Trigonometric Inequalities

Introduction to Trigonometric Inequalities

A trigonometric inequality is a mathematical statement that compares a trigonometric expression with a number or another expression using an inequality symbol.

Unlike trigonometric equations, which ask us to find specific angles that produce an exact value, trigonometric inequalities ask us to identify ranges of angles that satisfy a condition.

For example:

• sin θ > 0.5

• cos θ ≤ 0

• tan θ ≥ 1

• 2sin θ − 1 < 0

These inequalities are useful because many real-world situations involve conditions that must remain above or below a particular value rather than equal an exact value.

Examples include determining when a rotating object reaches a certain height, when daylight exceeds a particular duration, or when an alternating electrical signal is above a specified voltage.

Understanding Trigonometric Inequalities

The four main inequality symbols are:

The distinction between strict inequalities (> and <) and inclusive inequalities (≥ and ≤) is important.

For example:

• sin θ > 0.5 excludes angles where sin θ = 0.5.

• sin θ ≥ 0.5 includes angles where sin θ = 0.5.

These boundary angles determine whether the endpoints of solution intervals are included.

Why inequalities produce intervals

Consider:

sin θ > 0.5

Within one complete revolution, sin θ = 0.5 at:

θ = 30° and θ = 150°

Between these angles, the sine function has values greater than 0.5.

Therefore:

30° < θ < 150°

In interval notation:

(30°, 150°)

The endpoints are excluded because the inequality is strict.

Solving Sine Inequalities

Sine inequalities can be solved using the unit circle or the graph of y = sin θ.

The sine function has the following properties:

• Range: −1 ≤ sin θ ≤ 1

• Period: 360° or 2π radians

• Positive in Quadrants I and II

• Negative in Quadrants III and IV

Worked Example: Solving sin θ ≥ 0.5

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

sin θ = 0.5

θ = 30° and 150°

Step 2: Determine where the sine function satisfies the inequality.

Sine is greater than or equal to 0.5 between these angles.

Step 3: Write the solution.

30° ≤ θ ≤ 150°

Interval notation:

[30°, 150°]

Square brackets indicate that both endpoints are included.

Worked Example: Solving sin θ < −0.5

Solve for 0° ≤ θ < 360°.

The boundary angles are:

θ = 210° and 330°

The sine function is less than −0.5 between these angles.

Therefore:

210° < θ < 330°

Interval notation:

(210°, 330°)

Solving Cosine Inequalities

The cosine function has these properties:

• Range: −1 ≤ cos θ ≤ 1

• Period: 360° or 2π radians

• Positive in Quadrants I and IV

• Negative in Quadrants II and III

Cosine inequalities may produce two separate solution intervals within a complete revolution.

Worked Example: Solving cos θ > 0.5

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

cos θ = 0.5

θ = 60° and 300°

Step 2: Identify where cosine exceeds 0.5.

This occurs near the beginning and end of the revolution.

Step 3: Express the solution.

0° ≤ θ < 60° or 300° < θ < 360°

Interval notation:

[0°, 60°) ∪ (300°, 360°)

The symbol ∪ means union, indicating that either interval satisfies the inequality.

Notice that 0° is included because cos 0° = 1, while 360° is excluded because the original domain excludes it.

Solving Tangent Inequalities

The tangent function behaves differently from sine and cosine.

Important properties include:

• Period: 180° or π radians

• Range: all real numbers

• Undefined at θ = 90° + 180°k, where k is an integer

• Positive in Quadrants I and III

• Negative in Quadrants II and IV

Because tangent has vertical asymptotes, particular care is needed when determining solution intervals.

Worked Example: Solving tan θ ≥ 1

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

tan θ = 1

θ = 45° and 225°

Step 2: Identify where tangent is at least 1.

Tangent increases from 1 toward positive infinity in two intervals.

Step 3: Write the solution.

45° ≤ θ < 90° or 225° ≤ θ < 270°

Interval notation:

[45°, 90°) ∪ [225°, 270°)

The angles 90° and 270° must be excluded because tangent is undefined there.

Interpreting Solution Intervals

A solution interval describes a continuous range of values that satisfy an inequality.

There are four common forms of interval notation.

Multiple solution intervals

Some trigonometric inequalities have solutions in more than one interval.

For example:

cos θ < 0

For 0° ≤ θ < 360°:

90° < θ < 270°

Interval notation:

(90°, 270°)

However:

sin θ < 0.5

For the same domain:

0° ≤ θ < 30° or 150° < θ < 360°

Interval notation:

[0°, 30°) ∪ (150°, 360°)

Understanding these intervals allows us to describe precisely which angles satisfy a condition.

Representing Solutions Graphically

Graphs provide a visual method for solving trigonometric inequalities.

The general process is:

1. Graph the trigonometric function.

2. Draw a horizontal line representing the boundary value.

3. Identify the intersection points.

4. Determine where the graph lies above or below the boundary.

5. Express the corresponding x-values as intervals.

Worked Example: Graphical Solution

Solve:

2sin θ − 1 > 0

For 0° ≤ θ < 360°.

Step 1: Rearrange the inequality.

2sin θ > 1

sin θ > 0.5

Step 2: Identify the boundary angles.

θ = 30° and 150°

Step 3: Interpret the graph.

The sine curve lies above y = 0.5 between these angles.

Step 4: Write the solution.

30° < θ < 150°

Interval notation:

(30°, 150°)

The graphical and algebraic methods produce the same answer.

Solving Inequalities in Radians

Trigonometric inequalities can also be expressed in radians.

Important angle conversions include:

Worked Example: Solving sin x ≥ √2/2

For 0 ≤ x < 2π:

The boundary angles are:

x = π/4 and x = 3π/4

Sine is greater than or equal to √2/2 between these angles.

Therefore:

π/4 ≤ x ≤ 3π/4

Interval notation:

[π/4, 3π/4]

General Solutions and Periodicity

Trigonometric functions repeat their values at regular intervals.

This means that inequalities can have infinitely many solutions unless the domain is restricted.

For sine and cosine, the period is 360° or 2π.

For tangent, the period is 180° or π.

Worked Example: General Solution

Solve:

sin θ > 0.5

For all real values of θ.

Within one revolution:

30° < θ < 150°

Since sine repeats every 360°, the general solution is:

30° + 360°k < θ < 150° + 360°k

where k is any integer.

In radians:

π/6 + 2πk < θ < 5π/6 + 2πk

where k ∈ ℤ.

This notation represents all possible solution intervals.

Compound Trigonometric Inequalities

A compound inequality contains two conditions that must be satisfied simultaneously.

For example:

0 ≤ sin θ ≤ 0.5

This means:

sin θ ≥ 0

and

sin θ ≤ 0.5

Both conditions must be true.

Worked Example

Solve for 0° ≤ θ < 360°:

0 ≤ sin θ ≤ 0.5

Step 1: Identify where sin θ ≥ 0.

0° ≤ θ ≤ 180°

Step 2: Identify where sin θ ≤ 0.5.

0° ≤ θ ≤ 30° or 150° ≤ θ < 360°

Step 3: Find the intersection of the solution sets.

0° ≤ θ ≤ 30° or 150° ≤ θ ≤ 180°

Interval notation:

[0°, 30°] ∪ [150°, 180°]

The solution must satisfy both conditions, so we use the intersection of the two sets of possible values.

Real-World Applications

Trigonometric inequalities are especially useful when quantities change periodically.

Application: Height of a Ferris Wheel

A Ferris wheel has a radius of 10 m, and its centre is 12 m above the ground.

The height of a passenger is modeled by:

h(θ) = 12 + 10sin θ

where θ is measured in degrees.

Determine the angles for which the passenger is more than 17 m above the ground during one revolution.

Step 1: Write the inequality.

12 + 10sin θ > 17

Step 2: Rearrange.

10sin θ > 5

sin θ > 0.5

Step 3: Solve.

30° < θ < 150°

Interpretation:

The passenger is above 17 m between 30° and 150° of the wheel's rotation.

This represents one-third of a complete revolution.

Application: Alternating Electrical Voltage

An alternating voltage can be modeled by:

V(t) = 120sin(2πt)

where t is measured in seconds.

Determine when the voltage exceeds 60 V during the first second.

Step 1: Write the inequality.

120sin(2πt) > 60

Step 2: Simplify.

sin(2πt) > 0.5

Step 3: Solve the corresponding angle inequality.

π/6 < 2πt < 5π/6

Step 4: Divide by 2π.

1/12 < t < 5/12

Answer:

The voltage exceeds 60 V for:

0.0833 s < t < 0.4167 s

This method is useful when analyzing electrical signals and determining how long a voltage remains above a threshold.

Common Mistakes

Confusing equations with inequalities

An equation identifies boundary angles, while an inequality identifies intervals of angles.

For example, sin θ = 0.5 has two solutions in one complete revolution, but sin θ > 0.5 has an entire interval of solutions.

Using incorrect interval brackets

Strict inequalities exclude their boundary values.

• or < requires parentheses.

• ≥ or ≤ may require square brackets when the boundary is defined and included.

Forgetting the domain

A solution for 0° ≤ θ < 360° is different from a general solution covering all real angles.

Ignoring tangent asymptotes

Tangent is undefined at 90°, 270°, and equivalent angles. These values cannot be included in a solution.

Selecting the wrong side of a graph

For y > a, choose regions above the horizontal line y = a.

For y < a, choose regions below it.

Confusing union and intersection

• Union (∪) combines alternative solution intervals.

• Intersection (∩) identifies values satisfying multiple conditions simultaneously.

Forgetting to check the calculator mode

A calculator set to radians may give incorrect results when working with degree-based problems, and vice versa.

Key Terms

Key Takeaways

• Trigonometric inequalities describe ranges of angles, rather than only individual solutions.

• Solving an inequality usually begins by finding the angles where the corresponding equation is satisfied.

• The unit circle and trigonometric graphs help determine which intervals satisfy an inequality.

• Strict inequalities exclude boundary angles, while inclusive inequalities include valid boundary angles.

• Interval notation provides a concise and precise way to communicate solutions.

• A union combines separate solution intervals, while an intersection identifies values satisfying multiple conditions.

• Sine and cosine repeat every 360° or 2π radians, while tangent repeats every 180° or π radians.

• Tangent inequalities require special attention because tangent is undefined at its vertical asymptotes.

• General solutions use an integer parameter to describe repeated solution intervals.

• Trigonometric inequalities are useful for analyzing periodic phenomena such as rotating machinery, alternating current, oscillations, and seasonal cycles.

Check Your Understanding

Recall and Understanding

1. What is the difference between a trigonometric equation and a trigonometric inequality?

2. Explain why sin θ > 0.5 produces an interval rather than two individual angles.

3. What is the difference between (30°, 150°) and [30°, 150°]?

4. Why must 90° and 270° be excluded from solutions involving tan θ?

5. What does the union symbol ∪ mean in interval notation?

Application

6. Solve sin θ ≥ 0 for 0° ≤ θ < 360°.

7. Solve cos θ < 0 for 0° ≤ θ < 360°.

8. Solve tan θ > 1 for 0° ≤ θ < 180°.

9. Solve 2cos θ − 1 ≤ 0 for 0° ≤ θ < 360°.

10. Solve sin x < √3/2 for 0 ≤ x < 2π.

Analysis and Challenge

11. Explain why cos θ > 0.5 produces two separate intervals when the domain is 0° ≤ θ < 360°.

12. Solve the compound inequality 0 ≤ cos θ ≤ 0.5 for one complete revolution.

13. Find the general solution of sin θ ≤ −0.5, expressing your answer using an integer k.

14. A Ferris wheel passenger's height is modeled by h(θ) = 15 + 12sin θ. Determine the intervals when the passenger is at least 21 m above the ground.

15. An alternating electrical signal is modeled by V(t) = 80sin(4πt). Determine when the voltage is greater than 40 V during 0 ≤ t ≤ 1 second.

Final Summary

Trigonometric inequalities extend the methods used to solve trigonometric equations by identifying complete ranges of acceptable values. Rather than stopping at the angles where a function equals a particular value, we use graphs, unit-circle relationships, and interval notation to determine where the function satisfies the required condition.

Understanding how to interpret solution intervals, represent solutions graphically, and account for periodicity allows us to solve both mathematical inequalities and practical problems involving repeating patterns.

These techniques form an important foundation for further study of trigonometric functions, mathematical modeling, calculus, and periodic systems.