponds to angle θ\theta in standard position, find sin⁡θ\sin\theta, cos⁡θ\cos\theta, and tan⁡θ\tan\theta. c) Find sec⁡θ\sec\theta and csc⁡θ\csc\theta.

💡 Show Solution

Solution:

Part (a): The xx-coordinate is negative and the yy-coordinate is positive.

This means PP is in Quadrant II.

Part (b): On the unit circle, the coordinates of a point are (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).

Therefore:

  • cos⁡θ=−35\cos\theta = -\frac{3}{5}
  • sin⁡θ=45\sin\theta = \frac{4}{5}
  • tan⁡θ=sin⁡θcos⁡θ=4/5−3/5=−43\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{4/5}{-3/5} = -\frac{4}{3}

Part (c): Reciprocal functions:

sec⁡θ=1cos⁡θ=1−3/5=−53\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-3/5} = -\frac{5}{3}

csc⁡θ=1sin⁡θ=14/5=54\csc\theta = \frac{1}{\sin\theta} = \frac{1}{4/5} = \frac{5}{4}

5Problem 5medium

❓ Question:

A circle has radius 8 cm. Find the arc length and area of a sector with central angle 2π3\frac{2\pi}{3} radians.

💡 Show Solution

Solution:

Given:

  • Radius: r=8r = 8 cm
  • Central angle: θ=2π3\theta = \frac{2\pi}{3} radians

Arc Length:

Using s=rθs = r\theta:

s=8⋅2π3=16π3 cms = 8 \cdot \frac{2\pi}{3} = \frac{16\pi}{3} \text{ cm}

s≈16.76 cms \approx 16.76 \text{ cm}

Sector Area:

Using A=12r2θA = \frac{1}{2}r^2\theta:

A=12(8)2⋅2π3A = \frac{1}{2}(8)^2 \cdot \frac{2\pi}{3}

A=12⋅64⋅2π3A = \frac{1}{2} \cdot 64 \cdot \frac{2\pi}{3}

A=32⋅2π3=64π3 cm2A = 32 \cdot \frac{2\pi}{3} = \frac{64\pi}{3} \text{ cm}^2

A≈67.02 cm2A \approx 67.02 \text{ cm}^2

Answers:

  • Arc length: 16π3≈16.76\frac{16\pi}{3} \approx 16.76 cm
  • Sector area: 64π3≈67.02\frac{64\pi}{3} \approx 67.02 cm²
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