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Polar Coordinates: Graphs

Test polar equations for symmetry.

Testing Polar Equations for Symmetry

Just as a rectangular equation such as \(y={x}^{2}\) describes the relationship between \(x\) and \(y\) on a Cartesian grid, a polar equation describes a relationship between \(r\) and \(\theta\) on a polar grid. Recall that the coordinate pair \((r,\theta )\) indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of \(\theta ,\) and extend a ray from the pole (origin) \(r\) units in the direction of \(\theta .\) All points that satisfy the polar equation are on the graph.

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 \(r)\) to determine the graph of a polar equation.

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

\[\begin{array}{lllll}r=2\sin \ \theta & \\ -r=2\sin (-\theta )\begin{array}{llll} & & & \end{array} & \text{Replace}\ (r,\theta )\ \text{with }(-r,-\theta ). \\ -r=-2\sin \ \theta & \text{Identity: }\sin (-\theta )=-\sin \ \theta . \\ r=2\sin \ \theta & \text{Multiply both sides by}-1.\end{array}\]

This equation exhibits symmetry with respect to the line \(\theta =\frac{\pi }{2}.\)

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

\[\begin{array}{lllll}r=1-2\cos \ \theta & \\ r=1-2\cos (-\theta )\begin{array}{llll} & & & \end{array} & \text{Replace }(r,\theta )\ \text{with}\ (r,-\theta ). \\ r=1-2\cos \ \theta & \text{Even/Odd identity}\end{array}\]

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 \((r,\theta )\) with \((-r,\theta )\) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation \(r=2\sin (3\theta ).\)

\[\begin{array}{l}r=2\sin (3\theta ) \\ -r=2\sin (3\theta )\end{array}\]

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of \(x\) and \(y\) values. To graph in the polar coordinate system we construct a table of \(\theta\) and \(r\) values. We enter values of \(\theta\) into a polar equation and calculate \(r.\) However, using the properties of symmetry and finding key values of \(\theta\) and \(r\) means fewer calculations will be needed.

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

Example

Try it.

Sketch the graph of \(r=4\cos \ \theta .\)

Solution

First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum \(|r|\) for \(r=4\cos \ \theta .\) First, set \(r=0,\) and solve for \(\theta\). Thus, a zero occurs at \(\theta =\frac{\pi }{2}\pm k\pi .\) A key point to plot is \((0,\text{}\text{}\frac{\pi }{2})\ .\)

To find the maximum value of \(r,\) note that the maximum value of the cosine function is 1 when \(\theta =0\pm 2k\pi .\) Substitute \(\theta =0\) into the equation:

\[\begin{array}{l}r=4\cos \ \theta \\ r=4\cos (0) \\ r=4(1)=4\end{array}\]

The maximum value of the equation is 4. A key point to plot is \((4,\ 0).\)

As \(r=4\cos \ \theta\) is symmetric with respect to the polar axis, we only need to calculate r-values for \(\theta\) over the interval \([0,\) \(\pi ].\) Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to . The graph is shown in .

\(\theta\)	0	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{3\pi }{4}\)	\(\frac{5\pi }{6}\)	\(\pi\)
\(r\)	4	3.46	2.83	2	0	−2	−2.83	−3.46	−4

Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.

Summary of Curves

We have explored a number of seemingly complex polar curves in this section. and summarize the graphs and equations for each of these curves.

Key Concepts

It is easier to graph polar equations if we can test the equations for symmetry with respect to the line \(\theta =\frac{\pi }{2},\) the polar axis, or the pole.
There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See .
Polar equations may be graphed by making a table of values for \(\theta\) and \(r.\)
The maximum value of a polar equation is found by substituting the value \(\theta\) that leads to the maximum value of the trigonometric expression.
The zeros of a polar equation are found by setting \(r=0\) and solving for \(\theta .\) See .
Some formulas that produce the graph of a circle in polar coordinates are given by \(r=a\cos \ \theta\) and \(r=a\sin \ \theta .\) See .
The formulas that produce the graphs of a cardioid are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta ,\) for \(a>0,\) \(b>0,\) and \(\frac{a}{b}=1.\) See .
The formulas that produce the graphs of a one-loop limaçon are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta\) for \(1<\frac{a}{b}<2.\) See .
The formulas that produce the graphs of an inner-loop limaçon are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta\) for \(a>0,\) \(b>0,\) and \(a
The formulas that produce the graphs of a lemniscates are given by \({r}^{2}={a}^{2}\cos \ 2\theta\) and \({r}^{2}={a}^{2}\sin \ 2\theta ,\) where \(a\ne 0.\) See .
The formulas that produce the graphs of rose curves are given by \(r=a\cos \ n\theta\) and \(r=a\sin \ n\theta ,\) where \(a\ne 0;\) if \(n\) is even, there are \(2n\) petals, and if \(n\) is odd, there are \(n\) petals. See and .
The formula that produces the graph of an Archimedes’ spiral is given by \(r=\theta ,\) \(\theta \ge 0.\) See .

Testing Polar Equations for Symmetry

This equation exhibits symmetry with respect to the line \(\theta =\frac{\pi }{2}.\)

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

\[\begin{array}{l}r=2\sin (3\theta ) \\ -r=2\sin (3\theta )\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Graphing Polar Equations by Plotting Points

Example

Try it.

Sketch the graph of \(r=4\cos \ \theta .\)

Solution

To find the maximum value of \(r,\) note that the maximum value of the cosine function is 1 when \(\theta =0\pm 2k\pi .\) Substitute \(\theta =0\) into the equation:

\[\begin{array}{l}r=4\cos \ \theta \\ r=4\cos (0) \\ r=4(1)=4\end{array}\]

The maximum value of the equation is 4. A key point to plot is \((4,\ 0).\)

\(\theta\)	0	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{3\pi }{4}\)	\(\frac{5\pi }{6}\)	\(\pi\)
\(r\)	4	3.46	2.83	2	0	−2	−2.83	−3.46	−4

Condensed — the full section is in OpenStax Precalculus 2e.

Summary of Curves

We have explored a number of seemingly complex polar curves in this section. and summarize the graphs and equations for each of these curves.

Key Concepts

It is easier to graph polar equations if we can test the equations for symmetry with respect to the line \(\theta =\frac{\pi }{2},\) the polar axis, or the pole.
There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See .
Polar equations may be graphed by making a table of values for \(\theta\) and \(r.\)
The maximum value of a polar equation is found by substituting the value \(\theta\) that leads to the maximum value of the trigonometric expression.
The zeros of a polar equation are found by setting \(r=0\) and solving for \(\theta .\) See .
Some formulas that produce the graph of a circle in polar coordinates are given by \(r=a\cos \ \theta\) and \(r=a\sin \ \theta .\) See .
The formulas that produce the graphs of a cardioid are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta ,\) for \(a>0,\) \(b>0,\) and \(\frac{a}{b}=1.\) See .
The formulas that produce the graphs of a one-loop limaçon are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta\) for \(1<\frac{a}{b}<2.\) See .
The formulas that produce the graphs of an inner-loop limaçon are given by \(r=a\pm b\cos \ \theta\) and \(r=a\pm b\sin \ \theta\) for \(a>0,\) \(b>0,\) and \(a
The formulas that produce the graphs of a lemniscates are given by \({r}^{2}={a}^{2}\cos \ 2\theta\) and \({r}^{2}={a}^{2}\sin \ 2\theta ,\) where \(a\ne 0.\) See .
The formulas that produce the graphs of rose curves are given by \(r=a\cos \ n\theta\) and \(r=a\sin \ n\theta ,\) where \(a\ne 0;\) if \(n\) is even, there are \(2n\) petals, and if \(n\) is odd, there are \(n\) petals. See and .
The formula that produces the graph of an Archimedes’ spiral is given by \(r=\theta ,\) \(\theta \ge 0.\) See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Test the equation \(r=2\sin \ \theta\) for symmetry.

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Test for each of the three types of symmetry.

1) Replacing \((r,\theta )\) with \((-r,-\theta )\) yields the same result. Thus, the graph is symmetric with respect to the line \(\theta =\frac{\pi }{2}.\)	\(\begin{array}{ll}-r=2\sin (-\theta ) & \\ -r=-2\sin \ \theta & \text{Even-odd identity} \\ r=2\sin \ \theta & \text{Multiply}\ \text{by}\ -1 \\ \text{Passed} & \end{array}\)
2) Replacing \(\theta\) with \(-\theta\) 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.	\(\begin{array}{ll}r=2\sin (-\theta ) & \\ r=-2\sin \ \theta & \text{Even-odd identity} \\ r=-2\sin \ \theta \ne 2\sin \ \theta & \\ \text{Failed} & \end{array}\)
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.	\(\begin{array}{l}-r=2\sin \ \theta \\ r=-2\sin \ \theta \ne 2\sin \ \theta \\ \text{Failed}\end{array}\)

Test the equation for symmetry: \(r=-2\cos \ \theta .\)

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The equation fails the symmetry test with respect to the line \(\theta =\frac{\pi }{2}\) and with respect to the pole. It passes the polar axis symmetry test.
Using the equation in , find the zeros and maximum \(|r|\) and, if necessary, the polar axis intercepts of \(r=2\sin \ \theta .\)

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To find the zeros, set \(r\) equal to zero and solve for \(\theta .\)
\[\begin{array}{ll}2\sin \ \theta =0 & \\ \sin \ \theta =0 & \\ \theta ={\sin }^{-1}0 & \\ \theta =n\pi & \text{where }n\text{ is an integer}\end{array}\]
Substitute any one of the \(\theta\) values into the equation. We will use \(0.\)
\[\begin{array}{l}\begin{array}{l} \\ r=2\sin (0)\end{array} \\ r=0\end{array}\]
The points \((0,0)\) and \((0,\pm n\pi )\) are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.
To find the maximum value of the equation, look at the maximum value of the trigonometric function \(\sin \ \theta ,\) which occurs when \(\theta =\frac{\pi }{2}\pm 2k\pi\) resulting in \(\sin (\frac{\pi }{2})=1.\) Substitute \(\frac{\pi }{2}\) for \(\theta .\)
\[\begin{array}{l}r=2\sin (\frac{\pi }{2}) \\ r=2(1) \\ r=2\end{array}\]
Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of \(|r|:\) \(r=3\cos \ \theta .\)

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Tests will reveal symmetry about the polar axis. The zero is \((0,\frac{\pi }{2}),\) and the maximum value is \((3,0).\)

Sketch the graph of \(r=4\cos \ \theta .\)

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To find the maximum value of \(r,\) note that the maximum value of the cosine function is 1 when \(\theta =0\pm 2k\pi .\) Substitute \(\theta =0\) into the equation:

\[\begin{array}{l}r=4\cos \ \theta \\ r=4\cos (0) \\ r=4(1)=4\end{array}\]

The maximum value of the equation is 4. A key point to plot is \((4,\ 0).\)

\(\theta\)	0	\(\frac{\pi }{6}\)	\(\frac{\pi }{4}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{3\pi }{4}\)	\(\frac{5\pi }{6}\)	\(\pi\)
\(r\)	4	3.46	2.83	2	0	−2	−2.83	−3.46	−4

Sketch the graph of \(r=2+2\cos \ \theta .\)

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First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting \(r=0,\) we have \(\theta =\pi +2k\pi .\) The zero of the equation is located at \((0,\pi ).\) The graph passes through this point.
The maximum value of \(r=2+2\cos \ \theta\) occurs when \(\cos \ \theta\) is a maximum, which is when \(\cos \ \theta =1\) or when \(\theta =0.\) Substitute \(\theta =0\) into the equation, and solve for \(r.\)
\[\begin{array}{l}\begin{array}{l} \\ r=2+2\cos (0)\end{array} \\ r=2+2(1)=4\end{array}\]
The point \((4,0)\) is the maximum value on the graph.
We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval \([0,\ \pi ].\) The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in , and then we plot the points and draw the graph. See .
\(\theta\) \(0\) \(\frac{\pi }{4}\) \(\frac{\pi }{2}\) \(\frac{2\pi }{3}\) \(\pi\)
\(r\) 4 3.41 2 1 0

Graph the equation \(r=4-3\sin \ \theta .\)

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First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line \(\theta =\frac{\pi }{2},\) yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality.

Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting \(r=0\) results in \(\theta\) being undefined. What does this mean? How could \(\theta\) be undefined? The angle \(\theta\) is undefined for any value of \(\sin \ \theta >1.\) Therefore, \(\theta\) is undefined because there is no value of \(\theta\) for which \(\sin \ \theta >1.\) Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating \(r\) when \(\theta =0.\)

\[\begin{array}{l}r(0)=4-3\sin (0) \\ r=4-3⋅0=4\end{array}\]

So, there is at least one polar axis intercept at \((4,0).\)

Next, as the maximum value of the sine function is 1 when \(\theta =\frac{\pi }{2},\) we will substitute \(\theta =\frac{\pi }{2}\) into the equation and solve for \(r.\) Thus, \(r=1.\)

Make a table of the coordinates similar to .

\(\theta\)	\(0\)	\(\frac{\pi }{6}\)	\(\frac{\pi }{3}\)	\(\frac{\pi }{2}\)	\(\frac{2\pi }{3}\)	\(\frac{5\pi }{6}\)	\(\pi\)	\(\frac{7\pi }{6}\)	\(\frac{4\pi }{3}\)	\(\frac{3\pi }{2}\)	\(\frac{5\pi }{3}\)	\(\frac{11\pi }{6}\)	\(2\pi\)
\(r\)	4	2.5	1.4	1	1.4	2.5	4	5.5	6.6	7	6.6	5.5	4

The graph is shown in .

Sketch the graph of \(r=3-2\cos \ \theta .\)

Sketch the graph of \(r=2+5\text{cos}\ \theta .\)

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Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when \(r=0,\) \(\theta =1.98.\) The maximum \(|r|\) is found when \(\cos \ \theta =1\) or when \(\theta =0.\) Thus, the maximum is found at the point (7, 0).

Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.

See .

\(\theta\)

\(0\)

\(\frac{\pi }{6}\)

\(\frac{\pi }{3}\)

\(\frac{\pi }{2}\)

\(\frac{2\pi }{3}\)

\(\frac{5\pi }{6}\)

\(\pi\)

\(\frac{7\pi }{6}\)

\(\frac{4\pi }{3}\)

\(\frac{3\pi }{2}\)

\(\frac{5\pi }{3}\)

\(\frac{11\pi }{6}\)

\(2\pi\)

\(r\)

6.3

4.5

−0.5

−2.3

−3

−2.3

−0.5

4.5

6.3

As expected, the values begin to repeat after \(\theta =\pi .\) The graph is shown in .

Sketch the graph of \({r}^{2}=4\cos \ 2\theta .\)

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The equation exhibits symmetry with respect to the line \(\theta =\frac{\pi }{2},\) the polar axis, and the pole.
Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution \(u=2\theta .\)
\[\begin{array}{ll}0=4\cos \ 2\theta & \\ 0=4\cos \ u & \\ 0=\cos \ u & \\ {\cos }^{-1}0=\frac{\pi }{2} & \\ u=\frac{\pi }{2} & \text{Substitute }2\theta \text{ back in for }u. \\ 2\theta =\frac{\pi }{2} & \\ \theta =\frac{\pi }{4} & \end{array}\]
So, the point \((0,\frac{\pi }{4})\) is a zero of the equation.
Now let’s find the maximum value. Since the maximum of \(\cos \ u=1\) when \(u=0,\) the maximum \(\cos \ 2\theta =1\) when \(2\theta =0.\) Thus,
\[\begin{array}{l}{r}^{2}=4\cos (0) \\ {r}^{2}=4(1)=4 \\ r=\pm \sqrt{4}\ \pm 2\end{array}\]
We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line \(\theta =\frac{\pi }{2},\) and the polar axis, we only need to plot points in the first quadrant.
Make a table similar to .
\(\theta\) 0 \(\frac{\pi }{6}\) \(\frac{\pi }{4}\)
\(r\) \(\pm 2\) \(\pm \sqrt{2}\) 0
Plot the points on the graph, such as the one shown in .
Sketch the graph of \(r=2\cos \ 4\theta .\)

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Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line \(\theta =\frac{\pi }{2}\) and the pole.
Now we will find the zeros. First make the substitution \(u=4\theta .\)
\[\begin{array}{l}0=2\cos \ 4\theta \\ 0=\cos \ 4\theta \\ 0=\cos \ u \\ {\cos }^{-1}0=u \\ u=\frac{\pi }{2} \\ 4\theta =\frac{\pi }{2} \\ \theta =\frac{\pi }{8}\end{array}\]
The zero is \(\theta =\frac{\pi }{8}.\) The point \((0,\frac{\pi }{8})\) is on the curve.
Next, we find the maximum \(|r|.\) We know that the maximum value of \(\cos \ u=1\) when \(\theta =0.\) Thus,
\[\begin{array}{l}r=2\cos (4⋅0) \\ r=2\cos (0) \\ r=2(1)=2\end{array}\]
The point \((2,0)\) is on the curve.
The graph of the rose curve has unique properties, which are revealed in .
\(\theta\) 0 \(\frac{\pi }{8}\) \(\frac{\pi }{4}\) \(\frac{3\pi }{8}\) \(\frac{\pi }{2}\) \(\frac{5\pi }{8}\) \(\frac{3\pi }{4}\)
\(r\) 2 0 −2 0 2 0 −2
As \(r=0\) when \(\theta =\frac{\pi }{8},\) it makes sense to divide values in the table by \(\frac{\pi }{8}\) units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at \(r=0,\) each petal extends out a distance of \(r=2,\) and then turns back to zero \(2n\) times for a total of eight petals. See the graph in .
Sketch the graph of \(r=4\sin (2\theta ).\)

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The graph is a rose curve, \(n\) even
Sketch the graph of \(r=2\sin (5\theta ).\)

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The graph of the equation shows symmetry with respect to the line \(\theta =\frac{\pi }{2}.\) Next, find the zeros and maximum. We will want to make the substitution \(u=5\theta .\)
\[\begin{array}{l}0=2\sin (5\theta ) \\ 0=\sin \ u \\ {\sin }^{-1}0=0 \\ u=0 \\ 5\theta =0 \\ \theta =0\end{array}\]
The maximum value is calculated at the angle where \(\sin \ \theta\) is a maximum. Therefore,
\[\begin{array}{l}\begin{array}{l} \\ r=2\sin (5⋅\frac{\pi }{2})\end{array} \\ r=2(1)=2\end{array}\]
Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for \(n\) odd yields the same number of petals as \(n,\) there will be five petals on the graph. See .
Create a table of values similar to .
\(\theta\) 0 \(\frac{\pi }{6}\) \(\frac{\pi }{3}\) \(\frac{\pi }{2}\) \(\frac{2\pi }{3}\) \(\frac{5\pi }{6}\) \(\pi\)
\(r\) 0 1 −1.73 2 −1.73 1 0
Sketch the graph of \(r=3\cos (3\theta ).\)

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Rose curve, \(n\) odd
Sketch the graph of \(r=\theta\) over \([0,2\pi ].\)

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As \(r\) is equal to \(\theta ,\) the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.
Create a table such as .
\(\theta\) \(\frac{\pi }{4}\) \(\frac{\pi }{2}\) \(\pi\) \(\frac{3\pi }{2}\) \(\frac{7\pi }{4}\) \(2\pi\)
\(r\) 0.785 1.57 3.14 4.71 5.50 6.28
Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in .
Sketch the graph of \(r=-\theta\) over the interval \([0,4\pi ].\)
Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane.

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Symmetry with respect to the polar axis is similar to symmetry about the \(x\)-axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line \(\theta =\frac{\pi }{2}\) is similar to symmetry about the \(y\)-axis.
Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the x-axis, y-axis, and origin?
What are the steps to follow when graphing polar equations?

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Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at \(\theta =0,\ \frac{\pi }{2},\) \(\pi\) and \(\frac{3\pi }{2},\) and sketch the graph.
Describe the shapes of the graphs of cardioids, limaçons, and lemniscates.
What part of the equation determines the shape of the graph of a polar equation?

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The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.
\(r=5\cos \ 3\theta\)
\(r=3-3\cos \ \theta\)

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symmetric with respect to the polar axis
\(r=3+2\sin \ \theta\)
\(r=3\sin \ 2\theta\)

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symmetric with respect to the polar axis, symmetric with respect to the line \(\theta =\frac{\pi }{2},\) symmetric with respect to the pole
\(r=2\theta\)

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symmetric with respect to the line \(\theta =\frac{\pi }{2}\)
\(r=4\cos \ \frac{\theta }{2}\)
\(r=\frac{2}{\theta }\)

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Symmetric with respect to line \(\theta =\frac{\pi }{2}\) (y-axis)
\(r=3\sqrt{1-{\cos }^{2}\theta }\)
\(r=\sqrt{5\sin \ 2\theta }\)

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symmetric with respect to the pole
\(r=3\cos \ \theta\)
\(r=4\sin \ \theta\)

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circle
\(r=2+2\cos \ \theta\)
\(r=2-2\cos \ \theta\)

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cardioid
\(r=5-5\sin \ \theta\)
\(r=3+3\sin \ \theta\)

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cardioid
\(r=3+2\sin \ \theta\)
\(r=7+4\sin \ \theta\)

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one-loop/dimpled limaçon
\(r=4+3\cos \ \theta\)
\(r=5+4\cos \ \theta\)

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one-loop/dimpled limaçon

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Polar Coordinates: Graphs

Test polar equations for symmetry.
Graph polar equations by plotting points.
Substitute the appropriate combination of components for
If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.
Check equation for the three types of symmetry.
Find the zeros. Set
Find the maximum value of the equation according to the maximum value of the trigonometric expression.
Make a table of values for

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

Try your own

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

More in Trigonometry

The unit circle Trigonometric equations Trigonometric identities Degrees and radians Right-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosines Graphs of sine, cosine and tangent Inverse trigonometric functions

\(\theta\)	0	\(\frac{\pi }{8}\)	\(\frac{\pi }{4}\)	\(\frac{3\pi }{8}\)	\(\frac{\pi }{2}\)	\(\frac{5\pi }{8}\)	\(\frac{3\pi }{4}\)
\(r\)	2	0	−2	0	2	0	−2