Why Test a Polar Equation for Symmetry?
A polar equation describes a set of points, but its algebra can also make a geometric symmetry visible. A substitution such as \(\theta\mapsto-\theta\) may leave the equation unchanged, showing that reflecting its points across the polar axis preserves the curve.
Three symmetry tests are especially useful: symmetry about the polar axis, symmetry about the line \(\theta=\pi/2\), and symmetry about the pole. Each test uses a different coordinate transformation. The key is to match the transformation to the symmetry being tested, then decide whether the resulting equation describes the same set of points as the original.
The Three Substitutions
In polar coordinates, a point with coordinates \((r,\theta)\) has Cartesian coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\). The following transformations describe the three symmetries in the tests. The radius need not be positive: the transformations apply to signed polar coordinates as well.
| Symmetry being tested | Coordinate transformation | Substitution in the equation |
|---|---|---|
| Polar axis (the horizontal axis) | \((r,\theta)\mapsto(r,-\theta)\) | Replace \(\theta\) by \(-\theta\) |
| Line \(\theta=\pi/2\) (the vertical axis) | \((r,\theta)\mapsto(r,\pi-\theta)\) | Replace \(\theta\) by \(\pi-\theta\) |
| Pole | \((r,\theta)\mapsto(r,\theta+\pi)\), or equivalently \((r,\theta)\mapsto(-r,\theta)\) | Replace \(\theta\) by \(\theta+\pi\), or replace \(r\) by \(-r\) |
For a test to confirm symmetry, the equation after substitution must describe the same curve as the original equation. It does not have to look identical if algebraic rearrangement shows that the equations are equivalent. This applies both to equations written as \(r=f(\theta)\) and to implicit equations involving \(r\) and \(\theta\).
The transformations follow from the Cartesian coordinates. Reflection across the horizontal axis keeps \(x\) fixed and changes the sign of \(y\), which corresponds to changing \(\theta\) to \(-\theta\). Reflection across the vertical axis changes the sign of \(x\) but keeps \(y\) fixed, corresponding to \(\pi-\theta\). A half-turn about the pole changes both coordinate signs; keeping \(r\) and adding \(\pi\) to the angle has exactly that effect. Equivalently, keeping the angle and changing the sign of \(r\) reverses both coordinates.
Worked Example: Testing a Polar-Axis Symmetry
Worked Example: Testing a Polar-Axis Symmetry
Test the curve \(r=1+\cos\theta\) for symmetry about the polar axis, the line \(\theta=\pi/2\), and the pole.
Polar axis. Substitute \(-\theta\) for \(\theta\):
Since \(\cos(-\theta)=\cos\theta\), the equation is unchanged. The curve is symmetric about the polar axis.
Line \(\theta=\pi/2\). Substitute \(\pi-\theta\):
This is not the original equation \(r=1+\cos\theta\). To check that the symmetry really fails, take \(\theta=0\). The original equation gives \(r=1+\cos 0=2\), so \((2,0)\) is on the curve. Its reflection across the vertical axis is \((2,\pi)\), but at \(\theta=\pi\) the equation requires \(r=1+\cos\pi=0\), not \(2\). The reflected point is not on the curve.
Pole. Substitute \(\theta+\pi\):
This differs from the original equation. The point \((2,0)\) used above rotates about the pole to \((2,\pi)\), which is not on the curve, so the curve is not symmetric about the pole.
Check: At \(\theta=0\), the original relation gives \(r=2\). At the transformed angle \(\pi\), it gives \(r=0\). Thus the point and its vertical reflection or pole-rotated image do not both satisfy the equation.
Worked Example: Symmetry About the Vertical Axis
Worked Example: Symmetry About the Vertical Axis
Test \(r=2\sin\theta\) for the same three symmetries.
Polar axis. Substitute \(-\theta\):
The result differs from \(r=2\sin\theta\). To verify that symmetry about the polar axis fails, use \(\theta=\pi/6\). The original equation gives \(r=2\sin(\pi/6)=1\). Reflection across the polar axis gives \((1,-\pi/6)\), where the equation would require \(r=2\sin(-\pi/6)=-1\), not \(1\).
Vertical axis. Substitute \(\pi-\theta\):
The equation is unchanged, so the curve is symmetric about the line \(\theta=\pi/2\).
Pole. Substitute \(\theta+\pi\):
This differs from the original. The point \((1,\pi/6)\) on the curve rotates to \((1,7\pi/6)\). At the rotated angle, the equation requires \(r=2\sin(7\pi/6)=-1\), so the point with radius \(1\) is not on the curve. Pole symmetry fails.
Check: The substitutions for the vertical axis reproduce the original equation, while the polar-axis and pole transformations give a different required radius at the tested point.
Worked Example: All Three Symmetries
Worked Example: All Three Symmetries
Test the implicit polar equation \(r^2=4\cos(2\theta)\) for the three symmetries.
Polar axis. Replace \(\theta\) by \(-\theta\):
Because cosine is even, the equation is unchanged.
Vertical axis. Replace \(\theta\) by \(\pi-\theta\):
Cosine has period \(2\pi\), so \(\cos(2\pi-2\theta)=\cos(2\theta)\). The equation is unchanged.
Pole. Replace \(\theta\) by \(\theta+\pi\):
Again, the equation is unchanged because cosine has period \(2\pi\). The curve passes all three tests.
Check: For example, at \(\theta=\pi/6\), the equation gives \(r^2=4\cos(\pi/3)=2\). The reflected angles are \(-\pi/6\) and \(5\pi/6\), and the pole-rotated angle is \(7\pi/6\). At each angle, the double angle has cosine \(1/2\), so the equation still gives \(r^2=2\).
Why a Failed Substitution Needs Care
Passing a substitution test is strong evidence of symmetry: the transformed relation is the original relation, so points that satisfy it remain on the curve after the corresponding transformation. But a changed-looking equation is not, by itself, proof that symmetry is absent. Two different-looking equations may still describe the same set of points after simplification or use of polar coordinate aliases.
In particular, polar coordinates are not unique. Angles differing by \(2\pi\) represent the same direction, and a negative radius can represent the same point as a positive radius with the angle shifted by \(\pi\). When a substitution does not leave an equation visibly unchanged, compare the equations carefully. If needed, test a specific point and its reflected or rotated image, allowing for equivalent polar coordinates.
Common Mistake: Using the Wrong Reflection
A frequent error is to use \(\theta\mapsto-\theta\) when testing reflection across the vertical axis. That substitution tests reflection across the polar axis instead. For example, \(r=2\cos\theta\) is unchanged under \(\theta\mapsto-\theta\), since \(2\cos(-\theta)=2\cos\theta\). That establishes symmetry about the polar axis, not about the vertical axis.
For the vertical-axis test, use \(\theta\mapsto\pi-\theta\):
This is not the original relation. A direct check confirms the distinction: \((2,0)\) satisfies the original equation, but its reflection across the vertical axis, \((2,\pi)\), would require \(2=2\cos\pi=-2\). The reflected point is not on the curve.
A Reliable Testing Routine
Decide whether the target is the polar axis, the line \(\theta=\pi/2\), or the pole.
Use \(-\theta\), \(\pi-\theta\), or \(\theta+\pi\) (equivalently, \(r\mapsto-r\)) as appropriate.
Check whether the transformed relation is equivalent to the original, not merely whether it has the same visual form.
Substitute a point on the curve and test whether its reflected or rotated image also satisfies the equation.
These tests help identify structural symmetry without plotting every point. They are tests of the equation's invariance under a specific transformation; they do not, on their own, provide a complete sketch or classify the curve.
Check Your Understanding
For each question, state the substitution you would use and explain what the result would establish.
- Which substitution tests a polar equation for symmetry about the polar axis?
- Which substitution tests symmetry about the line \(\theta=\pi/2\)?
- Give the two equivalent substitutions used to test symmetry about the pole.
- For \(r=3+\sin\theta\), what equation results from replacing \(\theta\) by \(-\theta\)?
- Why does a substitution that changes the written equation not always prove that the curve lacks the symmetry?
