Problem 33 - Algebra and Trigonometry

Question

Involve equations with multiple angles. Solve each equation on the interval \([0,2 \pi)\) $$ \sin \frac{2 \theta}{3}=-1 $$

Step-by-Step Solution

Answer

The solution to the equation \( \sin\frac{2\theta}{3} = -1 \) on the interval [0, 2π) is \( \theta = \pi/4 \).

Step 1 - Identify Trigonometric Knowledge

We need to remember that sine equals -1 at an angle of \(3\pi/2\) or \(270^\circ\) in one full cycle of the sine function.

Step 2 - Solve for θ

We will start with the equation \( \sin\frac{2\theta}{3} = -1 \). From Step 1, we know \( \sin \beta = -1 \) where \( \beta = 3\pi/2 \). So we equate \( \frac{2\theta}{3} \) with \( 3\pi/2 \) which gives us \( 2\theta = 9\pi/2 \) after cross multiplying. Solving for \( \theta \) gives us \( \theta = 9\pi/4 \).

Step 3 - Check the solutions within the given interval

We have as solution \( \theta = 9\pi/4 \), but we need to ensure this falls within our interval [0, 2π). \( 9\pi/4 \) is greater than \( 2\pi \), so to get a θ within the given interval, we can subtract multiples of \( 2\pi \) (i.e., one full cycle) from our solution. \( 9\pi/4 - 2\pi = 9\pi/4 - 8\pi/4 = \pi/4 \), which falls within our interval [0, 2π).

Key Concepts

Multiple Angles

In trigonometry, equations can involve multiple angles, meaning the angle you work with is found within a trigonometric function. It's not just a simple angle like \(\theta\) but can be some multiple of it like \(\frac{2\theta}{3}\), as seen in the exercise. This can complicate equations because it stretches or compresses the graph of the trigonometric function.When dealing with multiple angles, it's important to understand:

• Each angle in a trigonometric function represents a rotation around the circle.
• A multiple angle indicates transforming the angle's movement through dilating the circle's rotation further.

Being multiple angles means our equation \( \sin \frac{2\theta}{3} = -1 \) needs to be handled by first recognizing the basic angle where sine hits -1, which is at \(3\pi/2\). After identifying this, we relate it back to our multiple angle expression \(\frac{2\theta}{3}\) and solve it like the following:

• Equate \(\frac{2\theta}{3}\) with \(3\pi/2\) to solve for \(\theta\).
• Perform cross-multiplication and simplify, where \(2\theta = 9\pi/2\), and then find \(\theta = 9\pi/4\).

Recognizing these transformations is key in addressing multiple angle equations effectively.

Sine Function

The sine function is one of the fundamental trigonometric functions that arise naturally in mathematics. It represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle but on the unit circle, it's the vertical coordinate of a point.Key characteristics of the sine function include:

• Its period is \(2\pi\), meaning it repeats itself every \(2\pi\).
• It oscillates between -1 and 1.
• The sine of an angle is -1 when the angle measures \(3\pi/2\) radians, or \(270^{\circ}\).

In the context of our exercise, understanding where the sine function equals -1 enables us to find the corresponding angle for \(\frac{2\theta}{3}\). This helps us know the necessary transformations we must make to solve \(\theta\). On the interval from \([0, 2\pi)\), it's important to remember:

• \(\sin \beta = -1\) only occurs at \(3\pi/2\) within a full cyclical rotation.

These elements help us recognize when and how to solve trigonometric problems involving the sine function effectively.

Interval Solutions

When solving trigonometric equations, a solution must often be found within a specified interval. For the problem provided, this interval is \([0, 2\pi)\), which requires us to identify solution angles that fit within this boundary.Steps to find solutions in this interval include:

• Calculate general angle solutions, as we found \(\theta = 9\pi/4\) from the angle equation.
• Because \(9\pi/4\) exceeds \(2\pi\), adjust by subtracting \(2\pi\) to bring the solution within the given interval.

The solution within the correct range is \(\theta = \pi/4\). This solution is valid because we've subtracted enough full cycles to fit the interval. Analyze multiple rotations of the circle keeping in mind: the circle's circumference is equivalent to one complete cycle of the sine function or \(2\pi\).When solving interval-based equations, double-checking boundaries is essential. Confirm if converted angles adhere to the specified interval to avoid inaccurate solutions.