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 $\theta$ in standard position:

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

Radian Measure

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

Key Conversions

$180° = \pi \text{ radians}$

To convert:

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

Common Angle Conversions

Special Angles on the Unit Circle

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

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 = $\theta$
• Quadrant II: reference angle = $180° - \theta$ or $\pi - \theta$
• Quadrant III: reference angle = $\theta - 180°$ or $\theta - \pi$
• Quadrant IV: reference angle = $360° - \theta$ or $2\pi - \theta$

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 $r$ and central angle $\theta$ (in radians):

Arc length: $s = r\theta$

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

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

$\frac{5\pi}{6} \cdot \frac{180}{\pi} = \frac{5 \cdot 180}{6} = \frac{900}{6} = 150°$

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

Reference angle: $\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}$

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

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

$\sin\frac{5\pi}{6} = +\sin\frac{\pi}{6} = \frac{1}{2}$

$\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$

$\tan\frac{5\pi}{6} = \frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

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

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

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

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

Quadrant III, reference angle: $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$

Cosine is negative in Quadrant III.

$\cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

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

Quadrant IV, reference angle: $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$

Tangent is negative in Quadrant IV.

$\tan\left(\frac{5\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$

Answers:

• a) $-\frac{1}{2}$
• b) $-\frac{\sqrt{2}}{2}$
• c) $-\sqrt{3}$

Therefore:

• $\cos\theta = -\frac{3}{5}$
• $\sin\theta = \frac{4}{5}$
• $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{4/5}{-3/5} = -\frac{4}{3}$

Part (c): Reciprocal functions:

$\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-3/5} = -\frac{5}{3}$

$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{4/5} = \frac{5}{4}$