AP Precalculus -3.10 Trigonometric Equations and Inequalities- Study Notes - Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- Study Notes – Effective Fall 2023
AP Precalculus -3.10 Trigonometric Equations and Inequalities- Study Notes – AP Precalculus- per latest AP Precalculus Syllabus.
LEARNING OBJECTIVE
Solve equations and inequalities involving trigonometric functions.
Key Concepts:
Using Inverse Trigonometric Functions to Solve Equations and Inequalities
Infinitely Many Solutions to Trigonometric Equations
Contextual Domain Restrictions in Trigonometric Problems
Using Inverse Trigonometric Functions to Solve Equations and Inequalities
Inverse trigonometric functions are commonly used to solve equations and inequalities involving trigonometric functions by isolating the trigonometric expression and then applying the appropriate inverse.
Because inverse trigonometric functions return values only from their restricted (principal) ranges, the solutions obtained directly from inverse functions may represent only one set of solutions.
As a result, solutions often need to be modified or extended to account for:
the periodic nature of trigonometric functions
the given domain of the problem
When solving inequalities, inverse trigonometric functions help determine boundary angles, but the final solution must be written as an interval or union of intervals that satisfies the inequality.
Example:
Solve the equation
\( \sin \theta = \dfrac{1}{2} \), for \( 0 \le \theta < 2\pi \).
▶️ Answer/Explanation
Apply the inverse sine function:
\( \theta = \sin^{-1}\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6} \)
This value lies in the principal range of \( \sin^{-1} \).
Since sine is positive in Quadrants I and II, there is another solution in the given interval:
\( \theta = \pi – \dfrac{\pi}{6} = \dfrac{5\pi}{6} \)
Final answer: \( \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6} \).
Example:
Solve the inequality
\( \cos \theta \ge \dfrac{1}{2} \), for \( 0 \le \theta \le 2\pi \).
▶️ Answer/Explanation
First, find the boundary angles using the inverse cosine function:
\( \cos^{-1}\!\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3} \)
Cosine is greater than or equal to \( \dfrac{1}{2} \) in Quadrants I and IV.
Thus, the solution intervals are:
\( 0 \le \theta \le \dfrac{\pi}{3} \)
\( \dfrac{5\pi}{3} \le \theta \le 2\pi \)
Final answer: \( \theta \in \left[0, \dfrac{\pi}{3}\right] \cup \left[\dfrac{5\pi}{3}, 2\pi\right] \).
Infinitely Many Solutions to Trigonometric Equations
Trigonometric functions such as sine, cosine, and tangent are periodic, meaning their values repeat at regular intervals.
Because of this periodicity, a trigonometric equation often has infinitely many solutions.
If an angle \( \theta \) is a solution to a trigonometric equation, then additional solutions can be found by adding or subtracting integer multiples of the function’s period.
For example:
Sine and cosine have period \( 2\pi \)
Tangent has period \( \pi \)
Thus, general solutions to trigonometric equations are often written using an integer parameter \( k \), which represents the number of full periods added or subtracted.
Example:
Solve the equation
\( \sin \theta = \dfrac{1}{2} \).
▶️ Answer/Explanation
One solution is
\( \theta = \dfrac{\pi}{6} \)
Another solution in one period is
\( \theta = \dfrac{5\pi}{6} \)
To represent all solutions, add multiples of \( 2\pi \):
\( \theta = \dfrac{\pi}{6} + 2k\pi \)
\( \theta = \dfrac{5\pi}{6} + 2k\pi \)
where \( k \) is any integer.
Example:
Solve the equation
\( \tan \theta = -1 \).
▶️ Answer/Explanation
One solution is
\( \theta = -\dfrac{\pi}{4} \)
Since tangent has period \( \pi \), all solutions are given by
\( \theta = -\dfrac{\pi}{4} + k\pi \)
where \( k \) is any integer.
Contextual Domain Restrictions in Trigonometric Problems
In trigonometric equations and inequalities that arise from a contextual scenario, the context often implies a domain restriction on the input values.
Although trigonometric equations may have infinitely many mathematical solutions due to periodicity, only the solutions that lie within the contextual domain are meaningful.
These domain restrictions commonly come from:
time intervals
physical limitations
measurement constraints
When solving such problems, it is essential to first find all mathematical solutions and then restrict the answer to the given or implied domain.
Example:
The height of a Ferris wheel is modeled by a trigonometric function. One full ride lasts 12 minutes.
Solve \( \sin t = 0 \) and explain which solutions are meaningful.
▶️ Answer/Explanation
Mathematically,
\( \sin t = 0 \Rightarrow t = k\pi \), where \( k \) is any integer.
The contextual domain is
\( 0 \le t \le 12 \).
Only values of \( t = k\pi \) within this interval are valid solutions.
Conclusion: The context limits the number of solutions.
Example:
A pendulum’s angle of motion is modeled by a cosine function over one oscillation.
Solve \( \cos \theta = \dfrac{1}{2} \) for one oscillation.
▶️ Answer/Explanation
Mathematically,
\( \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3} + 2k\pi \)
For one oscillation, the domain is
\( 0 \le \theta \le 2\pi \).
Thus, the valid solutions are
\( \theta = \dfrac{\pi}{3} \) and \( \theta = \dfrac{5\pi}{3} \).
Final answer: Only the solutions within the contextual domain apply.