Problem 34 - Precalculus: Functions and Graphs

Question

Use sum-to-product formulas to find the solutions of the equation. $$\sin 5 x-\sin x=2 \cos 3 x$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{6} + \frac{n\pi}{3}\) and \(x = \frac{\pi}{4} + k\pi\), with \(n, k \in \mathbb{Z}\).

Step 1: Apply the Sum-to-Product Formula for the Sine Difference

The sum-to-product formulas are used to express sums or differences of sine or cosine functions as products. Here, we can start with the left side of the equation involving the sine difference: \[ \sin 5x - \sin x = 2 \cos 3x. \]We will use the identity\[ \sin a - \sin b = 2 \cos \left( \frac{a+b}{2} \right) \sin \left( \frac{a-b}{2} \right). \]Applying this formula to \(\sin 5x - \sin x\), we have:\[ \sin 5x - \sin x = 2 \cos \left( \frac{5x + x}{2} \right) \sin \left( \frac{5x - x}{2} \right) = 2 \cos(3x) \sin(2x). \]

Step 2: Equate and Simplify

Substitute the result from Step 1 back into the equation:\[ 2 \cos(3x) \sin(2x) = 2 \cos(3x). \]We can divide both sides of the equation by 2, which simplifies to:\[ \cos(3x) \sin(2x) = \cos(3x). \]

Step 3: Factor Out Common Terms

We notice that \(\cos(3x)\) is a common factor in both terms on the left side:\[ \cos(3x) (\sin(2x) - 1) = 0. \]This equation implies two potential solutions based on the zero-product property:1. \(\cos(3x) = 0\)2. \(\sin(2x) - 1 = 0\) which simplifies to \(\sin(2x) = 1\).

Step 4: Solve \(\cos(3x) = 0\)

For \(\cos(3x) = 0\), the general solution is:\[ 3x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}. \]Divide by 3 to solve for \(x\):\[ x = \frac{\pi}{6} + \frac{n\pi}{3}, \quad n \in \mathbb{Z}. \]

Step 5: Solve \(\sin(2x) = 1\)

For \(\sin(2x) = 1\), the solutions occur when the angle is an odd multiple of \(\frac{\pi}{2}\):\[ 2x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}. \]Solve for \(x\) by dividing by 2:\[ x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}. \]

Step 6: Combine Solutions

To express the full solution set, we combine the results from Step 4 and Step 5:1. \(x = \frac{\pi}{6} + \frac{n\pi}{3}, \quad n \in \mathbb{Z}.\)2. \(x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}.\)

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental relationships between trigonometric functions that hold true for any angle. They are crucial tools for solving trigonometric equations and simplifying expressions. One of these identities is the sum-to-product formula, which converts sums or differences of sines and cosines into products. This conversion makes it easier to solve certain types of equations. In the provided exercise, the sum-to-product formula for the sine difference is utilized:

• Formula: \( \sin a - \sin b = 2 \cos \left( \frac{a+b}{2} \right) \sin \left( \frac{a-b}{2} \right) \)
• Example from exercise: \( \sin 5x - \sin x = 2 \cos(3x) \sin(2x) \)

Understanding and applying these identities helps in transforming complex trigonometric expressions into more manageable forms. This simplification plays a vital role in solving trigonometric equations as shown in the given problem.

Solving Trigonometric Equations

To solve trigonometric equations effectively, a strategic approach involving identities, factoring, and understanding periodic properties is necessary. In the problem, after applying the sum-to-product identity, the equation is simplified to:

• \( 2 \cos(3x) \sin(2x) = 2 \cos(3x) \)

By dividing both sides by 2, we further simplify to:

• \( \cos(3x) \sin(2x) = \cos(3x) \)

This can be factored by recognizing common terms, leading to the equation:

• \( \cos(3x) ( \sin(2x) - 1) = 0 \)

The zero-product property tells us that if a product is zero, at least one of the factors must be zero. This gives two separate equations to solve:

• \( \cos(3x) = 0 \)
• \( \sin(2x) = 1 \)

Each equation is then solved individually to find the general solutions, which are later combined for the complete solution set.

Sine and Cosine Functions

The sine and cosine functions are two of the most important trigonometric functions. They describe how the lengths of the sides of a right triangle relate to its angles. Importantly, these functions are periodic, which means they repeat their values in regular intervals. This property is key when solving trigonometric equations, as solutions will often repeat over their respective periods.The exercise highlights this with functions like \( \sin(2x) \) and \( \cos(3x) \). These functions demonstrate specific periodic properties:

• \( \sin(2x) \) has a period of \( \pi \), compared to the standard \( 2\pi \) for \( \sin x \).
• \( \cos(3x) \) has a period of \( \frac{2\pi}{3} \), compared to \( 2\pi \) for \( \cos x \).

When solving the exercise, understanding these periods allows one to determine all possible solutions within the desired interval. Recognizing when these functions reach zero or one, as seen in solving the equations \( \cos(3x) = 0 \) and \( \sin(2x) = 1 \), is crucial for finding all eligible solutions.