Problem 42 - Precalculus Mathematics for Calculus

Question

Use an Addition or Subtraction Formula to simplify the equation. Then find all solutions in the interval \([0,2 \pi)\) $$\sin 3 \theta \cos \theta-\cos 3 \theta \sin \theta=0$$

Step-by-Step Solution

Answer

The solutions are \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\).

Step 1: Recognize and Apply Trigonometric Identities

Notice the given equation \[ \sin 3\theta \cos \theta - \cos 3\theta \sin \theta = 0 \] resembles the form \(\sin A \cos B - \cos A \sin B\), which is the sine of a difference formula: \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). Thus, applying the identity, we can write the equation as \(\sin(3\theta - \theta) = \sin 2\theta = 0\).

Step 2: Solve for θ Using the Simplified Equation

Now that the equation is simplified to \(\sin 2\theta = 0\), solve for \(2\theta\). Since \(\sin x = 0\) at multiples of \(\pi\), set \(2\theta = n\pi\), where \(n\) is an integer.

Step 3: Find θ Values in the Specified Interval

Divide both sides by 2 to find \(\theta\): \(\theta = \frac{n\pi}{2}\). Now determine which values of \(\theta\) lie within the interval \([0, 2\pi)\). Considering integers \(n\), we have: \[\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\].

Step 4: Verify the Values Within the Interval

Ensure that each calculated \(\theta\) value satisfies the original equation and lies in the interval. Each of \(\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\) yields zero when substituted back into the original expression, confirming they are correct.

Key Concepts

Addition and Subtraction Formulas

When working with trigonometric identities, the addition and subtraction formulas provide an essential method for simplifying expressions.
These formulas relate the trigonometric functions of sum and difference of angles to products of trigonometric functions of individual angles.
In this exercise, the given equation \( \sin 3\theta \cos \theta - \cos 3\theta \sin \theta = 0 \) aligns with the subtraction formula for sine:

• \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)

Applying this identity transforms the equation into \( \sin(3\theta - \theta) = \sin 2\theta \).
This step simplifies solving significantly by reducing the complexity of the expression. With the formula, we effectively rewrite a difference of sines as a single sine function.
Understanding these formulas helps in both deriving solutions quickly and verifying given expressions within a problem context.

Sine Equation Solutions

Solving equations involving sine functions often revolves around finding the points where the function equals a specific value, like zero.
For our simplified equation \( \sin 2\theta = 0 \), we know that the sine function equals zero at integer multiples of \( \pi \).
Therefore, we set \( 2\theta = n\pi \), where \( n \) is any integer.
This forms the basis for determining the values of \( \theta \) that satisfy the equation.
Typically, solving trigonometric equations involves:

• Identifying the basic trigonometric identity
• Using known zero points of the sine function
• Applying the algebra needed to isolate the variable

After simplifying, dividing both sides by 2 gives \( \theta = \frac{n\pi}{2} \). This equation makes it straightforward to generate initial solutions before narrowing them down to those within a specified interval, like \([0, 2\pi)\).

Interval Solutions for Trigonometric Equations

Once a trigonometric equation is solved generally, the next step involves finding solutions within a particular interval.
The interval \( [0, 2\pi) \) represents one full cycle of the trigonometric functions, ensuring that we find all unique solutions in this period.
For \( \theta = \frac{n\pi}{2} \), consider integer values of \( n \) that yield \( \theta \) in the desired range:

• \( n = 0 \Rightarrow \theta = 0 \)
• \( n = 1 \Rightarrow \theta = \frac{\pi}{2} \)
• \( n = 2 \Rightarrow \theta = \pi \)
• \( n = 3 \Rightarrow \theta = \frac{3\pi}{2} \)

These \( \theta \) values are then verified by substitution back into the original equation to ensure they satisfy the condition.
Mastering how to navigate between finding the general solutions of a trigonometric equation and validating them within an interval can significantly enhance problem-solving skills.