Where two polar curves meet, and why solving the two equations simultaneously does not answer that question. Establishes the four-step guidelines — sketch and check the pole, solve directly, shift by a full turn, flip the sign and half-turn — through the circle-and-rose example whose eight intersection points are found four by solving and four only by the last step. Closes with a pair of equations that turn out to describe the same curve, and with describing regions of the plane by inequalities on r and theta.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.5 Graphs of Polar Equations, pp. 951-958
Objectives
Five outcomes, and the second is the one that turns four answers into eight.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 951-958 — the pages these objectives are drawn from
Warm-up
Two curves are drawn and they visibly cross. Finding where should be a matter of solving.
Discussion prompt
The circle r equals 2 sine theta and the cardioid r equals 2 minus 2 sine theta both pass through the pole. Set the two expressions equal and solve. Does the pole appear among your solutions?
Hint: At what angle is each curve at the pole?
Answer:
Equating gives sine theta equal to one half, so theta is pi over 6 or 5 pi over 6, and both give r equal to 1. Two points, and neither is the pole.
But both curves plainly pass through the pole. The circle is there when theta is a multiple of pi; the cardioid is there when theta is pi over 2 plus a full turn. There is no angle at which both are there at once, so no simultaneous solution can find it.
The pole is nonetheless a genuine point of intersection, because it lies on both graphs. Solving the equations answers a different question from where do the graphs meet, and this lesson is about closing that gap.
Concept
Two graphs meet at a point when the point lies on both. But a point has many names, and the name satisfying one equation need not satisfy the other. So the algebra must be run over the possible renamings as well.
Guidelines for Finding Points of Intersection — Sketch both graphs and check the pole. Solve the equations directly. Then substitute theta plus 2 pi k for theta in one equation and solve again. Then substitute negative r for r and theta plus an odd multiple of pi for theta in one equation, and solve again.
\[ P \text{ on both} \;\Longleftrightarrow\; \exists (r, \theta) \text{ satisfying } E_1 \text{ and } \exists (r', \theta') \text{ satisfying } E_2 \]
The two substitutions are exactly the two clauses of the equivalence rule from Lesson 11.4a. They are applied to one equation only, since applying them to both would just restate the original system.
Figure (svg): A circle and a cardioid crossing at three points, with the two points found by solving marked in one colour and the pole, which solving misses, marked in another
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 951-955
Section
Section 1
Concept
The pole is on a curve if r equals zero for some angle. Two curves may each reach it, at angles that never coincide, so no simultaneous solution can detect it.
The pole is the only point for which this can happen in this particular way, because it is the only point whose representations do not determine the angle at all.
Figure (svg): A circle and a cardioid crossing at three points, with the two points found by solving marked in one colour and the pole, which solving misses, marked in another
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 952-952
Picture it
The two green points come out of the algebra. The red one does not.
Figure (svg): A circle and a cardioid crossing at three points, with the two points found by solving marked in one colour and the pole, which solving misses, marked in another
Sketching both curves first is what makes the missing point obvious, which is why the guidelines put the sketch before the algebra.
Worked example
Example 11.5.3, part 1. The full answer, pole included.
\[ \text{Find where } r = 2\sin(\theta) \text{ and } r = 2 - 2\sin(\theta) \text{ meet.} \]
Equate the two expressions
Why: Collect the sine terms.
\[ \sin \theta = \frac{1}{2} \]
Solve on a full turn
Why: Two solutions.
\[ \theta = \frac{\pi}{6}, 5 \pi / 6 \]
Find r at each
Why: Both equations agree.
\[ (1, \frac{\pi}{6}), (1, 5 \pi / 6) \]
Check the pole separately
Why: The circle reaches it at multiples of pi; the cardioid at pi over 2.
Figure (svg): The solution to Worked example circle and cardioid shown as a ladder of expressions, one row per legal move
\[ \left(1, \tfrac{\pi}{6}\right), \quad \left(1, \tfrac{5\pi}{6}\right), \quad \text{and the pole} \]
Verify: confirm the pole is on each
Why: Setting 2 sine theta to zero gives theta equal to a multiple of pi, so the circle passes through the pole. Setting 2 minus 2 sine theta to zero gives sine theta equal to 1, so theta is pi over 2. Each curve reaches the pole, so the pole is on both graphs — even though no single angle serves both.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 952-952
Sorting
Ask whether r equals zero has any solution.
Sort into buckets
Sort each curve.
Worked example
The same check, giving the opposite answer.
\[ \text{Is the pole an intersection of } r = 2 \text{ and } r = 3\cos(\theta)? \]
Test the first curve
Why: Set the expression to zero.
\[ 2 = 0,\text{ impossible} \]
Conclude for the first
Why: The circle never reaches the pole.
\[ \text{not on } r = 2 \]
Test the second
Why: Set 3 cosine theta to zero.
\[ \theta = \frac{\pi}{2}, 3 \pi / 2 \]
Conclude overall
Why: One curve reaches the pole and the other does not.
Figure (svg): The solution to Worked example a case where the pole is not shared shown as a ladder of expressions, one row per legal move
\[ r = 2 \text{ has no solution } r = 0 \;\Longrightarrow\; \text{the pole is not on both} \]
Verify: check the real intersections
Why: Equating 2 and 3 cosine theta gives cosine theta equal to two thirds, with solutions the arccosine of two thirds and its reflection. Those are the only two intersection points, one in quadrant one and one in quadrant four, which matches the sketch of a circle crossing a circle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 952-953
Trap
\[ \text{circle at pole when } \theta = \pi k; \quad \text{cardioid when } \theta = \tfrac{\pi}{2} + 2\pi k \]
Note that no k makes these equal, and conclude the pole is not an intersection
Why: The two conditions genuinely cannot be satisfied at once.
But the pole is a single point of the plane and both curves pass through it. That the passages happen at different angles is a fact about the parametrisation, not about the geometry.
The question is whether both graphs contain the point, and both do.
Since the pole is 0 comma theta for every theta, whichever angle each curve arrives at is a legitimate name for the same point.
Hold the distinction firmly: a graph is a set of points, and two curves meet where their point sets overlap. When two curves reach the pole is a question about how they are traced, and it has no bearing on whether they meet there.
Prediction
Two polar curves each pass through the pole, but at completely different angles.
Predict first
Is the pole a point of intersection?
Correct: Yes, always.
Why: The pole is a single point of the plane, and the pair 0 comma theta names it for every theta whatsoever. So if each curve contains the pole, the pole is on both graphs and is an intersection point. The angles at which each curve happens to arrive there describe how the curves are traced, not where they are, and nothing about the geometry depends on them.
Faded example
Determine whether r equal to 2 minus 3 sine theta passes through the pole.
Fill in the blanks
2 - 3\sin\theta = 0 \;\Longrightarrow\; \sin\theta = 2/3, \textwithin ___ \text___
Why: Two thirds lies between negative 1 and 1, so solutions exist and the curve does reach the pole — twice on each turn, in fact. Had the equation demanded a sine larger than 1, no solution would exist and the curve would miss the pole entirely.
Socratic
For any other shared point, solving does eventually find it.
Discussion prompt
What is special about the pole that makes it invisible to simultaneous solving?
Hint: How much do the pole's names have in common?
Answer:
Every other point has names whose first coordinates are r and negative r and whose angles are determined up to full and half turns. So the renaming is constrained, and the two substitutions of steps 3 and 4 cover every case.
The pole's names have nothing in common but the zero. Its angle is entirely unconstrained, so the renaming is not a shift of a known amount — it is arbitrary.
That is why the pole gets its own step rather than being folded into the substitutions. The substitutions parametrise renaming by an integer k, and the pole's renaming has no such parameter. Recognising it as a genuinely different case, rather than an oversight, is the point.
Section
Section 2
Concept
After sketching and checking the pole, solve the two equations directly. Then rerun the algebra with each of the two renaming substitutions applied to one equation, and collect any new solutions.
The substitutions are applied to whichever equation makes the algebra easier, and never to both — doing both simply returns the original system.
Figure (svg): The four steps of the guidelines for finding points of intersection of polar graphs, in order
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 955-955
Picture it
Two of them are algebra you would have done anyway, and two are the renamings.
Figure (svg): The four steps of the guidelines for finding points of intersection of polar graphs, in order
In many problems steps 3 and 4 produce nothing new, and it is tempting to skip them. The next section is the example that punishes that.
Worked example
Example 11.5.3, part 3, first substitution. A common outcome.
\[ \text{Apply step 3 to } r = 6\cos(2\theta) \text{ against } r = 3. \]
Substitute theta plus 2 pi k
Why: Into the rose equation.
\[ r = 6 \cos(2(\theta + 2 \pi k)) \]
Expand the argument
Why: Two times 2 pi k is 4 pi k.
Simplify
Why: Four pi k is a whole number of full turns.
Compare with the original
Why: The equation is unchanged.
Figure (svg): The solution to Worked example step 3 producing nothing shown as a ladder of expressions, one row per legal move
\[ \cos(2\theta + 4\pi k) = \cos(2\theta) \;\Longrightarrow\; \text{same equation} \]
Verify: explain why
Why: The multiplier 2 turns a full turn in theta into two full turns inside the cosine, which the cosine cannot distinguish from none. Any curve whose equation involves an integer multiple of theta will behave this way, so step 3 is often vacuous — but it has to be checked, not assumed.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 953-953
Matching
Each of the four steps addresses a different way a point can hide.
Match the pairs
Why: Steps 3 and 4 are the two clauses of the equivalence rule from Lesson 11.4a, and step 1 handles the case that rule explicitly excludes. Together the four steps cover every way two names for one point can differ, which is why the list is complete.
Worked example
Example 11.5.3, part 3, second substitution. The step that doubles the answer.
\[ \text{Apply step 4 to } r = 6\cos(2\theta) \text{ against } r = 3. \]
Substitute negative r and a half-turn shift
Why: Into the rose equation.
\[ -r = 6 \cos(2(\theta + (2 k + 1) \pi)) \]
Simplify the argument
Why: Twice an odd multiple of pi is an even multiple.
Rewrite and pair with the circle
Why: Multiply through by negative one.
Solve
Why: The cosine of twice theta equals negative one half.
\[ \theta = \frac{\pi}{3} + \pi k, 2 \pi / 3 + \pi k \]
Figure (svg): The solution to Worked example step 4 producing everything shown as a ladder of expressions, one row per legal move
\[ \left(-3, \tfrac{\pi}{3}\right), \; \left(-3, \tfrac{2\pi}{3}\right), \; \left(-3, \tfrac{4\pi}{3}\right), \; \left(-3, \tfrac{5\pi}{3}\right) \]
Verify: count against the sketch
Why: The sketch of a circle of radius 3 crossing a four-petalled rose of maximum radius 6 shows eight crossings, two in each quadrant. Step 2 found four and step 4 found four more, which accounts for all eight.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 954-954
Error analysis
A student applies the sign-flip substitution to each equation in turn.
Annotate
On: \( -r = 6\cos(2\theta) \quad \text{and} \quad -r = 3 \;\Longrightarrow\; \text{solve this pair} \)
The asymmetry is the whole idea. Intersections are missed precisely when the point's name satisfying one equation differs from the name satisfying the other, so the substitution has to break the symmetry rather than preserve it.
Prediction
You apply step 3, replacing theta by theta plus 2 pi k, to the equation r equal to 5 sine of 3 theta.
Predict first
What happens?
Correct: The equation is unchanged.
Why: Three times 2 pi k is 6 pi k, a whole number of full turns, which the sine cannot distinguish from none. This is what happens whenever theta appears multiplied by an integer, so step 3 is vacuous for every rose and every lemniscate. It is not vacuous in general — an equation involving theta over 2, as in the last section of this lesson, behaves quite differently.
Fill the middle
Apply step 4 to r equal to 4 sine theta against r equal to 2.
Fill in the blanks
-r = 4\sin(\theta + (2k+1)\pi) = -4\sin\theta \;\Longrightarrow\; r = 4\sin\theta \;\textno ___ \text___
Why: The sine of theta plus an odd multiple of pi is the negative of the sine, so the two minus signs cancel and the equation returns to its original form. Step 4 is therefore vacuous here, and the direct solution already found every intersection. Vacuous is a legitimate outcome; it just has to be established rather than assumed.
Edge cases
The guidelines list four steps and claim to find every intersection.
Discussion prompt
Why are four enough, and what would a fifth step even be?
Hint: How many ways can two names for the same point differ?
Answer:
The equivalence rule from Lesson 11.4a lists exactly two ways two non-zero names can name one point: same r with coterminal angles, or opposite r with angles differing by an odd multiple of pi. Steps 3 and 4 are those two, and step 2 is the case where no renaming is needed at all.
Step 1 handles the pole, which the rule explicitly excludes. So the four steps are the three cases of the rule plus its one exception, and the rule is exhaustive.
A fifth step would have to correspond to a fifth way of renaming a point, and there is none. The completeness of the guidelines is inherited directly from the completeness of the equivalence rule, which is why the earlier lesson stated that rule as carefully as it did.
Section
Section 3
Concept
A circle and a four-petalled rose cross eight times. Solving the equations directly finds four of those crossings. The other four are found only by the sign-flip substitution.
Sketching first is what makes the shortfall visible. Eight crossings are plainly there in the picture, so an answer with four is known to be incomplete before the algebra is even reread.
Figure (svg): A circle crossing a four-petalled rose at eight points, with four found by direct solving and four found only by the sign-flipping substitution
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 953-954
Picture it
The green ones came from solving. The pink ones came from step 4.
Figure (svg): A circle crossing a four-petalled rose at eight points, with four found by direct solving and four found only by the sign-flipping substitution
The pink points are on the circle at distance 3 and on a petal, exactly like the green ones. Nothing distinguishes them geometrically — only the names they happen to carry.
Worked example
Example 11.5.3, part 3, step 2.
\[ \text{Solve } 6\cos(2\theta) = 3. \]
Divide by 6
Why: Isolate the cosine.
Solve for twice theta
Why: The reference angle is pi over 3.
Divide by 2
Why: Halve everything.
\[ \theta = +- \frac{\pi}{6} + \pi k \]
List the distinct points
Why: Four of them on a full turn.
Figure (svg): The solution to Worked example the direct solutions shown as a ladder of expressions, one row per legal move
\[ \left(3, \tfrac{\pi}{6}\right), \; \left(3, \tfrac{5\pi}{6}\right), \; \left(3, \tfrac{7\pi}{6}\right), \; \left(3, \tfrac{11\pi}{6}\right) \]
Verify: check one against both equations
Why: At theta equal to pi over 6, the circle gives r equal to 3 and the rose gives 6 cosine of pi over 3, which is 6 times one half, namely 3. Both agree, so the point is genuinely on both curves under this single name.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 953-953
Prediction
A sketch shows two polar curves crossing six times, but solving the equations gives only three solutions.
Predict first
What should you conclude?
Correct: Three intersections are hiding under different names.
Why: The direct solution finds only the crossings where a single name satisfies both equations. A shortfall against the sketch is the standard signal that the renaming substitutions are needed, and it is why the guidelines put the sketch first. Suspecting the sketch is the wrong instinct here — a hand sketch of a rose and a circle is very reliable about counting crossings.
Worked example
Collecting all eight, and checking against the sketch.
\[ \text{Find all intersections of } r = 3 \text{ and } r = 6\cos(2\theta). \]
Check the pole
Why: The circle of radius 3 never reaches it.
Step 2
Why: Four points with r equal to 3.
Step 3
Why: The equation is unchanged.
Step 4
Why: Four more, named with r equal to negative 3.
Figure (svg): The solution to Worked example the whole answer shown as a ladder of expressions, one row per legal move
\[ \left(3, \tfrac{\pi}{6}\right), \left(3, \tfrac{5\pi}{6}\right), \left(3, \tfrac{7\pi}{6}\right), \left(3, \tfrac{11\pi}{6}\right), \left(-3, \tfrac{\pi}{3}\right), \left(-3, \tfrac{2\pi}{3}\right), \left(-3, \tfrac{4\pi}{3}\right), \left(-3, \tfrac{5\pi}{3}\right) \]
Verify: count against the picture
Why: Each of the four petals crosses the circle twice, once on the way out and once on the way back, giving eight crossings. The algebra produced eight points and the geometry demands eight, so the answer is complete.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 954-954
Trap
\[ 6\cos(2\theta) = 3 \;\Longrightarrow\; \text{four points} \;\Longrightarrow\; \text{done} \]
Solve the system and report the solutions
Why: The algebra is correct and the four points are genuine.
But the sketch shows eight crossings. Four correct points is not a correct answer to a question asking for all of them, and nothing in the algebra signals the shortfall.
Sketch first and count the crossings. That gives a target the algebra must meet, and it is the only warning you will get.
Then run steps 3 and 4. Here step 3 is vacuous and step 4 supplies the missing four.
The general lesson: in polar coordinates the algebra does not know how many answers there should be, because it works with names and the question is about points. The sketch is the bridge, and skipping it removes the only check available.
Faded example
For r equal to 3 against r equal to 6 cosine of twice theta, apply step 4.
Fill in the blanks
-6\cos(2\theta) = 3 \;\Longrightarrow\; \cos(2\theta) = -1/2 \;\Longrightarrow\; 2\theta = \pm\tfracpi/3___ + 2\pi k \;\Longrightarrow\; \theta = \pm___ + \pi k
Why: The sign flip turns the equation into minus 6 cosine of twice theta equal to 3, so the cosine equals negative one half and the reference angle for twice theta is pi over 3, placed to give 2 pi over 3. Halving gives theta equal to plus or minus pi over 3 plus a multiple of pi, which is the four remaining points.
Sorting
For the circle of radius 3 and the rose r equal to 6 cosine of twice theta.
Sort into buckets
Sort each point by the step that produces it.
Explain it to yourself
The four points found by step 4 are named with a negative first coordinate.
Discussion prompt
Explain why those points are nonetheless on the circle r equal to 3, which fixes r at a positive value.
Hint: Does a point have only one name?
Answer:
The point named negative 3 comma pi over 3 is also named 3 comma pi over 3 plus pi, which is 3 comma 4 pi over 3. That name has first coordinate 3, so it satisfies the circle's equation.
So the point is on the circle. It is simply that the name which satisfies the rose's equation is the one with the negative r, and the name which satisfies the circle's is the other.
This is the whole phenomenon in one sentence: each equation is satisfied by a different name for the same point. Direct solving searches for a name satisfying both at once, and there is none, so it finds nothing — even though the point is plainly on both graphs.
Section
Section 4
Concept
Sometimes step 4 turns one equation into the other. That is not a new intersection point but a proof that every point of one curve lies on the other, so the two equations have the same graph.
The half-angle is what makes this possible. With theta divided by 2, a half turn in theta becomes a quarter turn inside the function, which the cofunction identity converts a sine into a cosine.
Figure (svg): Two columns showing that two visibly different polar equations can describe exactly the same set of points
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 954-955
Picture it
One point by direct solving; every point by the substitution.
Figure (svg): Two columns showing that two visibly different polar equations can describe exactly the same set of points
The direct solution was not wrong, only badly incomplete. It answered a question about names when the question was about points.
Worked example
Example 11.5.3, part 4, step 2. One point, and a division that has to be justified.
\[ \text{Solve } 3\sin\left(\tfrac{\theta}{2}\right) = 3\cos\left(\tfrac{\theta}{2}\right). \]
Check the divisor first
Why: If the cosine were zero the sine would have to be too.
Divide through
Why: By 3 cosine of half theta.
\[ \tan(\theta / 2) = 1 \]
Solve
Why: The tangent is 1 at pi over 4 plus multiples of pi.
\[ \theta / 2 = \frac{\pi}{4} + \pi k \]
Double
Why: And read off the one distinct point.
\[ \theta = \frac{\pi}{2} + 2 \pi k \]
Figure (svg): The solution to Worked example the direct solution shown as a ladder of expressions, one row per legal move
\[ \left(\tfrac{3\sqrt{2}}{2}, \tfrac{\pi}{2}\right) \]
Verify: justify the division
Why: No angle has both its sine and its cosine equal to zero, since their squares sum to 1. So the divisor cannot vanish on the solution set and nothing was lost. This is the same justification the book used in the previous lesson, and it is worth reusing rather than reinventing.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 954-955
Prediction
Applying step 4 to one of two equations turns it into the other equation exactly.
Predict first
What does that tell you?
Correct: The two graphs are identical.
Why: It means that whenever a name satisfies the first equation, the renamed version of it satisfies the second. Since a name and its renaming describe the same point, every point of the first graph is on the second, and the argument runs symmetrically in reverse. So the two equations, though different, have exactly the same set of points as their graph.
Worked example
Example 11.5.3, part 4, step 4. Where the real answer is.
\[ \text{Apply step 4 to } r = 3\cos\left(\tfrac{\theta}{2}\right). \]
Substitute
Why: Negative r, and a half-turn shift in theta.
\[ -r = 3 \cos(\frac{\theta + (2 k + 1) \pi}{2}) \]
Split the argument
Why: Half of an odd multiple of pi.
\[ \cos(\theta / 2 + (2 k + 1) \pi / 2) \]
Expand with the sum formula
Why: The cosine of that odd multiple is 0 and its sine is plus or minus 1.
\[ = +- \sin(\theta / 2) \]
Take k equal to zero
Why: The signs work out to give the other equation.
\[ r = 3 \sin(\theta / 2) \]
Figure (svg): The solution to Worked example the substitution that proves the identity shown as a ladder of expressions, one row per legal move
\[ r = 3\sin\left(\tfrac{\theta}{2}\right) \text{ and } r = 3\cos\left(\tfrac{\theta}{2}\right) \text{ have the same graph} \]
Verify: state the consequence precisely
Why: If a name r comma theta satisfies the sine equation, then the name negative r comma theta plus pi automatically satisfies the cosine equation. Since those two names describe the same point, every point of one curve is a point of the other. The graphs are equal as sets of points, though the equations are different.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 955-955
Error analysis
A student solves the two half-angle equations and reports the result.
Annotate
On: \( \tan\left(\tfrac{\theta}{2}\right) = 1 \;\Longrightarrow\; \text{one intersection point} \)
This is the most extreme version of the lesson's theme. The direct solution understated the answer by an infinite amount, and only the sketch and the substitution revealed it.
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about polar intersections is wrong.
Survives elimination: B
Why: Statement B is false, and refuting it is the entire purpose of this lesson. Direct solving finds only the points possessing a single name that satisfies both equations. The circle-and-rose example has eight intersections of which solving finds four, and the two half-angle curves share infinitely many points of which solving finds one.
Faded example
Why is dividing by 3 cosine of half theta safe in this problem?
Fill in the blanks
If the cosine were zero, the equation would force the sine to be zero as well, and no angle has both of its sine and cosine equal to zero.
Why: The Pythagorean Identity says the squares of the sine and cosine sum to 1, so they cannot both vanish. That rules out the divisor being zero anywhere on the solution set, which is exactly the condition needed before dividing.
Counterexample
A student proposes: if two polar equations look different, their graphs are different.
Discussion prompt
Give two counterexamples of different kinds, and say what actually decides whether two graphs agree.
Hint: One trivial example and one surprising one.
Answer:
The easy one: r equals 3 and r equals negative 3 both graph the circle of radius 3, since every point of one has a name satisfying the other.
The hard one: r equals 3 sine of half theta and r equals 3 cosine of half theta, whose graphs coincide entirely — a fact no amount of staring at the equations suggests, and which took the sum formula for cosine to establish.
What actually decides it is whether every point of one graph has some representation satisfying the other equation, and vice versa. That is a statement about sets of points, and the equations are merely one way of specifying them. Two specifications can be very different and pick out the same set, which is a general fact about definitions and not special to trigonometry.
Section
Section 5
Concept
An inequality on r says how far out the region extends at each angle; an inequality on theta says which angles are included. Together they describe a region, and the intersection points found earlier are what set the angular limits.
A union of two such sets describes a region bounded by one curve over part of its angular range and by another over the rest, which is how a region between two intersecting curves is usually written.
Figure (svg): Three polar regions described by inequalities: a single rose petal, the part of a petal outside a circle, and the inner loop of a limacon
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 956-958
Picture it
A leaf, a slice between two curves, and an inner loop.
Figure (svg): Three polar regions described by inequalities: a single rose petal, the part of a petal outside a circle, and the inner loop of a limacon
In each case the angular limits are read off the geometry, and in the middle one they came directly from the intersection calculation of the previous section.
Worked example
Example 11.5.4, part 1. The simplest kind of region.
\[ \text{Sketch } \left\{(r, \theta) : 0 \le r \le 5\sin(2\theta), \; 0 \le \theta \le \tfrac{\pi}{2}\right\}. \]
Identify the curve
Why: The four-petalled rose from the previous lesson.
Identify the angular range
Why: Zero to pi over 2 traces the first-quadrant leaf.
Read the radial inequality
Why: From the pole out to the curve.
\[ 0 \le r \le\text{ curve} \]
Describe the region
Why: Everything between the pole and that leaf.
Figure (svg): The solution to Worked example one leaf of a rose shown as a ladder of expressions, one row per legal move
\[ \text{the leaf of } r = 5\sin(2\theta) \text{ lying in Quadrant I} \]
Verify: check the endpoints
Why: At theta equal to 0 and pi over 2 the curve is at the pole, so the region pinches to a point at each end. In between it reaches out to 5 at theta equal to pi over 4. That is exactly the shape of a leaf.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 956-956
Sorting
It depends on the sign of the bounding expression over the interval.
Sort into buckets
Sort each situation.
Worked example
Example 11.5.4, part 2. The angular limit comes from the intersection.
\[ \text{Sketch } \left\{(r, \theta) : 3 \le r \le 6\cos(2\theta), \; 0 \le \theta \le \tfrac{\pi}{6}\right\}. \]
Identify both curves
Why: The circle of radius 3 and the four-petalled rose.
Note where the upper limit comes from
Why: The two meet at theta equal to pi over 6.
Read the radial inequality
Why: At least 3, so outside the circle; at most the rose.
Describe the region
Why: The part of the leaf lying outside the circle.
Figure (svg): The solution to Worked example a region between two curves shown as a ladder of expressions, one row per legal move
\[ \text{outside } r = 3 \text{ and inside } r = 6\cos(2\theta), \text{ for } 0 \le \theta \le \tfrac{\pi}{6} \]
Verify: check the upper limit
Why: At theta equal to pi over 6 the rose gives 6 cosine of pi over 3, which is 3 — exactly the circle's radius. So the two bounds coincide there and the region closes to a point, which is why pi over 6 is the right upper limit and not an arbitrary choice.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 956-956
Trap
\[ 0 \le r \le 2 + 4\cos(\theta), \quad \tfrac{2\pi}{3} \le \theta \le \tfrac{4\pi}{3} \]
Write the region from the pole out to the curve, as usual
Why: That form worked for every positive region.
But on that interval the expression is negative, so the inequality demands r be both at least zero and at most a negative number. No r satisfies it and the region is empty.
\[ 2 + 4\cos(\theta) \le r \le 0, \quad \tfrac{2\pi}{3} \le \theta \le \tfrac{4\pi}{3} \]
Put the negative expression on the smaller side
Why: Then the inequality has solutions and describes the inner loop.
The habit: check the sign of the bounding expression on the interval before writing the inequality. An empty region produces no error message, so this is a failure that has to be caught by inspection.
Prediction
A region is described by an inequality on r together with the angular range from 0 to pi over 6.
Predict first
Where does that angular limit most likely come from?
Correct: The point where the two bounding curves intersect.
Why: A region between two curves closes where the curves meet, so the angular limits of such a region are the angles of intersection. That is why finding intersections and describing regions are treated together: the first supplies the limits the second needs. Occasionally a limit is an axis or a chosen boundary, but the intersection is the usual source.
Fill the middle
Describe the inner loop of r equal to 2 plus 4 cosine theta as a region.
Fill in the blanks
\left\2 + 4 cos th} \le r \le 0, \; \tfrac4pi/3___ \le \theta \le ___\right\}
Why: The two angular limits are the zeros of r, found in the previous lesson, and between them the expression is negative — so it goes on the left of the inequality with 0 on the right. That describes everything between the pole and the loop, which is the loop's interior.
Real world
A lighthouse's light is visible over a sector: not in every direction, because the land behind it blocks the beam, and not at every distance, because the beam is invisible beyond about 20 nautical miles and is obscured within half a mile by the tower itself.
Discussion prompt
Write that visibility area as a polar region, and say what the two kinds of inequality correspond to physically.
Hint: One inequality bounds the distance and one bounds the direction.
Answer:
The region is a set of pairs r comma theta with r between 0.5 and 20 and theta running over the unobstructed sector, say from one bearing to another. Two inequalities, exactly as in this lesson.
The radial inequality is about the physics of the light: too close and the structure blocks it, too far and the intensity falls below what the eye can detect. The angular inequality is about the geometry of the site: headlands and buildings define which bearings are served.
On a real chart this appears as an arc-and-sector symbol, and the angular limits are printed as bearings. Those bearings are intersection angles — the directions in which the beam's edge meets the obstructing terrain — which is the same relationship this lesson keeps drawing between intersections and the limits of a region.
It is worth noticing how naturally the polar description fits. In rectangular coordinates the same region would require an annulus intersected with a wedge, described by inequalities that mix squares and arctangents. The region is simple because the coordinate system matches the way the constraints were generated, which is the recurring lesson of this whole section.
Comparison
Fill the blanks from memory. The completeness of the list is inherited from the equivalence rule.
Comparison matrix
| Step | What it does | What it finds |
|---|---|---|
| 1. sketch and check the pole | ask whether r = 0 has a solution for each curve | the pole, whose angle is unconstrained |
| 2. solve directly | equate the two expressions for r | points sharing a single name |
| 3. shift theta by 2 pi k | in one equation only | points whose names differ by full turns |
| 4. flip r, shift by an odd multiple of pi | in one equation only | points whose names differ in the sign of r |
Steps 3 and 4 are frequently vacuous, and that has to be established rather than assumed. When step 4 is not vacuous it can double the answer, or reveal that the two curves were the same curve all along.
Pattern
Five moves, covering both intersections and regions.
If a substitution turns one equation into the other, the two curves are identical and every point is an intersection.
OpenStax Algebra and Trigonometry 2e, §10.4 Polar Coordinates: Graphs §10.4
Check
The pole.
Check your understanding
The curves r equal to 5 sine theta and r equal to 5 cosine theta both pass through the pole, at theta equal to 0 and pi over 2 respectively. Is the pole an intersection point?
Answer: B
Why: An intersection point is a point lying on both graphs, and both graphs contain the pole. The angles at which each curve arrives there describe how it is traced, not where it is, and the pole is named 0 comma theta for every theta.
Check
The guidelines.
Check your understanding
Why is the sign-flip substitution applied to only one of the two equations?
Answer: B
Why: The point of the substitution is to let the two equations be satisfied by different names for the same point. Renaming both equations the same way renames the whole picture, leaving a system equivalent to the original, which by construction finds nothing new.
Check
Regions.
Check your understanding
On an interval where the curve r equal to f of theta is negative throughout, how should the region between the pole and the curve be written?
Answer: B
Why: Where the expression is negative, zero is the larger of the two bounds, so it goes on the right. Writing it the other way round demands that r be simultaneously at least zero and at most a negative number, which no r satisfies, and the region comes out empty.
Real world
Two radar installations each report the region they can see: a range-and-bearing sector, obstructed differently by the terrain around each site. A planner needs the area covered by both, so that a target there is tracked twice and can be located precisely.
Discussion prompt
Explain what makes this harder than intersecting two ordinary regions, and which parts of this lesson apply.
Hint: Are the two sectors described from the same pole?
Answer:
Each sector is a polar region of exactly the form in this lesson — an inequality on range and an inequality on bearing. But each is described from its own pole, so the two descriptions are not directly comparable, exactly as in the air traffic problem of Lesson 11.4b.
The overlap is bounded by arcs from both sectors, and the corners of the overlap are intersection points of the two boundary curves. Finding them is the intersection problem of this lesson, done after both regions have been brought into a common frame.
What carries over most directly is the discipline: sketch first and count the corners, then check that the algebra produced that many. A coverage map with a missing corner describes a region with a gap in it, and a gap in radar coverage is precisely the thing the planner was trying to eliminate.
The half-turn phenomenon has an analogue too. A sector described with a bearing measured from one reference and another measured from the opposite reference can produce boundary curves that coincide without looking alike — the same trap as the two half-angle equations, and the same fix: check whether one description transforms into the other before assuming they describe different things.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
You sketch two polar curves, count six crossings, and your direct solution gives four points. What is the most likely explanation?
Correct: Two crossings have names differing in the sign of r.
\[ (r, \theta) \text{ on } E_1 \quad \text{and} \quad (-r, \theta + \pi) \text{ on } E_2 \]
Why: That is the standard shortfall, and it is what step 4 exists to recover. The sketch is generally reliable for counting crossings, and the direct solution is generally correct as far as it goes — it simply cannot find points whose two names disagree. Suspecting the sketch or the arithmetic is the wrong first instinct here.
Explain it
They solved two polar equations simultaneously, got four points, and were marked down for missing four more. They think the marking scheme has invented extra answers.
Discussion prompt
In no more than five sentences, explain where the other four points come from.
Hint: How many names does each of those points have?
Answer:
A usable answer: solving finds the crossings where one single pair of coordinates satisfies both equations. But a point has many names, and a crossing can be such that the name satisfying the first equation is different from the name satisfying the second.
The four missing points are exactly those. Each lies on the circle under the name 3 comma something and on the rose under the name negative 3 comma something else, and no single name serves both — so the algebra never sees them.
The fix is to redo the algebra with negative r substituted into one equation. Sketch first and count the crossings, and you will know before you start how many answers to expect.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The pole is fixed by asking, of each curve alone, whether r equals zero has a solution. The substitutions are fixed by applying each to one equation only and simplifying the trigonometric argument carefully. Recognising a coincident curve is fixed by noticing when a substitution returns the other equation exactly. Regions are fixed by checking the sign of the bounding expression before writing the inequality. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page write the four guidelines in order, and beside each write in one line what it finds and which clause of the equivalence rule it comes from. Underneath, sketch the circle r equal to 2 sine theta together with the cardioid r equal to 2 minus 2 sine theta, mark the three intersection points, and note beside the pole why solving cannot find it. In the middle of the page sketch the circle of radius 3 with the rose r equal to 6 cosine of twice theta, mark all eight crossings, and colour the four found by direct solving differently from the four found by step 4. In the bottom left, write both half-angle equations from the last section and one line saying what step 4 reveals about them. In the bottom right, write the inner loop of r equal to 2 plus 4 cosine theta as a region, with both inequalities and the correct orientation. Finally, circle the one step of the guidelines that turned four answers into eight.
The circled step is the fourth. It is also the one most often skipped, because it produces nothing at all in the majority of problems — and then, in the problem that matters, it produces half the answer.
Recap
Five things, and the second is what the whole lesson exists for.
| If the question says | Your first move is |
|---|---|
| Find all points of intersection | Sketch both and count the crossings |
| Is the pole one of them? | Check r = 0 on each curve separately |
| Your count falls short of the sketch | Run the sign-flip substitution |
| A substitution gave the other equation | The two graphs are the same curve |
| Sketch the region described by | Check the sign of the bound before writing it |
The chapter has now built polar coordinates, graphed them, and found where their curves meet. The next section puts the same machinery to a different use: giving complex numbers a polar form, where multiplication becomes a rotation and roots become evenly spaced points on a circle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.5 Graphs of Polar Equations §11.5, pp. 951-958 — everything on these slides traces back here
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