This section covers:
 Plotting Points Using Polar Coordinates
 PolarRectangular Point Conversions
 Drawing Polar Graphs
 Converting Equations from Polar to Rectangular
 Converting Equations from Rectangular to Polar
 Polar Graph Points of Intersection
 More Practice
Note that we talk about converting back and forth from Polar Complex Form to Rectangular Complex form here in the Trigonometry and the Complex Plane section.
So far, we’ve plotted points using rectangular (or Cartesian) coordinates, since the points since we are going back and forth \(x\) units, and up and down \(y\) units.
Plotting Points Using Polar Coordinates
In the Polar Coordinate System, we go around the origin or the pole a certain distance out, and a certain angle from the positive \(x\)–axis:
The ordered pairs, called polar coordinates, are in the form \(\left( {r,\theta } \right)\), with \(r\) being the number of units from the origin or pole (if \(r>0\)), like a radius of a circle, and \(\theta \) being the angle (in degrees or radians) formed by the ray on the positive \(x\)–axis (polar axis), going counterclockwise. If \(r<0\), the point is units (like a radius) in the opposite direction (across the origin or pole) of the angle \(\theta \). If \(\theta <0\), you go clockwise with the angle, starting with the positive \(x\)–axis.
To plot a point, you typically circle around the positive \(x\)–axis \(\theta \) degrees first, and then go out from the origin or pole \(r\) units (if \(r\) is negative, go the other way (180°) \(r\) units).
Here a polar graph with some points on it. Note that we typically count in increments of 15°, or \(\displaystyle \frac{\pi }{{12}}\).
For a point \(\left( {r,\theta } \right)\), do you see how you always go counterclockwise (or clockwise, if you have a negative angle) until you reach the angle you want, and then out from the center \(r\) units, if \(r\) is positive? If \(r\) is negative, you go in the opposite direction from the angle \(r\) units. If both \(r\) and the angle \(\theta \) are negative, you have to make sure you go clockwise to get the angle, but in the opposite direction \(r\) units.
You may be asked to rename a point in several different ways, for example, between \(\left[ {2\pi ,2\pi } \right)\) or \(\left[ {360{}^\circ ,360{}^\circ } \right)\). For example, if we wanted to rename the point \(\left( {6,240{}^\circ } \right)\) three other different ways between \(\left[ {360{}^\circ ,360{}^\circ } \right)\), by looking at the graph above, we’d get \(\left( {6,60{}^\circ } \right)\)(make \(r\) negative and subtract 180°), \(\left( {6,120{}^\circ } \right)\) (subtract 360°), and \(\left( {6,240{}^\circ } \right)\) (make both negative). (Remember that 240° and –120°, and 60° and –240° are coterminal angles). To get these, if the first number (\(r\)) is negative, you want to go in the opposite direction, and if the angle is negative, you want to go clockwise instead of counterclockwise from the positive \(x\)–axis.
PolarRectangular Point Conversions
You will probably be asked to convert coordinates between polar form and rectangular form.
Converting from Polar to Rectangular Coordinates
Let’s first convert from polar to rectangular form; to do this we use the following formulas, as we can see this from the graph:
This conversion is pretty straightforward:
\(\begin{array}{l}x\,\,=\,\,r\,\cos \,\theta \\y\,\,=\,\,r\,\sin \,\theta \end{array}\)
Converting from Rectangular to Polar Coordinates
Converting from rectangular coordinates to polar coordinates can be a little trickier since we need to check the quadrant of the rectangular point to get the correct angle; the quadrants must match. Here are the formulas:
\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\text{(this will be positive)}\)
\(\displaystyle \theta ={{\tan }^{{1}}}\left( {\frac{y}{x}} \right)\,\,\,\,\,\,\text{(check for correct quadrant)}\)
One word of caution: If \(x=0\), we’ll get an error when trying to obtain \(\theta \). In these cases, graph the point (it will be on the \(y\)axis) to get the angle.
Again, we need to check quadrants when using the calculator to get \({{\tan }^{{1}}}\). We’ll have to add the following degrees or radians when the point is in the following quadrants (this is because the \({{\tan }^{{1}}}\) function on the calculator only gives answers back in the interval \(\displaystyle \left( {\frac{\pi }{2},\,\,\frac{\pi }{2}} \right)\)):
Note that there can be multiple “answers” when converting from rectangular to polar, since polar points can be represented in many different ways (coterminal angles, positive or negative “\(r\)”, and so on). Thus, it is typically easier to convert from polar to rectangular.
Examples
Here are some examples of conversions both way; note you may be asked to convert back to polar into degrees or radians. For converting back to polar, make sure answers are either between 0 and 360° for degrees or 0 to \(2\pi \) for radians. (And again, note that when we convert back to polar coordinates, we may not always get the same representation as the polar point we started out with.)
Polar Coordinate  Convert from Polar to Rectangular  Position  Convert from Rectangular to Polar 
\(\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)\)  \(\displaystyle x=2\cos \frac{{4\pi }}{3}=2\left( {\frac{1}{2}} \right)=1\)
\(\displaystyle y=2\sin \frac{{4\pi }}{3}=2\left( {\frac{{\sqrt{3}}}{2}} \right)=\sqrt{3}\)
\(\left( {1,\sqrt{3}} \right)\) 
\(r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+3}}=2\)
\(\displaystyle \theta ={{\tan }^{{1}}}\left( {\frac{{\sqrt{3}}}{{1}}} \right)=\frac{{4\pi }}{3}\text{ (3rd quadrant)}\) \(\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)\) 

\(\left( {4,\pi } \right)\)  \(\begin{array}{c}x=4\cos \pi =4\left( {1} \right)=4\\y=4\sin \pi =4\left( 0 \right)=0\end{array}\)
\(\left( {4,0} \right)\) 
\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{16+0}}=4\)
\(\displaystyle \begin{array}{c}\theta ={{\tan }^{{1}}}\left( {\frac{0}{{4}}} \right)=0\text{ (between 2nd }\\\text{ and 3rd quadrants)}\\\left( {4,\pi } \right)\end{array}\) 

\(\left( {7,45{}^\circ } \right)\) 
\(\displaystyle x=7\cos \left( {45{}^\circ } \right)=\frac{{7\sqrt{2}}}{2}\) \(\displaystyle y=7\sin \left( {45{}^\circ } \right)=\frac{{7\sqrt{2}}}{2}\)
\(\displaystyle \left( {\frac{{7\sqrt{2}}}{2},\frac{{7\sqrt{2}}}{2}} \right)\) 
\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{49\left( 2 \right)}}{4}+\frac{{49\left( 2 \right)}}{4}}}=7\)
\(\theta ={{\tan }^{{1}}}\left( 1 \right)=225{}^\circ \text{ (3rd quadrant)}\) \(\left( {7,225{}^\circ } \right),\text{ which is the same as }\left( {7,45{}^\circ } \right)\) 

\(\left( {5,150{}^\circ } \right)\)  \(\displaystyle x=5\cos \left( {150{}^\circ } \right)=5\left( {\frac{{\sqrt{3}}}{2}} \right)=\frac{{5\sqrt{3}}}{2}\)
\(\displaystyle y=5\sin \left( {150{}^\circ } \right)=5\left( {\frac{1}{2}} \right)=\frac{5}{2}\)
\(\displaystyle \left( {\frac{{5\sqrt{3}}}{2},\frac{5}{2}} \right)\) 
\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{25\left( 3 \right)}}{4}+\frac{{25}}{4}}}=5\)
\(\displaystyle \theta ={{\tan }^{{1}}}\left( {\frac{{\frac{5}{2}}}{{\frac{{5\sqrt{3}}}{2}}}} \right)=\text{150}{}^\circ \text{ (2nd quadrant)}\) \(\displaystyle \left( {5,150{}^\circ } \right)\) 

\(\displaystyle \left( {3,\frac{{3\pi }}{2}} \right)\)  \(\displaystyle x=3\cos \left( {\frac{{3\pi }}{2}} \right)=3\left( 0 \right)=0\)
\(\displaystyle y=3\sin \left( {\frac{{3\pi }}{2}} \right)=3\left( 1 \right)=3\)
\(\displaystyle \left( {0,3} \right)\) 
\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{0+9}}=3\)
\(\displaystyle \begin{array}{c}\theta ={{\tan }^{{1}}}\left( {\frac{3}{0}} \right)=\frac{\pi }{2}\text{ (we want where tan is }\\\text{ undefined, between 1st and 2nd quadrants)}\end{array}\) \(\displaystyle \left( {3,\frac{\pi }{2}} \right),\text{ which is the same as }\left( {3,\frac{{3\pi }}{2}} \right)\)

Here’s one where we go from Rectangular to Polar and we can’t get the angle from the Unit Circle. Note that we had to add \(\boldsymbol {\pi }\) to our answer since we want Quadrant II.
Rectangular Point  Convert from Rectangular to Polar 
\(\left( {1,5} \right)\)

\(\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+25}}=\sqrt{{26}}\)
\(\displaystyle \theta ={{\tan }^{{1}}}\left( {\frac{5}{{1}}} \right)=1.373+\pi =1.768\text{ (2nd quadrant)}\) \(\displaystyle \left( {\sqrt{{26}},1.768} \right)\) 
Note that you can also use “2^{nd} APPS (ANGLE)” on your graphing calculator to do these conversions, but you won’t get the answers with the roots in them (you’ll get decimals that aren’t “exact”). And you have to solve for the \(x\) and \(y\), or \(r\) and \(\theta \) separately, and use the “,” above the 7 for the comma.
Make sure you have your calculator either in DEGREES or RADIANS (in MODE), depending on what you’re working with.
Drawing Polar Graphs
I find that drawing polar graphs is a combination of part memorizing and part knowing how to create polar tcharts.
First, here is a table of some of the more common polar graphs. I included tcharts in both degrees and radians.
(Note that you can also put these in your graphing calculator, as an example, with radians: MODE: RADIAN, POLAR and WINDOW: θ = [0, 2π], θstep = π/12 or π/6, X = [–10, 10], Y = [–6, 6], and then using “Y=” to put in the equation, or just put in graph and use ZOOM ZTRIG (option 7). By putting in smaller values of θstep, such as .1, the graph is drawn more slowly and more accurately; to redraw graph, you can turn the graph off and back on by going to “=” and unhighlighting and highlighting it back again before hitting “graph”.
Here is an example:
Let’s start with polar equations that result in circle graphs:
Type of Polar Function  TChart  Graph  
Circle \(r=5\)
Note: This makes sense since \(r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\), and the equation of a circle is \({{x}^{2}}+{{y}^{2}}={{5}^{2}}\).
Note that \(r=5\) would produce the same graph. 


Shifted Circle \(r=4\cos \theta \)
Circle symmetric with \(x\)axis with diameter 4.
If negative (such as \(r=4\cos \theta \)), reflect over \(y\)axis so it’s on lefthand side. 


Shifted Circle \(r=6\sin \theta \)
Circle symmetric with \(y\) axis with diameter 6.
Since it’s negative, reflect over \(x\)axis so it’s on the “bottom”. (Positive would be above \(x\)axis). 

Here are some polar equations that result in lines:
Type of Polar Function  TChart  Graph  
Line \(\displaystyle \theta =\frac{\pi }{4}\)
Note: There is no “\(r\)” in the equation; just draw line at \(\displaystyle \frac{\pi }{4}\). These graphs always go through the pole (center).
Note that \(\displaystyle \theta =\frac{{5\pi }}{4}\) and \(\displaystyle \theta =\frac{{3\pi }}{4}\) would produce the same graph. 
n/a = not applicable; can be anything 

Vertical Line \(r=3\sec \theta \)
I remember that this line is vertical since it’s the same as \(r\cos \theta =3\). This is the same as \(x=3\) in rectangular form, which is a vertical line.
If negative (such as \(r=3\sec \theta \)), reflect over \(y\)axis so it’s on lefthand side. 
und = undefined 

Horizontal Line \(r=4\csc \theta \)
I remember that this line is horizontal since it’s the same as \(r\sin \theta =4\). This is the same as is \(y=4\) in rectangular form, which is a horizontal line.
Since it’s negative, reflect over \(x\)axis so it’s on the “bottom”. (Positive would be above \(x\)axis). 


Here are graphs that we call Cardioids and Limacons. They are in the form \(r=a+b\cos \theta \) or \(r=a+b\sin \theta \).
Note: Unlike their rectangular equivalents, \(r=a\pm b\cos \theta \) and \(r=a\pm b\cos \theta \) (same with \(r=a\pm b\sin \theta \) and \(r=a\pm b\sin \theta \)) are the same polar graph! Try it!
First, the Cardioids (Hearts); note that these and the Limecon “Loops” touch the pole (origin), while the Limecon “Beans” do not:
Type of Polar Function  TChart  Graph  
Cardioid or Heart \(\begin{array}{l}r=a+b\cos \theta ,\,\,\,a=b\\r=a+b\sin \theta ,\,\,\,a=b\end{array}\)
\(r=3+3\cos \theta \)
Note: I remember that when \(a=b\), things are in harmony, like a heart.
The heart goes out to \(3+3=6\) on the \(x\)axis and hits 3 and –3 on the \(y\).
If cos is negative (such as \(r=33\cos \theta \)), reflect over \(y\)axis so it’s on lefthand side. 
Note that \(r=3+3\cos \theta \) would make same graph. 

Cardioid or Heart \(\begin{array}{l}r=a+b\cos \theta ,\,\,\,a=b\\r=a+b\sin \theta ,\,\,\,a=b\end{array}\)
\(r=44\sin \theta \)
Note: I remember that when \(a=b\), things are in harmony, like a heart.
The heart goes out to \(4+4=8\) on the \(y\)axis and hits 4 and –4 on the \(x\).
Since sin is negative, reflect over \(x\)axis so it’s on the “bottom”. (Positive would be above \(x\)axis). 
Note that \(r=44\sin \theta \) would make same graph. 
Here are the Limacons:
Type of Polar Function  TChart  Graph  
Limacon (Loop) \(\begin{array}{l}r=a+b\cos \theta ,\,\,\,a<b\\r=a+b\sin \theta ,\,\,\,a<b\end{array}\)
\(r=35\cos \theta \)
Note: I remember that when \(a\) is less than \(b\) (\(a<b\)), it’s a loop (the l’s match).
The limacon goes out to \(3+5=8\) on the negative \(x\)axis and hits 3 and –3 on the \(y\)axis. It also hits \(53=2\) in the loop.
Since cos is negative, reflect over \(y\)axis so it’s on the lefthand side. (Positive would be on the right side). 
Note that \(r=35\cos \theta \) would make same graph. 

Limacon (Bean) \(\begin{array}{l}r=a+b\cos \theta ,\,\,\,a>b\\r=a+b\sin \theta ,\,\,\,a>b\end{array}\)
\(r=4+3\sin \theta \)
Note: I remember that when b is the smallest, it’s a “bean”.
The bean goes up to \(4+3=7\) on the \(y\)axis and hits 4 and –4 on the \(x\). It also hits \(43=1\) on the negative \(y\)axis.
If sin is negative (such as \(r=43\sin \theta \)), reflect over \(x\)axis so it’s on the bottom. 
Note that \(r=4+3\sin \theta \) would make same graph. 
Graphs of Roses produce “petals” and are in the form \(r=a\cos \left( {b\theta } \right)\) or \(r=a\sin \left( {b\theta } \right)\). Note that since we have the starting point for these graphs, and the distance between the petals, the tchart isn’t that helpful. (In the tcharts, I made the \(\Delta \) angle the same as the distance between petals).
Let’s start with the cos Rose graphs:
Type of Polar Function  TChart  Graph  
Rose (“\(\boldsymbol {b}\)” is even) \(r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)\)
\(r=7\cos \left( {4\theta } \right)\)
Since \(b\) (4) is even, we have \(2b\) petals, or 8 petals.
With positive cos, they start at the positive \(\boldsymbol { x}\)axis and they are \(\displaystyle \frac{{360}}{8}\), or 45° apart, going counterclockwise.
The length of each petal is \(a\) (7). 


Rose (“\(\boldsymbol {b}\)” is odd) \(r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)\)
\(r=6\cos \left( {5\theta } \right)\)
Since \(b\) (5) is odd, we have \(b\) petals, or 5 petals (we don’t multiply by 2).
With negative cos, they start at the negative positive \(\boldsymbol { x}\)axis (reflect over \(\boldsymbol { y}\)axis) and are \(\displaystyle \frac{{360}}{5}\), or 72° apart, going counterclockwise. (Note that since the tchart starts on the positive \(\boldsymbol { x}\)axis, the \(r\)’s are negative in the chart).
The length of each petal is \(a\) (6). 

And here are some sin Rose graphs:
Type of Polar Function  TChart  Graph  
Rose (“\(\boldsymbol {b}\)” is even) \(r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)\)
\(r=8\sin \left( {4\theta } \right)\)
Since \(b\) (4) is even, we have \(2b\) petals, or 8 petals.
With positive sin, they start at \(\displaystyle \frac{{90}}{b}=\frac{{90}}{4}=22.5\) degrees from the positive \(\boldsymbol { x}\)axis (memorize this) and they are \(\displaystyle \frac{{360}}{8}\), or 45° apart, going counterclockwise.
The length of each petal is \(a\) (8). 


Rose (“\(\boldsymbol {b}\)” is odd) \(r=a\cos \left( {b\theta } \right),\,\,r=a\sin \left( {b\theta } \right)\)
\(r=6\cos \left( {5\theta } \right)\)
Since \(b\) (5) is odd, we have \(b\) petals, or 5 petals (we don’t multiply by 2).
With negative sin, they start at \(\displaystyle \frac{{90}}{b}=\frac{{90}}{5}=18\) degrees down from the positive \(\boldsymbol { x}\)axis (reflect over \(x\)axis) and are \(\displaystyle \frac{{360}}{5}\), or 72° apart, going counterclockwise. (Note that since the tchart starts on the positive \(x\)axis, the \(r\)’s are negative in the chart). The length of each petal is \(a\) (6). 

Note: For a rose graph, you may be asked to name the order that petals are drawn. One way to do this is to use the angle measurements \(\displaystyle 0,\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4},\,\frac{{7\pi }}{4}\), solve for \(r\), and observe the order of the petals. You can also use the graphing calculator as shown above, but make the θstep smaller to slow down the drawing of the graph.
Note: You may also see a combination of a rose and a limacon in the form \(r=a+b\cos \left( {k\theta } \right),\,\,r=a+b\sin \left( {k\theta } \right),\,\,k>1\). In these cases, you may see graphs that don’t meet at the origin; try these on your calculator!
Here are a couple more polar graphs (Spirals and Lemniscates) that you might see:
Type of Polar Function  TChart  Graph  
Spiral \(r=a\theta \)
\(r=2\theta \)
When filling in the tchart, use radians.
The smaller the \(r\), the tighter the spiral. 


Lemniscate (Figure “8”) \({{r}^{2}}={{a}^{2}}\cos \left( {2\theta } \right),\,\,{{r}^{2}}={{a}^{2}}\sin \left( {2\theta } \right)\)
\(r=\sqrt{{{{4}^{2}}\cos \left( {2\theta } \right)}}=\sqrt{{16\cos \left( {2\theta } \right)}}\)
With cos, graph is horizontal across the \(\boldsymbol {x}\)axis. The graph has 2 petals and the length of each petal is \(a\) (4). 
und = undefined 

Lemniscate (Figure “8”) \({{r}^{2}}={{a}^{2}}\cos \left( {2\theta } \right),\,\,{{r}^{2}}={{a}^{2}}\sin \left( {2\theta } \right)\)
\(r=\sqrt{{{{7}^{2}}\sin \left( {2\theta } \right)}}=\sqrt{{49\sin \left( {2\theta } \right)}}\)
With sin, graph is along \(\displaystyle \frac{\pi }{4}\). The graph has 2 petals and the length of each petal is \(a\) (7). 


You might be asked to obtain the equation of a polar function from a graph:
Graph  Get Polar Equation From Graph  
.  We see that the graph is a Rose pattern with an odd number of petals, so we don’t have to divide the number of petals by 2 to get \(b\) in the polar graph rose formula \(r=a\cos \left( {b\theta } \right)\) or \(r=a\sin \left( {b\theta } \right)\).
We also see that the petals are 4 units long (\(a\)) and don’t start on the \(x\)axis, so we have \(r=4\sin \left( {5\theta } \right)\) or \(r=a\sin \left( {b\theta } \right)\).
With 5 petals and with sin, they should start \(\displaystyle \frac{{90}}{b}=\frac{{90}}{5}=18\) degrees up from the positive \(x\)axis; since they start 18 degrees below (reflected) over the \(x\)axis, the graph would have to be \(r=4\sin \left( {5\theta } \right)\). 

The graph is a Bean (Limacon – no loop) and not a Cardioid or Heart, since it doesn’t go through the origin.
The graph is horizontal and is reflected over the \(\boldsymbol {y}\)axis, so we have \(r=ab\cos \theta ,\,\,\,a>b\).
Since the graph hits 5 (\(a\)) on the \(\boldsymbol {y}\)axis and goes out to 9 to the left, we have \(r=54\cos \theta \), since \(b=9a\). Also, since \(ab \,\,(54=1)\) is where it hits the \(x\)axis, it looks good! 


Here are more types of questions you may get asked when studying polar graphs:
Polar Equation Question  Solution 
For the rose polar graph \(5\sin \left( {10\theta } \right)\):
Find the length of each petal, number of petals, spacing between each petal, and the tip of the 1^{st} petal in Quadrant I. 
The length of each petal in the rose polar graph is \(a\), so this length is 5. Since \(b\) (10) is even, we need to double it to get the number of petals, so we get 20. The spacing between each petal is \(\displaystyle \frac{{360{}^\circ }}{{20}}=18{}^\circ \), since there are 360° in a circle. The tip of the 1^{st} petal in Quadrant I will be at \(\displaystyle \frac{{90{}^\circ }}{b}=\frac{{90{}^\circ }}{{10}}=9{}^\circ \). 
Write a polar equation for a spiral that goes through the polar point \(\left( {6,60{}^\circ } \right)\).  Since the equation of a polar spiral graph is \(r=a\theta \) and we know a point on the graph is \(\left( {6,60{}^\circ } \right)\), we can solve for \(a\): \(\displaystyle r=a\theta ;\,\,6=a60;\,\,a=\frac{1}{{10}}\). Thus, the polar equation is \(\displaystyle r=\frac{1}{{10}}\theta \) or \(\displaystyle r=\frac{\theta }{{10}}\). 
Can a rose have 14 petals? Explain why or why not.  No. Since, for a rose, if the number of pedals is even, you will have twice that number of petals. If the number of petals is odd, you have exactly that number of petals. Since twice 7 is 14, but 7 is odd, this can’t happen. 
Graph the rectangular and polar equation \(r=6\cos \left( {5\theta } \right)\).  We learned from the Transformations of Trig Functions section that a rectangular graph in the form \(y=a\cos b\left( {xc} \right)+d\) has an amplitude of \(\left a \right\) (and is flipped over the \(x\)axis is there’s a negative), has a period of \(\displaystyle \frac{{2\pi }}{b}\), a horizontal shift of \(c\), and a vertical shift of \(d\). In our case, the amplitude is 6, it’s flipped over the \(x\)axis, and the period is \(\displaystyle \frac{{2\pi }}{b}=\frac{{2\pi }}{5}\).
We already saw that the polar graph is a rose with an odd number of petals (5), each 6 units long, and starts at \(\displaystyle \frac{{90}}{b}=\frac{{90}}{5}=18{}^\circ \). You can see how different these graphs look:

Converting Equations from Rectangular to Polar
To convert Rectangular Equations to Polar Equations, we want to get rid of the \(x\)’s and \(y\)’s and only have \(r\)’s and/or \(\theta \)’s in the answer. We can do this with the following equations:
\(\begin{array}{l}x=r\cos \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\\y=r\,\sin \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\end{array}\)
Here are some examples; note that we want to solve for \(r\) if we can; in the case of quadratics or higher degrees, this may involve moving everything to one side and factoring.
Note that when we also get \(r=0\) (the pole) for the answers, this is one point only, and in these cases, the pole is included in the other part of the answers. Thus, we can discard \(r=0\).
Rectangular Equation  Convert to Polar Equation 
\(x=3\)  \(\displaystyle \begin{array}{c}r\cos \theta =3\\r=\frac{{3}}{{\cos \theta }};\,\,\,\,\,\,\,\,\,\,\underline{{r=3\sec \theta }}\end{array}\) 
\(y={{x}^{4}}\)  \(\begin{array}{c}r\sin \theta ={{\left( {r\cos \theta } \right)}^{4}}\\r\sin \theta ={{r}^{4}}{{\cos }^{4}}\theta \\r\sin \theta {{r}^{4}}{{\cos }^{4}}\theta =0\\r\left( {\sin \theta {{r}^{3}}{{{\cos }}^{4}}\theta } \right)=0\end{array}\) \(\require {cancel} \displaystyle \begin{array}{c}\xcancel{{r=0}}\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\sin \theta {{r}^{3}}{{\cos }^{4}}\theta =0\\\,r=\sqrt[3]{{\frac{{\sin \theta }}{{{{{\cos }}^{4}}\theta }}}}\\\underline{{r=\sqrt[3]{{\tan \theta {{{\sec }}^{3}}\theta }}}}\end{array}\) 
\({{x}^{2}}+{{y}^{2}}=3x\)  \(\displaystyle \begin{array}{c}\text{Note that }{{x}^{2}}+{{y}^{2}}={{r}^{2}}:\\{{r}^{2}}=3\left( {r\cos \theta } \right)\\{{r}^{2}}+3r\cos \theta =0\\r\left( {r+3\cos \theta } \right)=0\end{array}\) \(\xcancel{{r=0\,}}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\underline{{r=3\cos \theta }}\) 
\(2x+y=3\)  \(\begin{array}{c}2\left( {r\cos \theta } \right)+r\sin \theta =3\\r\left( {2\cos \theta +\sin \theta } \right)=3\end{array}\) \(\displaystyle \underline{{r=\frac{3}{{2\cos \theta +\sin \theta }}}}\) 
\({{x}^{2}}+{{y}^{2}}=49\) 
\(\begin{array}{c}\text{Use }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\,\,\text{identity:}\\{{\left( {r\cos \theta } \right)}^{2}}+{{\left( {r\sin \theta } \right)}^{2}}=49\\{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =49\end{array}\) \(\begin{array}{c}{{r}^{2}}\left( {{{{\cos }}^{2}}\theta +{{{\sin }}^{2}}\theta } \right)=49\\{{r}^{2}}\left( 1 \right)=49\\\underline{{r=\pm 7}}\end{array}\) 
\(y=x\) 
\(\begin{array}{c}r\sin \theta =r\cos \theta \\r\sin \theta +r\cos \theta =0\\r\left( {\sin \theta +\cos \theta } \right)=0\end{array}\) \(\displaystyle \begin{array}{c}\xcancel{{r=0}}\,\,\,\,\,\text{or}\,\,\,\,\sin \theta =\cos \theta \\\,\tan \theta =1\\\,\underline{{\theta =\frac{{3\pi }}{4}}}\end{array}\) 
Converting Equations from Polar to Rectangular
To convert Polar Equations to Rectangular Equations, we want to get rid of the \(r\)’s and \(\theta \)’s and only have \(x\)’s and/or \(y\)’s in the answer. We can do this with the following equations, depending on what we have in the polar equation:
\(\displaystyle r=\sqrt{{{{x}^{2}}+\,\,\,{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta \,\,=\,\,\frac{x}{r}=\frac{x}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}\)
\(\displaystyle \theta ={{\tan }^{{1}}}\frac{y}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin \theta =\frac{y}{r}=\frac{y}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}\)
Here are some examples. Note that sometimes we may be asked to Complete the Square to get the equation in a circle form; we learned how to do this in the Factoring Quadratics and Completing the Square section here.
Polar Equation  Convert to Rectangular Equation 
\(r=4\sin \theta \)  \(\displaystyle \begin{array}{c}\sqrt{{{{x}^{2}}+{{y}^{2}}}}=4\left( {\frac{y}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}} \right)\\{{\left( {\sqrt{{{{x}^{2}}+{{y}^{2}}}}} \right)}^{2}}=4y\\{{x}^{2}}+{{y}^{2}}=4y\end{array}\) \(\begin{array}{c}\text{You can complete the square to get circle:}\\{{x}^{2}}+{{y}^{2}}=4y\\{{x}^{2}}+{{y}^{2}}4y=0\\{{x}^{2}}+\left( {{{y}^{2}}4y+4} \right)=0+4\\\underline{{{{x}^{2}}+{{{\left( {y2} \right)}}^{2}}=4}}\end{array}\) 
\(\theta =45{}^\circ \)  \(\displaystyle {{\tan }^{{1}}}\left( {\frac{y}{x}} \right)=45{}^\circ ;\,\,\,\,\,\frac{y}{x}=1;\,\,\,\,\,\,\,\,\,\,\,\,\underline{{y=x}}\) 
\(\displaystyle \theta =\frac{\pi }{6}\)  \(\displaystyle {{\tan }^{{1}}}\left( {\frac{y}{x}} \right)=\frac{\pi }{6};\,\,\,\,\,\frac{y}{x}=\frac{1}{{\sqrt{3}}};\,\,\,\,\,\,\,\,\,\,\,\,\underline{{y=\frac{x}{{\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}}}=\frac{{x\sqrt{3}}}{3}\) 
\(r=5\)  \(\sqrt{{{{x}^{2}}+{{y}^{2}}}}=5;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{{{x}^{2}}+{{y}^{2}}=25}}\) 
\(\displaystyle r=\frac{4}{{2+\cos \theta }}\)  \(\displaystyle \begin{array}{c}\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\frac{4}{{2+\frac{x}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}}}\\\sqrt{{{{x}^{2}}+{{y}^{2}}}}\left( {2+\frac{x}{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}} \right)=4\\\underline{{2\sqrt{{{{x}^{2}}+{{y}^{2}}}}+x=4}}\end{array}\) \(\begin{array}{c}\text{It }\!\!’\!\!\text{ s probably easier to leave }r\text{ in the}\\\text{equations and substitute }\sqrt{{{{x}^{2}}+{{y}^{2}}}}\text{ later}\text{.}\end{array}\) 
\({{r}^{2}}\sin \left( {2\theta } \right)=4\)  \(\require {cancel} \displaystyle \begin{array}{c}{{r}^{2}}\cdot 2\sin \theta \cos \theta =4\\\cancel{{{{r}^{2}}}}\cdot 2\left( {\frac{y}{{\cancel{r}}}} \right)\left( {\frac{x}{{\cancel{r}}}} \right)=4\\2xy=4\\\underline{{xy=2}}\end{array}\) \(\displaystyle \begin{array}{c}\text{Note that we used the identity}\\\sin \left( \theta \right)=2\sin \theta \cos \theta .\\\\\text{Note also that we left }r\,\text{in the equation and}\\\text{we could simplify before substituting}\text{.}\end{array}\) 
\(r=2\sin \theta +3\cos \theta \) 
\(\displaystyle \begin{array}{c}r=2\left( {\frac{y}{r}} \right)+3\left( {\frac{x}{r}} \right)\\{{r}^{2}}=2y+3x\\{{x}^{2}}+{{y}^{2}}=2y+3x\end{array}\) \(\begin{array}{c}\text{You can complete the square to get circle:}\\{{x}^{2}}3x+{{y}^{2}}2y=0\\\left( {{{x}^{2}}3x+\frac{9}{4}} \right)+\left( {{{y}^{2}}2y+1} \right)=0+\frac{9}{4}+1\\\underline{{{{{\left( {x\frac{3}{2}} \right)}}^{2}}+{{{\left( {y1} \right)}}^{2}}=\frac{{13}}{4}}}\end{array}\) 
Polar Graph Points of Intersection
To find the intersection points for sets of polar curves, it’s helpful to draw the curves and also to solve algebraically. To solve algebraically, we just set the \(r\)’s together and solve for the \(\theta \).
Note also that after we solve for one variable (like \(\theta \)), we have to plug it back in either equation to get the other coordinate (like \(r\)).
We also have to be careful since there are “phantom” or “elusive” points, typically at the pole. The reason these points are “phantom” is because, although we don’t necessarily get them algebraically, we can see them on a graph. This is because, with an “\(r\)” of 0, the \(\theta \) could really be anything, since we aren’t going out any distance.
We will also see phantom points when one of the equations is “\(r=\) constant”, since another way to write this is “\(r=\) the negative of that constant”.
Note that with “phantom” points, both equations do not have to work; I know, it’s weird. To get all these elusive points, you put in the r value in both curves to see what additional points you get.
Find the intersection points for the following sets of polar curves (algebraically) and also draw a sketch. Find the intersections when \(\theta \) is between 0 and \(\boldsymbol {2\pi} \).
Polar Equations  Intersecting Points  Graph 
\(\begin{array}{l}r=\sin \theta \text{ }\\r=\cos \theta \end{array}\)  \(\displaystyle \begin{array}{c}\sin \theta =\cos \theta ;\,\,\,\,\,\,\tan \theta =1\\\theta =\frac{{3\pi }}{4}\,\,\left( {r=\cos \left( {\frac{{3\pi }}{4}} \right)=\frac{{\sqrt{2}}}{2}} \right)\text{ (duplicate)},\,\,\,\,\frac{{7\pi }}{4}\,\,\left( {r=\frac{{\sqrt{2}}}{2}} \right)\\\underline{{\left( {\frac{{\sqrt{2}}}{2},\frac{{7\pi }}{4}} \right),\,\,\left( {0,0} \right)\text{ (”phantom” point)}}}\end{array}\)  
\(\begin{array}{l}r=\cos \theta \\r=\cos 2\theta \end{array}\)  \(\displaystyle \begin{array}{c}\cos \theta =\cos 2\theta \\\cos \theta =2{{\cos }^{2}}\theta 1\,\,\text{(identity)}\\2{{\cos }^{2}}\theta \cos \theta 1=0;\,\,\,\,\,\,\left( {2\cos +1} \right)\left( {\cos \theta 1} \right)=0\\\cos \theta =\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\cos \theta =1\\\theta =\frac{{2\pi }}{3}\,\,\left( {r=\cos \left( {\frac{{2\pi }}{3}} \right)=\frac{1}{2}} \right),\,\,\frac{{4\pi }}{3}\,\,\left( {r=\frac{1}{2}} \right),\,\,\,\theta =0\,\,\left( {r=1} \right)\\\underline{{\left( {\frac{1}{2},\frac{{2\pi }}{3}} \right),\,\,\left( {\frac{1}{2},\frac{{4\pi }}{3}} \right),\,\,\,\left( {1,0} \right),\,\,\left( {0,0} \right)\text{ (”phantom” point) }}}\end{array}\)  
\(\begin{array}{c}r=3\\r=6\sin \theta \end{array}\) 
\(\displaystyle \begin{array}{c}6\sin \theta =3;\,\,\,\,\,\,\sin \theta =\frac{1}{2}\\\theta =\frac{{7\pi }}{6}\,\,\left( {r=3} \right),\,\,\,\,\frac{{11\pi }}{6}\,\,\left( {r=3} \right)\\\\\text{”Phantom” points: Use}\,\,r=3\text{, since that }\!\!’\!\!\text{ s the same circle}\\\text{ as}\,\,r=3.\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=6\sin \theta ;\,\,\,\,\theta =\frac{\pi }{6},\,\,\frac{{5\pi }}{6},\\\text{so ”phantom points” are}\,\,\left( {3,\frac{\pi }{6}} \right),\,\,\left( {3,\frac{{5\pi }}{6}} \right).\\\underline{{\left( {3,\frac{{7\pi }}{6}} \right),\,\,\left( {3,\frac{{11\pi }}{6}} \right),\,\,\left( {3,\frac{\pi }{6}} \right)\,\,\,\left( {3,\frac{{5\pi }}{6}} \right)}}\end{array}\) 

\(\begin{array}{l}r=\sin 2\theta \\r=\cos \theta \end{array}\)  \(\displaystyle \begin{array}{c}\sin 2\theta =\cos \theta \\2\sin \theta \cos \theta \,\,\text{(identity)}=\cos \theta \\2\sin \cos \theta \cos \theta =0;\,\,\,\cos \theta \left( {2\sin \theta 1} \right)=0\\\cos \theta =0\,\,\,\,\,\,\,\,\,\,\sin \theta =\frac{1}{2}\\\theta =\frac{\pi }{2}\,\,\left( {r=\cos \left( {\frac{\pi }{2}} \right)=0} \right),\,\,\,\,\frac{{3\pi }}{2}\,\,\left( {r=0} \right)\,\,\,\text{(duplicate)},\,\\\theta =\frac{\pi }{6}\,\,\left( {r=\frac{{\sqrt{3}}}{2}} \right),\,\,\,\,\frac{{5\pi }}{6}\,\,\left( {r=\frac{{\sqrt{3}}}{2}} \right)\\\underline{{\left( {0,\frac{\pi }{2}} \right),\,\,\,\left( {\frac{{\sqrt{3}}}{2},\frac{\pi }{6}} \right),\,\,\left( {\frac{{\sqrt{3}}}{2},\frac{{5\pi }}{6}} \right)}}\end{array}\) 

It also might be good to know the sequence in which the polar graphs are drawn; in other words, from 0 to \(2\pi \), which parts of the graphs are drawn before the other graphs. (Check it out on a graphing calculator, where you can see it!)
You can use a tchart, or set the polar equation to 0 if the graph crosses the pole, and test points in between. Here are some examples:
Problem and Solution  TChart  Graph  
The inner loop for \(2+4\cos \theta \) is formed between what two values of \(\theta \)?
Solution: Find out the two cases when \(r=0\), since that’s before and after the graph draws its inner loop: \(\displaystyle 0=2+4\cos \theta ;\,\,\,\,\,\cos \theta =\frac{1}{2}\) \(\displaystyle \theta =\frac{{2\pi }}{3};\,\,\,\,\,\theta =\frac{{4\pi }}{3}\) Since the end of the inner loop is at \(\left( {2,\pi } \right)\) (same as \(\displaystyle \left( {2,0{}^\circ } \right)\)), and this is between \(\displaystyle \frac{{2\pi }}{3}\) and \(\displaystyle \frac{{4\pi }}{3}\), the inner loop is formed when \(\displaystyle \frac{{2\pi }}{3}<\theta <\frac{{4\pi }}{3}\) 


The leftmost petal for \(\cos 2\theta \) is drawn between what two values of \(\theta \)? In what order are the petals drawn?
Solution: With this problem, we can create the following tchart and see the sequence of petals being drawn. To determine which values to use in the tchart, set \(\cos 2\theta \) to 0 to see what \(\theta \) values are between each petal: \(\displaystyle \begin{array}{c}0=\cos 2\theta \\0=2{{\cos }^{2}}\theta 1\,\,\,\text{(identity)}\,\,\,\,\,\end{array}\) \(\displaystyle \text{cos}\theta \,\,\text{=}\,\,\pm \sqrt{{\frac{1}{2}}}=\pm \frac{{\sqrt{2}}}{2}\) \(\displaystyle \theta =\frac{\pi }{4};\,\,\,\,\,\theta =\frac{{3\pi }}{4};\,\,\,\,\,\theta =\frac{{5\pi }}{4};\,\,\,\,\,\theta =\frac{{7\pi }}{4}\)
Thus, the leftmost (3^{rd}) petal is formed when \(\displaystyle \frac{{3\pi }}{4}<\theta <\frac{{5\pi }}{4}\) 

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