Problem 15 - Algebra 2 and Trigonometry

Question

In \(15-20\) , find, to the nearest hundredth of a radian, the values of \(\theta\) in the interval \(0 \leq \theta<2 \pi\) that satisfy the equation. $$ \tan ^{2} \theta-5 \tan \theta+6=0 $$

Step-by-Step Solution

Answer

The values of \(\theta\) are approximately 1.11, 4.25, 1.25, and 4.39 radians.

Step 1: Identify Substitution

Recognize that we can use substitution to simplify the given equation. Let \( x = \tan \theta \). Now our equation is in terms of \( x \): \( x^2 - 5x + 6 = 0 \).

Step 2: Solve Quadratic Equation

Solve the quadratic equation \( x^2 - 5x + 6 = 0 \). Factor the quadratic: \( (x - 2)(x - 3) = 0 \). Thus, \( x = 2 \) and \( x = 3 \).

Step 3: Solve for \( \theta \) with \( x = 2 \)

Find \( \theta \) when \( \tan \theta = 2 \). We need to find \( \theta \) such that \( \theta = \tan^{-1}(2) \). Using a calculator, \( \theta \approx 1.107 \text{ or } \theta \approx 1.107 + \pi \), giving \( \theta \approx 1.107 \) and \( \theta \approx 4.249 \).

Step 4: Solve for \( \theta \) with \( x = 3 \)

Find \( \theta \) when \( \tan \theta = 3 \). Solve for \( \theta \) so that \( \theta = \tan^{-1}(3) \). Using a calculator, \( \theta \approx 1.249 \text{ or } \theta \approx 1.249 + \pi \), giving \( \theta \approx 1.249 \) and \( \theta \approx 4.391 \).

Step 5: Verify the Range of Solutions

Ensure all values of \( \theta \) are within the specified interval \(0 \leq \theta < 2\pi\). All calculated \( \theta \approx 1.107, 4.249, 1.249, 4.391\) fall within the interval. Round each to the nearest hundredth: \( \theta \approx 1.11, 4.25, 1.25, 4.39 \).

Key Concepts

Tangent Function

The tangent function, expressed as \( \tan(\theta) \), is one of the basic trigonometric functions. It's defined as the ratio of the sine to the cosine functions: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] This ratio is helpful in trigonometry to study angles and their relationships in triangles, particularly right-angled triangles.

• The tangent function is periodic, with a period of \(\pi \), meaning it repeats every \(\pi\) radians.
• Unlike sine and cosine, the tangent function can take on any real number value.
• It has vertical asymptotes (places where it becomes undefined) wherever the cosine function is zero:

This feature of the tangent function is particularly useful when solving trigonometric equations, as we can find multiple solutions within a given interval by using its periodic nature.

Solving Quadratic Equations

A quadratic equation is an equation that can be written in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations is a fundamental skill in algebra, important for finding unknown values that satisfy a particular condition. In this exercise, the equation \( x^2 - 5x + 6 = 0 \) is a quadratic, where:

• "\(x\)" is substituted by the tangent function \(\tan(\theta)\).
• "\(a\)" equals 1, "\(b\)" equals -5, and "\(c\)" equals 6.

To solve this equation, we can use factoring as a method: 1. Write the equation as \((x - 2)(x - 3) = 0\)2. Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:

• \( x = 2 \)
• \( x = 3 \)

Thus, the two solutions for \(x\) when using factoring are 2 and 3. This practical approach can often simplify solving equations that initially seem complex.

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial when we need to find an angle given a trigonometric value. For the tangent function, the inverse or arc function would be written as \( \tan^{-1}(x) \) or \( \arctan(x) \).

• This function helps compute the angle \( \theta \) for a given value of \( \tan(\theta) = x \).
• For example, with \( x = 2 \) and \( x = 3 \) from the original equation:
  • We find \( \theta = \tan^{-1}(2) \approx 1.107 \).
  • Similarly, \( \theta = \tan^{-1}(3) \approx 1.249 \).

Both the regular and the periodic solutions consider the repetitive nature of the tangent function, using \( \theta + \pi \) to account for all solutions in the interval \( 0 \leq \theta < 2\pi \). In problems, using a calculator is essential for determining these angles accurately to a specific decimal point, as it enhances precision in trigonometry.