Problem 27 - Algebra 2 and Trigonometry

Question

In \(25-28,\) find, to the nearest hundredth, the radian measures of all \(\theta\) in the interval \(0 \leq \theta<2 \pi\) that make the equation true. $$ 15 \tan \theta-7=5 \tan \theta-3 $$

Step-by-Step Solution

Answer

The radian measures are approximately \(0.38\) and \(3.52\).

Step 1: Understand the Equation

The problem gives the equation \( 15 \tan \theta - 7 = 5 \tan \theta - 3 \). Our task is to solve for \( \theta \) in the interval \( 0 \leq \theta < 2\pi \).

Step 2: Simplify the Equation

Subtract \( 5 \tan \theta \) from both sides to get \( 10 \tan \theta - 7 = -3 \).

Step 3: Solve for \( \tan \theta \)

Add 7 to both sides of the equation to obtain \( 10 \tan \theta = 4 \). Then divide by 10 to isolate \( \tan \theta \): \( \tan \theta = \frac{2}{5} \).

Step 4: Use Inverse Tangent Function

Apply the inverse tangent function to find \( \theta \): \( \theta = \tan^{-1}\left(\frac{2}{5}\right) \). Calculate \( \tan^{-1}\left(\frac{2}{5}\right) \) in radians and approximate it to the nearest hundredth.

Step 5: Consider the Tangent Periodicity

The tangent function has a period of \( \pi \). Thus, the general solutions are \( \theta = \tan^{-1}\left(\frac{2}{5}\right) + n\pi \), where \( n \) is an integer.

Step 6: Find All Solutions in the Interval

Calculate possible values by adding multiples of \( \pi \) to \( \theta = \tan^{-1}\left(\frac{2}{5}\right) \) and ensure each value stays within \( 0 \leq \theta < 2\pi \).

Step 7: Calculate Specific Solutions

Using a calculator, \( \tan^{-1}\left(\frac{2}{5}\right) \approx 0.38 \) radians. The potential solutions are \( \theta = 0.38 \) and \( \theta = 0.38 + \pi \approx 3.52 \). Both values are within the specified interval.

Key Concepts

Tangent Function

The tangent function, often represented as \( \tan \theta \), is one of the primary trigonometric functions in mathematics. It relates the ratio of the side opposite to an angle \( \theta \) to the adjacent side in a right triangle. It is mathematically expressed as:

• \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)

The tangent function has some unique properties:

• Periodicity: The tangent function repeats its values in regular intervals, called periods. The period of the tangent function is \( \pi \).
• Vertical Asymptotes: It has vertical asymptotes at odd multiples of \( \frac{\pi}{2} \), meaning the function is undefined at these points.
• Range: Unlike sine and cosine functions, which are bounded, the tany function ranges from \(-\infty\) to \(+\infty\).

These characteristics should be kept in mind when solving equations involving \( \tan \theta \), as they influence how and where solutions occur.

Inverse Trigonometric Functions

Inverse trigonometric functions provide the angles (or arc measures) given trigonometric values. For tangent, the inverse function is represented as \( \tan^{-1}(x) \) (or \( \arctan(x) \)). It outputs an angle whose tangent is \( x \).

• Purpose: It's used to find the angle \( \theta \) when you know the value of \( \tan \theta \).
• Domain and Range: For \( \arctan(x) \), the domain is all real numbers, while the range is \(-\frac{\pi}{2} < \theta < \frac{\pi}{2} \).
• Applications: It's widely used in solving trigonometric equations, like in determining the exact angles satisfying certain conditions.

In our problem, we saw the application of the inverse tangent to solve for \( \theta \): \( \theta = \tan^{-1}\left(\frac{2}{5}\right) \), somewhat like finding which angle gives \( \tan \theta = \frac{2}{5} \).

Trigonometric Periodicity

Periodicity is a fundamental trait of trigonometric functions. For the tangent function, the period is \( \pi \). This means:

• The function repeats its values every \( \pi \) units.
• If \( \theta \) is a solution, then \( \theta + n\pi \) (where \( n \) is an integer) will also be a solution.

This concept is particularly useful when determining multiple solutions within a given interval. For example, once you find an initial solution using \( \tan^{-1}\left(\frac{2}{5}\right) \), subsequent solutions can be found by adding or subtracting multiples of \( \pi \), ensuring they lie within a specified range (e.g., \( 0 \leq \theta < 2\pi \)).

Solving Equations in Radians

When solving trigonometric equations, it's common to work in radians instead of degrees. Radians provide a natural measure for angles based on the radius of the circle.

• Why Radians? Using radians simplifies many mathematical expressions and is the standard in calculus and higher-level mathematics.
• Conversion: 1 radian equals approximately 57.2958 degrees, and \( 2\pi \) radians is equal to 360 degrees.
• Example Application: In our exercise, we converted the solution to radians to find precise values for \( \theta \), specifically highlighting the periodicity by considering angles like \( 0.38 \) and adding \( \pi \) to get 3.52.
• Interval Considerations: When finding solutions in an interval, always ensure the results are adjusted correctly to fit within the specified radian range, such as \( 0 \leq \theta < 2\pi \).