Problem 40 - Precalculus with Calculus Previews

Question

Find the points of intersection of the graphs of the given pair of polar equations. $$ r=\sin \theta, r=\sin 2 \theta $$

Step-by-Step Solution

Answer

Intersection points are: \((0, n\pi)\), \(\left(\frac{{\sqrt{3}}}{2}, \frac{\pi}{3} + 2n\pi\right)\), and \(-\left(\frac{{\sqrt{3}}}{2}, -\frac{\pi}{3} + 2n\pi\right)\)."

Step 1: Equating the Polar Equations

To find the points of intersection, we equate the given polar equations: \( ext{Equation 1: } r = ext{sin } \theta\)\( ext{Equation 2: } r = ext{sin } 2\theta\)Thus, we have:\( ext{sin } \theta = ext{sin } 2\theta\)

Step 2: Using Trigonometric Identity

We utilize the double-angle identity for sine: \( ext{sin } 2\theta = 2 ext{sin } \theta ext{cos } \theta\)Substitute it into the equation:\( ext{sin } \theta = 2 ext{sin } \theta ext{cos } \theta\)

Step 3: Solving the Trigonometric Equation

Rearrange the equation obtained:\( ext{sin } \theta - 2 ext{sin } \theta ext{cos } \theta = 0\)Factor out \( ext{sin } \theta\):\( ext{sin } \theta (1 - 2 ext{cos } \theta) = 0\)This results in two possible solutions:1. \( ext{sin } \theta = 0\)2. \(1 - 2 ext{cos } \theta = 0\)

Step 4: Finding Solutions for \(\text{sin } \theta = 0\)

Solve \( ext{sin } \theta = 0\), which indicates \(\theta = n\pi\), where \(n\) is an integer.

Step 5: Finding Solutions for \(1 - 2 \text{cos } \theta = 0\)

Solve \(1 - 2 \text{cos } \theta = 0\) by rearranging to find \( ext{cos } \theta = \frac{1}{2}\).This gives \(\theta = \frac{\pi}{3} + 2n\pi\) or \(\theta = -\frac{\pi}{3} + 2n\pi\), where \(n\) is an integer.

Step 6: Converting \(\theta\) to Polar Coordinates

Convert the \(\theta\) values into the original polar equations to find \(r\). 1. For \(\theta = n\pi\), \(r = ext{sin } n\pi = 0\). Intersection points are at \((0, n\pi)\).2. For \(\theta = \frac{\pi}{3} + 2n\pi\), \(r = \text{sin } \frac{\pi}{3} = \frac{{\sqrt{3}}}{2}\). Intersection points are at \(\left(\frac{{\sqrt{3}}}{2}, \frac{\pi}{3} + 2n\pi\right)\).3. For \(\theta = -\frac{\pi}{3} + 2n\pi\), \(r = \text{sin } -\frac{\pi}{3} = -\frac{{\sqrt{3}}}{2}\). Intersection points are at \(-\left(\frac{{\sqrt{3}}}{2}, -\frac{\pi}{3} + 2n\pi\right)\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics. They help simplify and solve equations involving angles and trigonometric functions like sine, cosine, and tangent.
Identities can transform expressions to make complex equations more manageable. In this exercise, we employ the double-angle identity for sine, which is:

• \( \sin 2\theta = 2 \sin \theta \cos \theta \)

This identity is crucial as it helps convert the polar equation \( r = \sin 2\theta \) into a form that's easier to compare directly with \( r = \sin \theta \).
Understanding these identities allows you to solve mathematical problems involving various trigonometric functions more effectively. Mastery of identities like the double-angle ones expands your problem-solving toolkit, letting you approach equations in unique and insightful ways.
By applying trigonometric identities in these exercises, students recognize patterns and relationships, ultimately making complex problems simpler to solve.

Solving Polar Equations

Solving polar equations involves finding the values of \( r \) and \( \theta \) that satisfy given conditions.
Polar coordinates represent points in a plane using a radius and angle rather than x-y coordinates.
This method requires substituting and manipulating equations to find where two or more polar graphs intersect. In our case, we started by equating the two polar equations:

• \( r = \sin \theta \)
• \( r = \sin 2\theta \)

Next, using the trigonometric identity, we replaced \( \sin 2\theta \) with \( 2 \sin \theta \cos \theta \).
This allowed us to rewrite the equations as:

• \( \sin \theta = 2 \sin \theta \cos \theta \)

The next step involves solving for \( \theta \). We set up the equation \( \sin \theta (1 - 2 \cos \theta) = 0 \), which unfolds into two scenarios:

• \( \sin \theta = 0 \)
• \( 1 - 2 \cos \theta = 0 \)

Hence, solving polar equations is about breaking the problem into smaller equations that are simpler to handle, making intersections easier to identify.

Points of Intersection

Finding points of intersection in polar coordinates requires determining where two or more graphs meet.
These intersections reveal common solutions between equations.
In our case, these are the points where \( r = \sin \theta \) and \( r = \sin 2\theta \) overlap.
Through our solution, we found:

• When \( \sin \theta = 0 \), it gives us angles \( \theta = n\pi \) with \( r = 0 \). Intersection points are \( (0, n\pi) \), representing points at the origin.
• For \( 1 - 2 \cos \theta = 0 \), solving \( \cos \theta = \frac{1}{2} \) results in \( \theta = \frac{\pi}{3} + 2n\pi \) and \( \theta = -\frac{\pi}{3} + 2n\pi \).

Converting \( \theta \) into polar coordinates reveals specific intersection points, such as:

• \( \left(\frac{\sqrt{3}}{2}, \frac{\pi}{3} + 2n\pi\right) \)
• \(-\left(\frac{\sqrt{3}}{2}, -\frac{\pi}{3} + 2n\pi\right) \)

These illustrate how the polar system displays intersections, especially when recognizing symmetry and periodicity in trigonometric functions.
Recognizing intersections in polar graphs improves understanding of how these equations depict specific geometric relationships.