Testing a Polar Equation for Symmetry

Test the equation $r = 2 \sin θ$ for symmetry.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Test the equation for symmetry: $r = - 2 \cos θ .$

How did it go?

Test the equation $r = 2 \sin θ$ for symmetry.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Test the equation for symmetry: $r = - 2 \cos θ .$

How did it go?

1) Replacing $(r, θ)$ with $(- r, - θ)$ yields the same result. Thus, the graph is symmetric with respect to the line $θ = \frac{π}{2} .$	$\begin{array}{l} - r = 2 \sin (- θ) \\ - r = −2 \sin θ & Even-odd identity \\ r = 2 \sin θ & Multiply by −1 \\ Passed \end{array}$
2) Replacing $θ$ with $- θ$ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.	$\begin{array}{l} r = 2 \sin (- θ) \\ r = −2 \sin θ & Even-odd identity \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$
3) Replacing $r$ with $- r$ changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.	$\begin{array}{l} - r = 2 \sin θ \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$

1) Replacing $(r, θ)$ with $(- r, - θ)$ yields the same result. Thus, the graph is symmetric with respect to the line $θ = \frac{π}{2} .$	$\begin{array}{l} - r = 2 \sin (- θ) \\ - r = −2 \sin θ & Even-odd identity \\ r = 2 \sin θ & Multiply by −1 \\ Passed \end{array}$
2) Replacing $θ$ with $- θ$ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.	$\begin{array}{l} r = 2 \sin (- θ) \\ r = −2 \sin θ & Even-odd identity \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$
3) Replacing $r$ with $- r$ changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.	$\begin{array}{l} - r = 2 \sin θ \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$