Introduction

Trigonometric equations and inequalities are essential topics for high school students tackling Precalculus and those preparing to study 3.10 Trigonometric Equations and Inequalities in AP® Precalculus. Understanding the process of solving trig equations not only aids in mathematical comprehension but also boosts problem-solving skills across various disciplines. This article explores trigonometric functions, equations, and inequalities, breaking them down into manageable segments with clear examples for easy understanding.

Understanding Trigonometric Functions

Trigonometric functions form the foundation of trigonometry. These include sine, cosine, and tangent, which are closely related to the unit circle—a circle with a radius of one. The unit circle helps describe how these functions behave, especially noting their periodicity. Periodicity means these functions repeat their values at regular intervals.

Example 1: Basic Function Values

To find basic function values like sine and cosine using the unit circle, consider an angle of 30^\circ.

• Step 1: Locate 30^\circ on the unit circle.
• Step 2: Understand that in the unit circle, sine corresponds to the y-coordinate, and cosine corresponds to the x-coordinate.
• Step 3: Read the coordinates: (\frac{\sqrt{3}}{2}, \frac{1}{2}) .

Therefore, \sin(30^\circ) = \frac{1}{2} and \cos(30^\circ) = \frac{\sqrt{3}}{2}.

Solving Trig Equations

A trigonometric equation is an equation involving trigonometric functions set equal to a value. Solving trig equations can involve inverse trigonometric functions and may require attention to domain restrictions.

Example 2: Solving a Simple Trigonometric Equation

Solve the equation: \sin(x) = 0.5

• Step 1: Use the inverse sine function to find x. This gives x = \sin^{-1}(0.5).
• Step 2: Calculate x. \sin^{-1}(0.5) = 30^\circ or \frac{\pi}{6} radians. However, we must use our knowledge of the unit circle as well to know that \sin x is also positive in Quadrant II and therefore \frac{5\pi}{6} is another valid solution
• Step 3: Consider the periodicity of sine. Sine repeats every 360^\circ or 2\pi. Thus, x = \frac{\pi}{6} + 2k\pi and x = \frac{5\pi}{6} + 2k\pi, where k is an integer.

Infinite Solutions and Periodicity

Trigonometric functions are periodic, which means solutions to trigonometric equations often repeat indefinitely. It is crucial to express these infinite solutions correctly.

Example 3: Finding All Solutions

Solve \tan(x) = 1.

• Step 1: Use the inverse tangent function: x = \tan^{-1}(1).
• Step 2: Calculate x. \tan^{-1}(1) = 45^\circ or \frac{\pi}{4}.
• Step 3: Tangent has a period of 180^\circ or \pi, so x = \frac{\pi}{4} + k\pi.

Solving Trigonometric Inequalities

Trigonometric inequalities involve expressions with greater than or less than symbols. Solving them requires methods like testing intervals or graphing the function.

Example 4: Solving a Trigonometric Inequality

Challenging Trigonometric Inequality Problem

Solve the inequality: 2\cos(x) + 1 \leq 0 for 0 \leq x < 2\pi

Step 1: Isolate the Trigonometric Function

Rearrange the inequality:

2\cos(x) \leq -1
\cos(x) \leq -\frac{1}{2}

Step 2: Determine Reference Angles

The reference angle for \cos^{-1} \left(\frac{1}{2}\right) is: \frac{\pi}{3}

Since \cos(x) is negative in quadrants II and III, the general solutions are:

x = \pi - \frac{\pi}{3}, \quad x = \pi + \frac{\pi}{3}
x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}

Step 3: Express the Solution Set

Since we are looking for \cos(x) \leq -\frac{1}{2} , we include all x values in the intervals where cosine is less than or equal to -\frac{1}{2} , which are:

\frac{2\pi}{3} \leq x \leq \frac{4\pi}{3}

Thus, the solution is: \left[\frac{2\pi}{3}, \frac{4\pi}{3}\right]

Contextual Scenarios and Domain Restrictions

Real-world problems often impose domain restrictions. Understanding these contexts helps apply trigonometric solutions practically.

Example 5: Contextual Problem – Ferris Wheel Height Restriction

A Ferris wheel has a radius of 25 meters and takes 40 seconds to complete one full revolution. A rider’s height ( h(t) ) above the ground at time ( t ) (in seconds) is given by:

h(t) = 25\sin\left(\frac{\pi}{20} t - \frac{\pi}{2} \right) + 30

We want to find when the rider is at least 40 meters above the ground:

25\sin\left(\frac{\pi}{20} t - \frac{\pi}{2} \right) + 30 \geq 40

1. Solve for ( t ):
   25\sin\left(\frac{\pi}{20} t - \frac{\pi}{2} \right) \geq 10
   \sin\left(\frac{\pi}{20} t - \frac{\pi}{2} \right) \geq \frac{2}{5}
2. Domain Restriction:

• Normally, this equation would have multiple solutions.
• However, since the Ferris wheel completes a cycle in 0 to 40 seconds, we only consider solutions in this time frame.

Thus, the real-world context limits the number of solutions compared to a general trigonometric equation.

Quick Reference Chart

Conclusion

Mastering trigonometric equations and inequalities can seem daunting, but breaking the process into steps demystifies the topic. With practice, these concepts reinforce mathematical skills essential for AP® Precalculus and beyond.

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