Problem 14 - Algebra 2 and Trigonometry

Question

In \(9-14,\) find, to the nearest tenth of a degree, the values of \(\theta\) in the interval \(0^{\circ} \leq \theta<360^{\circ}\) that satisfy each equation. $$ \sec ^{2} \theta-7 \sec \theta+12=0 $$

Step-by-Step Solution

Answer

\( \theta \approx 75.5^\circ, 284.5^\circ, 70.5^\circ, 289.5^\circ \).

Step 1: Understand the Equation

The given equation is \( \sec^2 \theta - 7 \sec \theta + 12 = 0 \). The secant function is the reciprocal of the cosine function, so \( \sec \theta = \frac{1}{\cos \theta} \). This equation is quadratic in form with \( \sec \theta \) being the variable.

Step 2: Substitute a Variable

Let \( x = \sec \theta \). Then the equation becomes \( x^2 - 7x + 12 = 0 \). We can solve for \( x \) using the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -7 \), and \( c = 12 \).

Step 3: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \] which simplifies to \[ x = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2} \].

Step 4: Solve for x

Calculate the two possible values of \( x \):1. \( x = \frac{7 + 1}{2} = 4 \)2. \( x = \frac{7 - 1}{2} = 3 \)Thus, \( \sec \theta = 4 \) or \( \sec \theta = 3 \).

Step 5: Find Cosine Values

Since \( \sec \theta = \frac{1}{\cos \theta} \), solve for \( \cos \theta \):1. If \( \sec \theta = 4 \), then \( \cos \theta = \frac{1}{4} \).2. If \( \sec \theta = 3 \), then \( \cos \theta = \frac{1}{3} \).

Step 6: Determine Angles with Inverse Cosine

Use the inverse cosine to find \( \theta \):1. For \( \cos \theta = \frac{1}{4} \), \( \theta = \cos^{-1}(\frac{1}{4}) \approx 75.5^\circ \).2. For \( \cos \theta = \frac{1}{3} \), \( \theta = \cos^{-1}(\frac{1}{3}) \approx 70.5^\circ \).

Step 7: Find All Angles in Given Interval

Cosine has a positive value in the first and fourth quadrants:- For \( \cos \theta = \frac{1}{4} \), solutions are \( 75.5^\circ \) and \( 360^\circ - 75.5^\circ = 284.5^\circ \).- For \( \cos \theta = \frac{1}{3} \), solutions are \( 70.5^\circ \) and \( 360^\circ - 70.5^\circ = 289.5^\circ \).

Key Concepts

Quadratic Equations

Quadratic equations are crucial in solving trigonometric problems that resemble polynomial forms. In this exercise, the equation \( \sec^2 \theta - 7 \sec \theta + 12 = 0 \) is quadratic in the variable \( \sec \theta \). To solve this, we treat \( \sec \theta \) like a variable, let's call it \( x \). The equation thus simplifies to \( x^2 - 7x + 12 = 0 \). We can use the quadratic formula to find the values of \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula tells us the solutions for \( x \) based on the coefficients \( a \), \( b \), and \( c \). Here, \( a = 1 \), \( b = -7 \), and \( c = 12 \). After substituting in these values, the formula provides two potential solutions. Quadratic equations are powerful as they allow us to approach problems that initially seem too complex by simplifying them into polynomial forms.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools when finding angle measures from known trigonometric values. In the provided exercise, after solving the quadratic equation, we must find the angles corresponding to the solutions \( \cos \theta = \frac{1}{4} \) and \( \cos \theta = \frac{1}{3} \). This is done using the inverse cosine function, also denoted as \( \cos^{-1} \). The inverse cosine function provides angles whose cosine is a given value:

• \( \theta = \cos^{-1}(\frac{1}{4}) \approx 75.5^\circ \)
• \( \theta = \cos^{-1}(\frac{1}{3}) \approx 70.5^\circ \)

Inverse trigonometric functions allow us to transition from evaluating the function itself to understanding the angles that produce those trigonometric values. This shift is vital for determining all relevant solutions within a specific interval.

Trigonometric Identities

Trigonometric identities serve as foundational tools in solving trigonometric equations like the one in this exercise. One key identity utilized here is the reciprocal identity: \( \sec \theta = \frac{1}{\cos \theta} \). This identity relates the secant and cosine functions, allowing us to convert the problem into a more manageable form.
Identities help solve equations by simplifying expressions and transforming complex trigonometric problems into more accessible algebraic or polynomial forms. Recalling essential trigonometric identities is an instrumental skill in mathematics, providing the foundation for analyzing and solving a wide range of problems. By using identities effectively, we break down challenging trigonometric equations into simpler components, easing the path to finding solutions.