Problem 38 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$\sin \theta-\sin 2 \theta=0$$

Step-by-Step Solution

Answer

In degrees: \( 0, 60, 180, 300 \). In radians: \( 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).

Step 1: Write the equation in terms of a single trigonometric function

We need to express \( \sin(2\theta) \) in terms of \( \sin(\theta) \). Use the double-angle identity for sine: \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]},

Step 2: Substitute the double-angle identity into the original equation

Replace \( \sin(2\theta) \) in the equation with \( 2\sin(\theta)\cos(\theta) \). The equation becomes: \[ \sin(\theta) - 2\sin(\theta)\cos(\theta) = 0 \]

Step 3: Factorize the equation

Factor out \( \sin(\theta) \) from the equation: \[ \sin(\theta)(1 - 2\cos(\theta)) = 0 \]. For this product to be zero, either \( \sin(\theta) = 0 \) or \( 1 - 2\cos(\theta) = 0 \) must hold.

Step 4: Solve each factor separately

First, solve \( \sin(\theta) = 0 \):- \( \theta = 0 + 180k \) where \( k \) is an integer.Next, solve \( 1 - 2\cos(\theta) = 0 \):- \( 2\cos(\theta) = 1 \) leads to \( \cos(\theta) = \frac{1}{2} \).- \( \theta = 60 + 360k \) or \( \theta = 300 + 360k \) where \( k \) is an integer.

Step 5: Determine the non-negative solutions

Consider the range \( 0 \leq \theta < 360\) degrees:- From \( \sin(\theta) = 0 \): \( \theta = 0, 180 \).- From \( \cos(\theta) = \frac{1}{2} \): \( \theta = 60, 300 \).So, the non-negative solutions in degrees are \( \theta = 0, 60, 180, 300 \).

Step 6: Convert the solutions to radians

Convert the angles in degrees to radians:- \( 0^\circ = 0 \text{ radians} \)- \( 60^\circ = \frac{\pi}{3} \text{ radians} \)- \( 180^\circ = \pi \text{ radians} \)- \( 300^\circ = \frac{5\pi}{3} \text{ radians} \)Hence, the solutions in radians are \( 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).

Key Concepts

Double-Angle Identity

The double-angle identity is pivotal in trigonometry. It helps express angles that are multiples of a basic angle, such as two times an angle. Specifically, the double-angle identity for sine is \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \].
To solve the original trigonometric equation \( \sin \theta - \sin 2\theta = 0 \), we first replace \( \sin(2\theta) \) with its identity form.
This substitution reduces the problem to dealing with simpler trigonometric functions of \( \theta \).
Using identities in this manner simplifies solving complex equations and is fundamental when working with trigonometric expressions in homework or exams. By recognizing and applying the double-angle identity, you are able to transform and simplify trigonometric equations. This is especially useful when aiming to solve them effectively.

Factoring

Factoring is a method used to simplify expressions and solve equations. It involves rewriting an expression as a product of its factors. In trigonometric equations, it often helps find solutions by breaking them into simpler parts.
In the context of our example equation \( \sin(\theta) - 2\sin(\theta)\cos(\theta) = 0 \), factoring out the common term \( \sin(\theta) \) yields \[ \sin(\theta)(1 - 2\cos(\theta)) = 0 \]. This tells us that the equation is satisfied if either \( \sin(\theta) = 0 \) or \( 1 - 2\cos(\theta) = 0 \) is true. This technique is extremely effective and commonly used in solving polynomial and trigonometric equations.

• Look for common factors in each term.
• Factor them out and simplify.

This breaks down complex expressions into manageable pieces, enabling easier solving of equations.

Sine and Cosine Functions

Sine and cosine functions are fundamental in trigonometry. They describe the relationships between angles and side lengths in right triangles and are periodic functions that repeat their values in regular intervals.
The sine function \( \sin(\theta) \) takes values between -1 and 1, reaching zero at \( \theta = 0, \pi, 2\pi,\) and so on. In our problem, solving \( \sin(\theta) = 0 \) gives immediate solutions: \( \theta = 0, \pi, 2\pi \ldots \).Cosine, on the other hand, is similar yet shifted, with the cosine function reaching zero at different intervals.

• For sine: Zero where sine's graph cuts the horizontal axis.
• For cosine: Zero where the graph of \( \cos(\theta) \) cuts the horizontal line at zero points.

Knowing these values helps us when solving equations involving these functions.

Angle Conversion

Converting angles between degrees and radians is a vital skill in trigonometry. Degrees measure angles based on dividing a circle into 360 equal parts, while radians measure angles using the radius of the circle.
To convert from degrees to radians, multiply by \( \frac{\pi}{180} \). This is crucial because many mathematical functions in calculus use radians.

• Degrees to radians: multiply by \( \frac{\pi}{180} \).
• Radians to degrees: multiply by \( \frac{180}{\pi} \).

Using our example, the solutions in degrees \(0, 60, 180, 300\) become \(0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}\) in radians. This conversion ensures solutions are in the form needed for various applications in mathematics and physics.