Problem 56 - Precalculus Mathematics for Calculus

Question

Solve the equation by first using a Sum-to-Product Formula. $$\sin 5 \theta-\sin 3 \theta=\cos 4 \theta$$

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{6} + 2k\pi \) or \( \theta = \frac{5\pi}{6} + 2k\pi \), \( k \in \mathbb{Z} \).

Step 1: Apply Sum-to-Product Formula

We start with the expression on the left side of the equation: \( \sin 5\theta - \sin 3\theta \). The sum-to-product identity for the difference of sines is: \( \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) \). Applying this to our terms, \( A = 5\theta \) and \( B = 3\theta \), we get: \[ 2 \cos \left( \frac{5\theta + 3\theta}{2} \right) \sin \left( \frac{5\theta - 3\theta}{2} \right) = 2 \cos(4\theta) \sin(\theta). \]

Step 2: Substitute into the Original Equation

Substitute the transformed expression from Step 1 back into the original equation: \[ 2 \cos(4\theta) \sin(\theta) = \cos 4\theta. \]

Step 3: Simplify the Equation

Divide both sides of the equation by \( \cos 4\theta \), assuming \( \cos 4\theta eq 0 \):\[ 2 \sin \theta = 1. \]This simplifies to:\[ \sin \theta = \frac{1}{2}. \]

Step 4: Solve for \( \theta \)

The equation \( \sin \theta = \frac{1}{2} \) is satisfied by \( \theta = \frac{\pi}{6} + 2k\pi \) or \( \theta = \frac{5\pi}{6} + 2k\pi \), where \( k \) is any integer. These solutions represent general solutions for \( \theta \) within the domain of the sine function.

Key Concepts

Sum-to-Product Identities

Sum-to-product identities are useful tools in trigonometry that help simplify expressions involving the sum or difference of sines or cosines. These identities convert such sums or differences into products, thereby making the equations easier to solve. In the given problem, we apply the sum-to-product identity for the difference of sines:

• The formula is: \( \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) \)
• This formula is particularly helpful when dealing with angles that are multiples or transformations of one another, like \( 5\theta \) and \( 3\theta \).

Using this identity, the expression \( \sin 5\theta - \sin 3\theta \) becomes \( 2\cos(4\theta)\sin(\theta) \). Thus, it simplifies the equation and brings it to a format where solving becomes more straightforward.
The same identity can help in many different trigonometric problems, streamlining complex calculations to products.

Sine Function

The sine function, one of the fundamental trigonometric functions, relates the angles of a triangle to the lengths of its sides in the unit circle representation. Therefore, it becomes crucial in solving trigonometric equations involving angles.

• It oscillates between -1 and 1 as the input, usually an angle measured in radians, goes from \( 0 \) to \( 2\pi \).
• Key angles, such as \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), provide simple and recognizable sine values, here being \( \frac{1}{2} \).

In this problem, after transforming and simplifying the equation using the sum-to-product identity, we are left with \( \sin \theta = \frac{1}{2} \).
This equation results in specific solutions based on the well-known angles of the sine function. Understanding these intrinsic properties of the sine function helps in quickly identifying solutions.

General Solutions in Trigonometry

General solutions in trigonometry allow us to find all possible solutions for a given trigonometric equation, taking into account the periodic nature of the functions involved. Trigonometric functions repeat their values in a predictable manner, most commonly every \( 2\pi \) radians.

• The general solution of \( \sin \theta = \frac{1}{2} \) is found by identifying primary angles \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).
• Write down each solution with \( +2k\pi \), where \( k \) is an integer, to account for the sine wave's periodicity:
  \( \theta = \frac{\pi}{6} + 2k\pi \),
  \( \theta = \frac{5\pi}{6} + 2k\pi \).

This approach ensures we capture all potential solutions along the trigonometric function's complete cycle.
General solutions are vital for comprehensive problem-solving in trigonometry, ensuring no possible solution is overlooked.