Question

can a polar graph exhibit symmetry, even if it fails algebraic symmetry test? If so, give an example.

Answer 1

Yes, polar graph exhibit symmetry even it fails algebraic symmetry test. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of

\displaystyle rr)
to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line \displaystyle \theta =\frac{\pi }{2}θ=​2​​π​​ (y-axis). We replace \displaystyle \left(r,\theta \right)(r,θ) with \displaystyle \left(-r,-\theta \right)(−r,−θ) to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation \displaystyle r=2\sin \thetar=2sinθ;

r=2sinθ−r=2sin(−θ)Replace(r,θ)with (−r,−θ).−r=−2sinθIdentity: sin(−θ)=−sinθ.r=2sinθMultiply both sides by−1.r=2sin⁡θ−r=2sin⁡(−θ)Replace(r,θ)with (−r,−θ).−r=−2sin⁡θIdentity: sin⁡(−θ)=−sin⁡θ.r=2sin⁡θMultiply both sides by−1.

This equation exhibits symmetry with respect to the line \displaystyle \theta =\frac{\pi }{2}θ=​2​​π​​.

In the second test, we consider symmetry with respect to the polar axis ( \displaystyle xx -axis). We replace \displaystyle \left(r,\theta \right)(r,θ) with \displaystyle \left(r,-\theta \right)(r,−θ) or \displaystyle \left(-r,\pi -\theta \right)(−r,π−θ) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation \displaystyle r=1 - 2\cos \thetar=1−2cosθ.

r=1−2cosθr=1−2cos(−θ)Replace (r,θ)with(r,−θ).r=1−2cosθEven/Odd identityr=1−2cos⁡θr=1−2cos⁡(−θ)Replace (r,θ)with(r,−θ).r=1−2cos⁡θEven/Odd identity

The graph of this equation exhibits symmetry with respect to the polar axis.

In the third test, we consider symmetry with respect to the pole (origin). We replace \displaystyle \left(r,\theta \right)(r,θ) with \displaystyle \left(-r,\theta \right)(−r,θ) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation \displaystyle r=2\sin \left(3\theta \right)r=2sin(3θ).

\displaystyle \begin{array}{c}r=2\sin \left(3\theta \right)\\ -r=2\sin \left(3\theta \right)\end{array}​r=2sin(3θ)​−r=2sin(3θ)​​

The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line \displaystyle \theta =\frac{\pi }{2}θ=​2​​π​​, the polar axis, or the pole.