Problem 50 - Precalculus Mathematics for Calculus

Question

Use a Double- or Half-Angle Formula to solve the equation in the interval \([0,2 \pi)\) $$\sin 3 \theta-\sin 6 \theta=0$$

Step-by-Step Solution

Answer

Solutions are \(\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{\pi}{9}, \frac{7\pi}{9}, \frac{5\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}\).

Step 1: Use the Double Angle Formula

We start by using the double angle identity for sine: \( \sin 6\theta = 2 \sin 3\theta \cos 3\theta \). This simplifies the equation to \( \sin 3\theta - 2 \sin 3\theta \cos 3\theta = 0 \).

Step 2: Factor Out Common Terms

Factor out \(\sin 3\theta\) from the equation: \( \sin 3\theta (1 - 2\cos 3\theta) = 0 \). This results in two separate equations: \(\sin 3\theta = 0\) and \(1 - 2\cos 3\theta = 0\).

Step 3: Solve \(\sin 3\theta = 0\)

To solve \(\sin 3\theta = 0\), we find the general solutions for the sine function: \(3\theta = n \pi\) where \(n\) is an integer. This simplifies to \(\theta = \frac{n \pi}{3}\).

Step 4: Solve \(1 - 2\cos 3\theta = 0\)

For the second equation, solve for \(\cos 3\theta = \frac{1}{2} \). The general solutions for \(\cos 3\theta = \frac{1}{2}\) are \(3\theta = \frac{\pi}{3} + 2k\pi\) and \(3\theta = -\frac{\pi}{3} + 2k\pi\) where \(k\) is an integer. This results in \(\theta = \frac{\pi}{9} + \frac{2k\pi}{3}\) and \(\theta = -\frac{\pi}{9} + \frac{2k\pi}{3}\).

Step 5: Find Solutions in the Given Interval

For \(\sin 3\theta = 0\), the solutions in \([0,2\pi)\) are \(\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}\). For \(\cos 3\theta = \frac{1}{2}\), find values of \(k\) such that \(\theta\) within \(\theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{5\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}\).

Key Concepts

Double Angle Formula

The double angle formula is an essential tool in trigonometry. It helps simplify equations by expressing trigonometric functions of double angles in terms of single angles. For sine, the formula is given by \[ \sin 2\theta = 2 \sin \theta \cos \theta \] In our exercise, this formula was similarly adapted for our need: \[ \sin 6\theta = 2 \sin 3\theta \cos 3\theta \]

• This transformation helps break down complex trigonometric expressions into manageable parts.
• By converting multiples of angles, the problem solving becomes much simpler.

Understanding and applying the double angle formula allows us to simplify intricate expressions, making factors easier to isolate and solve.

Half-Angle Identities

Half-angle identities play a vital role just like their double angle counterparts by providing another way to express trigonometric functions. Although not directly used in the provided exercise, they often pair with double angle formulas to simplify problems further. For example:

• \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \)
• \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \)

When solving complex trigonometric equations, these identities can reduce the angle, helping find solutions or roots where double angle formulas appear initially cumbersome. Though it's not applied here, understanding half-angle identities reinforces how trigonometric functions relate at various scaled angles.

Solution Interval

The term solution interval refers to the range in which we find answers for given equations. Here, the interval is \[ [0, 2\pi) \] This basically means that we are finding solutions that fit within one complete cycle of a sine or cosine wave:

• The notation \([0, 2\pi)\) indicates that all angles from 0 up to but not including 2\pi are considered.
• For periodic functions like sine and cosine, this is the typical domain for one complete rotation on the unit circle.

By examining this interval, we can ensure our solutions are valid and represent all possible angles within one full trigonometric cycle.

Factoring Trigonometric Expressions

Factoring trigonometric expressions is similar to factoring in algebra. It helps in breaking down complex expressions into products simpler to solve. Here, we factored out common terms: \[ \sin 3\theta (1 - 2 \cos 3\theta) = 0 \]

• This step reduces the original expression into a multiplication of simpler terms: \( \sin 3\theta = 0 \) and \( 1 - 2 \cos 3\theta = 0 \).
• Each factor can be set equal to zero, leading us to the solutions for \( \theta \).

Factoring is crucial here, especially when trigonometric functions are involved, as it can separate the equation into solvable units, helping identify all possible solutions.