Problem 17 - 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. $$ 5 \sin ^{2} \theta+2 \sin \theta=0 $$

Step-by-Step Solution

Answer

The values of \( \theta \) are approximately \( 0, \pi, 3.55, \text{and}\ 5.87 \).

Step 1: Recognize the Type of Equation

The given equation \( 5\sin^2 \theta + 2\sin \theta = 0 \) is quadratic in form with respect to \( \sin \theta \). We will treat \( \sin \theta \) as a variable, say \( x \), and rewrite the equation as \( 5x^2 + 2x = 0 \).

Step 2: Factor the Quadratic Equation

Factor the quadratic equation \( 5x^2 + 2x = 0 \) by taking \( x \) common. This gives \( x(5x + 2) = 0 \).

Step 3: Apply the Zero-Product Property

According to the Zero-Product Property, if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \). Therefore, set each factor equal to zero: \( x = 0 \) or \( 5x + 2 = 0 \).

Step 4: Solve for \( x \)

For \( x(=\sin \theta) = 0 \), the solutions are \( \theta = 0 \) and \( \theta = \pi \) within the interval \( 0 \leq \theta < 2\pi \). For \( 5x + 2 = 0 \), solve for \( x \): \( 5x = -2 \Rightarrow x = -\frac{2}{5} \).

Step 5: Solve for \( \theta \) when \( \sin \theta = -\frac{2}{5} \)

Since \( \sin \theta = -\frac{2}{5} \), find the reference angle \( \theta_{ref} \) such that \( \sin \theta_{ref} = \frac{2}{5} \). Compute \( \theta_{ref} = \sin^{-1}\left(\frac{2}{5}\right) \approx 0.4115 \) radians. The solutions for \( \theta \) will be in the third and fourth quadrants, thus \( \theta = \pi + \theta_{ref} \approx 3.5531 \) and \( \theta = 2\pi - \theta_{ref} \approx 5.8717 \).

Step 6: Round to the Nearest Hundredth

Round each solution to the nearest hundredth as instructed: \( \theta \approx 0 \), \( \theta \approx \pi \), \( \theta \approx 3.55 \), and \( \theta \approx 5.87 \).

Key Concepts

Quadratic Form

Quadratic form is a way of expressing a mathematical equation that follows a specific structure similar to the general quadratic equation, which is written as:

• \( ax^2 + bx + c = 0 \)

In the problem at hand, the equation \(5 \sin^2 \theta + 2 \sin \theta = 0\) adopts a quadratic form where \( \sin \theta \) is treated as the variable \( x \). The idea is to temporarily replace trigonometric expressions that are squared with a variable to simplify solving the equation. By doing this, the equation transforms to:

• \( 5x^2 + 2x = 0 \)

At this point, it becomes easier to apply algebraic strategies such as factoring to solve for \( x \), which after solving is translated back to the trigonometric function to find \( \theta \). This strategy simplifies complex trigonometric equations and is a handy tool for students tackling higher-level algebra and trigonometry problems.

Zero-Product Property

The zero-product property is a fundamental rule in algebra. This principle states that if a product of two numbers (or expressions) is zero, then at least one of the numbers must be zero. Mathematically, if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \). In solving the transformed quadratic equation, \( 5x^2 + 2x = 0 \), we factor it to \( x(5x + 2) = 0 \). According to the zero-product property, we then set each factor equal to zero, providing two simpler equations:

• \( x = 0 \)
• \( 5x + 2 = 0 \)

Solving these equations individually gives the specific roots or solutions for \( x \). In this context, it enables us to find the corresponding angles \( \theta \) by considering each \( x \) value as the sine of an angle. Understanding and applying the zero-product property is crucial when dealing with equations that need to be factored, providing a clear path to finding the solution quickly and efficiently.

Reference Angle

The reference angle is used in trigonometry to find the angle whose sine, cosine, or tangent is the positive version of a given value. It is always found in the first quadrant of the unit circle, where all trigonometric values are positive. In the problem provided, we needed to find \( \theta \) such that \( \sin \theta = -\frac{2}{5} \). First, find the angle \( \theta_{ref} \) by calculating:

• \( \theta_{ref} = \sin^{-1}\left(\frac{2}{5}\right) \approx 0.4115 \text{ radians} \)

The reference angle helps locate the other angles \( \theta \) in the different quadrants where the original trigonometric identity holds true. Given that sine is negative in both the third and fourth quadrants:

• For the third quadrant: \( \theta = \pi + \theta_{ref} \)
• For the fourth quadrant: \( \theta = 2\pi - \theta_{ref} \)

Thus, these calculations reveal the possible radian measures for \( \theta \) that satisfy the given trigonometric equation. The reference angle is essential for translating and rotating angled values and for solving trigonometric equations across different quadrants.

Radian Measure

Radian measure is a way of expressing angles, where the angle is determined by the arc length it subtends on a unit circle. One full rotation around a circle is \( 2\pi \) radians, equating to 360 degrees. In trigonometry, radian measure offers a natural and direct way of defining angles and working through equations involving them. For the given problem, solutions were explicitly requested in radians, specifically:

• The solutions for \( \theta \) fall within the interval \( 0 \leq \theta < 2\pi \).

Radian measure simplifies the mathematics of circles and periodic functions, and it's particularly useful because it maintains consistency in the linear relationships present in circular motion. When working with trigonometric functions like sine and cosine, keeping measurements in radians often leads to cleaner and more straightforward equations. For students, understanding radian measure is crucial as it frequently appears in higher-level mathematics, physics, and engineering problems and aids in simplifying computational efforts involving periodic phenomena.