Problem 12 - Algebra 2 and Trigonometry

Question

In \(11-18,\) find all radian measures of \(\theta\) in the interval \(0 \leq \theta \leq 2 \pi\) that make each equation true. Express your answers in terms of \(\pi\) when possible; otherwise, to the nearest hundredth. $$ 2 \sin 2 \theta+\sin \theta=0 $$

Step-by-Step Solution

Answer

The solutions are \( \theta = 0, \pi, 2\pi, 1.82, \) and \( 4.47 \).

Step 1 - Understand the Problem

We need to find values of \( \theta \) in the interval \( 0 \leq \theta \leq 2\pi \) that satisfy the equation \( 2 \sin 2\theta + \sin \theta = 0 \). This involves trigonometric identities and solving for \( \theta \).

Step 2 - Use Trigonometric Identity

Apply the double angle identity: \( \sin 2\theta = 2\sin\theta\cos\theta \). Substitute into the equation: \( 2(2\sin\theta\cos\theta) + \sin\theta = 0 \) which simplifies to \( 4\sin\theta\cos\theta + \sin\theta = 0 \).

Step 3 - Factor the Equation

Factor out \( \sin\theta \) from the equation: \( \sin\theta(4\cos\theta + 1) = 0 \). This gives us two separate equations to solve: \( \sin\theta = 0 \) and \( 4\cos\theta + 1 = 0 \).

Step 4 - Solve \( \sin\theta = 0 \)

The sine of an angle is zero at \( \theta = 0, \pi, 2\pi \) within the interval \( 0 \leq \theta \leq 2\pi \).

Step 5 - Solve \( 4\cos\theta + 1 = 0 \)

Solve the equation by isolating \( \cos\theta \): \( 4\cos\theta = -1 \), which gives \( \cos\theta = -\frac{1}{4} \). We find \( \theta \) by taking the arccos: \( \theta = \cos^{-1}\left(-\frac{1}{4}\right) \) and its equivalent \( 2\pi - \theta \).

Step 6 - Determine Radian Measures

\( \cos^{-1}\left(-\frac{1}{4}\right) \approx 1.82 \) radians and the other angle \( 2\pi - 1.82 \approx 4.47 \) radians. Both angles need to be checked to ensure they lie within the given interval.

Key Concepts

Radian Measure

Radian measure is one of the ways to express angles, often used in mathematics because of its direct relationship with the radius of a circle. In the unit circle, the circumference of a full circle is \(2\pi\) times the radius. This translates to there being \(2\pi\) radians in a full circle. Therefore:

• Quarter circle: \(\frac{\pi}{2}\) radians
• Semi-circle: \(\pi\) radians
• Three-quarters of a circle: \(\frac{3\pi}{2}\) radians
• Full circle: \(2\pi\) radians

To convert between degrees and radians, we use the equivalence \(\pi\) radians = 180 degrees. Thus, 1 radian equals roughly 57.3 degrees. This conversion is essential when working with trigonometric equations in radian measure, ensuring correct calculation of angles within given intervals.

Trigonometric Identities

Trigonometric identities are formulas that relate different trigonometric functions to each other. They are crucial for simplifying complex trigonometric expressions and solving equations. An essential identity needed for this problem is the double angle identity:

• \(\sin 2\theta = 2\sin\theta\cos\theta\)

By recognizing this identity within the terms of the equation, you can simplify and transform the trigonometric equation into a product form, which is easier to solve. Understanding these identities can dramatically reduce complexity and facilitate solving real-world problems.

Solving Equations

Solving trigonometric equations often involves several strategies:

• Applying identities to simplify expressions.
• Factoring or using algebraic methods to break complex equations into simpler parts.

In the equation \(2 \sin 2\theta + \sin \theta = 0\), applying the double angle identity was our first simplification step. This allowed us to factor out \(\sin \theta\), turning the equation into \(\sin\theta (4\cos\theta + 1) = 0\). This sets up the equation into two smaller equations, \(\sin\theta = 0\) and \(4\cos\theta + 1 = 0\), making it possible to solve for \(\theta\). The process of breaking down and factorization is a powerful method in solving not only trigonometric but also other algebraic equations.

Interval Solutions

Interval solutions refer to finding specific values within a defined range that satisfy an equation. For the exercise provided, the interval in question is \(0 \leq \theta \leq 2\pi\). Solving within an interval means recognizing when solutions should repeat or when values need to be adjusted back into the interval range:

• For \(\sin\theta = 0\), the solutions are \(\theta = 0, \pi, \,\text{and} \,2\pi\), all within the interval.
• For the equation \(4\cos\theta + 1 = 0\), solving gives \(\theta = \cos^{-1}\left(-\frac{1}{4}\right)\) and its symmetrical point \(2\pi - \theta\). The solutions \(\approx 1.82 \,\text{and} \,4.47 \,\text{radians}\) must also be checked to make sure the angles remain within \(0 \leq \theta \leq 2\pi\).

Working within an interval ensures your solutions are applicable to the problem at hand, often necessitating the careful use of inverse trigonometric functions and periodic properties of trig functions to find all possible solutions.