Problem 35 - Algebra and Trigonometry

Question

Test for symmetry and then graph each polar equation. $$r=\cos \frac{\theta}{2}$$

Step-by-Step Solution

Answer

The polar equation \(r = \cos(\frac{\theta}{2})\) is symmetrical about the polar axis. However, it isn't symmetrical about the pole or the line \(\theta = \pi/2\). The graph can be obtained by plotting a few points derived from the equation and joining them smoothly.

Step 1: Testing for Symmetry About the Polar Axis

Replace \(\theta\) with \(-\theta\) in the original equation. If the altered equation is equivalent to the original equation, the equation is symmetrical about the polar axis. Substituting \(-\theta\) for \(\theta\) in the given polar equation, we get: \(r = \cos(\frac{-\theta}{2})\), which simplifies to \(r= \cos(\frac{\theta}{2})\), the same as the original, demonstrating that the graph is symmetrical about the polar axis.

Step 2: Testing for Symmetry About the Pole

Replace \(r\) with \(-r\) in the original equation. If the altered equation is equivalent to the original equation, the equation is symmetric about the pole. Substituting \(-r\) for \(r\) in the given polar equation, we get: \(-r = \cos(\frac{\theta}{2})\), which isn't the same as the original. Thus, the graph isn't symmetrical about the pole.

Step 3: Testing for Symmetry About the Line \(\theta = \pi/2\)

Replace \(\theta\) with \(\pi - \theta\) in the original equation. If the altered equation is equivalent to the original equation, the graph is symmetric about the line \(\theta = \pi/2\). Substituting \(\pi - \theta\) for \(\theta\) in the given polar equation, we get: \(r = \cos(\frac{\pi-\theta}{2})\), which simplifies to \(r = \cos(\frac{\pi}{2} - \frac{\theta}{2})\), which is not the same as the original. Thus, the graph isn't symmetrical about the line \(\theta = \pi/2\) as well.

Step 4: Graphing the Polar Equation

First, we create a table of values for the equation. Then we plot these values and connect the points smoothly. The graph will showcase the symmetry about the polar axis, but not symmetry about the pole or the line \(\theta = \pi/2\).