Trigonometric Functions and Applications: Signs, Quadrants, and Circular Motion

Trigonometric Functions: Signs and Quadrants

Determining the Sign of Trigonometric Functions

Understanding the sign of trigonometric functions in different quadrants is essential for solving problems involving angles and their trigonometric values.

• Quadrant I (0° to 90°): All trigonometric functions are positive.

• Quadrant II (90° to 180°): Only sine and cosecant are positive.

• Quadrant III (180° to 270°): Only tangent and cotangent are positive.

• Quadrant IV (270° to 360°): Only cosine and secant are positive.

Example:

• tan 320°: 320° is in the 4th quadrant, so tangent is negative.

• csc 115°: 115° is in the 2nd quadrant, so cosecant is positive.

Determining Quadrants for Given Conditions

To determine the quadrant(s) in which an angle lies based on the sign of trigonometric functions:

• cot θ < 0: Cotangent is negative in the 2nd and 4th quadrants.

• tan θ > 0, cos θ < 0: Tangent is positive in the 1st and 3rd quadrants; cosine is negative in the 2nd and 3rd quadrants. Therefore, θ is in the 3rd quadrant.

• csc θ < 0, sec θ > 0: Cosecant is negative in the 3rd and 4th quadrants; secant is positive in the 1st and 4th quadrants. Therefore, θ is in the 4th quadrant.

Examples with Specific Values

• sin θ = 0.34050: Sine is positive in the 1st and 2nd quadrants.

• cos θ = 0.2399 and sin θ > 0: Cosine is positive in the 1st and 4th quadrants; sine is positive in the 1st and 2nd quadrants. Therefore, θ is in the 1st quadrant.

Trigonometric Functions of Any Angle

Finding Angles for Given Trigonometric Values

To find all angles θ in a given interval (e.g., 0° ≤ θ ≤ 360°) that satisfy a trigonometric equation, use the reference angle and consider the sign of the function.

• Reference Angle (θref): The acute angle formed by the terminal side of θ and the x-axis.

• General Solution: Use the reference angle and the sign of the function to determine all possible solutions within the interval.

Examples:

• sin θ = 0.7561: θref = sin-1(0.7561) = 49.1°. Sine is positive in the 1st and 2nd quadrants, so θ = 49.1°, 180° - 49.1° = 130.9°.

• cos θ = -0.4025: θref = cos-1(0.4025) = 66.3°. Cosine is negative in the 2nd and 3rd quadrants, so θ = 180° - 66.3° = 113.7°, 180° + 66.3° = 246.3°.

• tan θ = -1.366: θref = tan-1(1.366) = 53.8°. Tangent is negative in the 2nd and 4th quadrants, so θ = 180° - 53.8° = 126.2°, 360° - 53.8° = 306.2°.

• sec θ = -1.784: θref = sec-1(1.784) = cos-1(1/1.784) = 55.9°. Secant is negative in the 2nd and 3rd quadrants, so θ = 180° - 55.9° = 124.1°, 180° + 55.9° = 235.9°.

• sin θ = 0.621 (in radians): θref = sin-1(0.621) = 0.670 rad. Sine is positive in the 1st and 2nd quadrants, so θ = 0.670 rad, π - 0.670 = 2.47 rad.

Applications: Circular Motion and Sectors

Arc Length and Area of a Sector

For a circle of radius r and central angle θ (in radians):

• Arc Length: $s = r\theta$

• Area of Sector: $A = \frac{1}{2} r^2 \theta$

Examples:

• Arc Length: If r = 3.30 cm and θ = 820°, first convert θ to radians: $\theta = 820^\circ \times \frac{\pi}{180^\circ} = 14.31$ radians. Then, $s = (3.30)(14.31) = 47.2$ cm.

• Area of Sector: For r = 1.25 m, θ = 50.6°, $\theta = 0.8831$ radians. $A = \frac{1}{2}(1.25^2)(0.8831) = 0.690$ m2.

Finding Central Angle from Area

Given area A and radius r, solve for θ:

• $A = \frac{1}{2} r^2 \theta \implies \theta = \frac{2A}{r^2}$

Example: If A = 75.5 m2 and r = 12.2 m, $\theta = \frac{2 \times 75.5}{12.2^2} = 1.01$ radians.

Linear and Angular Velocity

Definitions and Formulas

• Linear Velocity (v): The rate at which a point moves along a circular path. $v = r\omega$

• Angular Velocity (\omega): The rate at which the central angle changes. $\omega = \frac{\theta}{t}$

• Relationship: $v = r\omega$

Examples

• DVD Example: Diameter = 12.1 cm, r = 6.05 cm = 0.0605 m, ω = 360 rev/min = $360 \times 2\pi = 2261.95$ rad/min. $v = (0.0605)(2261.95) = 136.8$ m/min.

• Vehicle Tire Example: Diameter = 3.66 m, r = 1.83 m, v = 5.6 km/h = 93.33 m/min. $\omega = \frac{v}{r} = \frac{93.33}{1.83} = 51$ rad/min.

• Pulley Example: s = 4.00 m, t = 2.00 s, r = 0.200 m. $v = \frac{s}{t} = 2.0$ m/s, $\omega = \frac{v}{r} = 10$ rad/s.

Summary Table: Trigonometric Function Signs by Quadrant

Quadrant	Positive Functions	Negative Functions
I	sin, cos, tan, csc, sec, cot	None
II	sin, csc	cos, tan, sec, cot
III	tan, cot	sin, cos, csc, sec
IV	cos, sec	sin, tan, csc, cot

Additional info: The notes also include step-by-step worked examples for finding angles and applying arc length and area formulas, which are standard in Precalculus trigonometry.