Theorem 11.7 and what it does and does not settle. Converts points in both directions, showing why the arctangent alone cannot determine the angle and why every conversion into polar form starts with a sketch. Then converts equations both ways, contrasting the mechanical rectangular-to-polar substitution with the awkward reverse, and examining what multiplying by r and squaring both sides do to a solution set.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.4 Polar Coordinates, pp. 924-930
Objectives
Five outcomes, and the second is where nearly all the lost marks are.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-930 — the pages these objectives are drawn from
Warm-up
The previous lesson built the polar system alongside the rectangular one without connecting them.
Discussion prompt
A point is 5 units from the origin on the terminal side of an angle of 53 degrees. Write down its rectangular coordinates as accurately as you can, and say what you used.
Hint: Drop a perpendicular to the x-axis. What are the legs?
Answer:
Dropping a perpendicular gives a right triangle with hypotenuse 5 and angle 53 degrees, so the horizontal leg is 5 cosine 53 and the vertical leg is 5 sine 53. That is about 3 and about 4.
Which is to say: x is r cosine theta and y is r sine theta. Nothing new was needed; the definition of the circular functions already said this back in Lesson 10.2.
The theorem in this lesson is precisely that observation, stated carefully enough to survive negative values of r and the pole itself.
Concept
Identify the pole with the origin and the polar axis with the positive x-axis. Then the rectangular coordinates of a point are its distance times the cosine and sine of its angle, and the relations run both ways.
Theorem 11.7 — For a point with rectangular coordinates x, y and polar coordinates r, theta: x equals r cosine theta and y equals r sine theta, and x squared plus y squared equals r squared with the tangent of theta equal to y over x provided x is not zero.
\[ x = r\cos(\theta), \quad y = r\sin(\theta), \quad x^2 + y^2 = r^2, \quad \tan(\theta) = \frac{y}{x} \]
The first pair of relations is a formula that always works. The second pair determines r up to a sign and theta only up to its reference angle, so it must be finished by looking at the picture.
Figure (svg): The two coordinate systems overlaid, with the pole at the origin and the polar axis along the positive x-axis, and a single point carrying both sets of coordinates
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-924
Section
Section 1
Concept
For positive r the relations follow directly from the definition of the circular functions. For negative r the alternate representation with the opposite sign supplies the proof, and at the pole the statement reduces to zero equals zero.
The negative case matters because the whole point of directed distance was to allow it. A theorem that quietly assumed r positive would undo the flexibility the previous lesson built.
Figure (svg): The four conversion relations of Theorem 11.7, grouped by which direction each is used in
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-924
Picture it
The identification of pole with origin and polar axis with positive x-axis is the whole content.
Figure (svg): The two coordinate systems overlaid, with the pole at the origin and the polar axis along the positive x-axis, and a single point carrying both sets of coordinates
If a problem places the pole somewhere other than the origin, none of these relations apply until the shift is accounted for.
Worked example
Verifying that the theorem survives a negative first coordinate.
\[ \text{Show } x = r\cos(\theta) \text{ holds when } r < 0. \]
Replace the pair with an equivalent one
Why: Flip the sign and add a half turn.
\[ (-r, \theta + \pi) \]
Apply the positive case to it
Why: The first coordinate is now positive.
\[ x = (-r) \cos(\theta + \pi) \]
Use the half-turn identity
Why: The cosine of theta plus pi is the negative of the cosine.
\[ \cos(\theta + \pi) = -\cos \theta \]
Simplify the two minus signs
Why: They cancel.
\[ x = r \cos \theta \]
Figure (svg): The solution to Worked example the negative case shown as a ladder of expressions, one row per legal move
\[ x = (-r)\left(-\cos\theta\right) = r\cos(\theta) \]
Verify: test a case
Why: The pair negative 2 comma 0 names the point 2 units out along the negative x-axis, so x should be negative 2. The formula gives negative 2 times cosine 0, which is negative 2. Correct.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-924
Faded example
Convert (6, 2pi/3) to rectangular coordinates.
Fill in the blanks
x = 6\cos\left(\tfrac-1/23 sqrt 3\right) = 6\left(___\right) = -3, \quad y = 6\sin\left(\tfrac______\right) = ___
Why: Two thirds of pi has reference angle pi over 3 in quadrant two, so the cosine is negative one half and the sine is positive root three over two. That gives x equal to negative 3 and y equal to three root three, about 5.20, which is indeed quadrant two.
Worked example
The direction that needs no thought.
\[ \text{Convert } \left(-4, \tfrac{5\pi}{6}\right) \text{ to rectangular coordinates.} \]
Write both formulas
Why: Substitute r and theta straight in.
\[ x = -4 \cos(5 \pi / 6) \]
Evaluate the cosine
Why: Five sixths of pi has reference angle pi over 6 in quadrant two.
\[ \cos = -\sqrt{3} / 2 \]
Compute x
Why: Two minus signs give a positive.
\[ x = 2 \sqrt{3} \]
Compute y the same way
Why: The sine of 5 pi over 6 is one half.
\[ y = -2 \]
Figure (svg): The solution to Worked example polar to rectangular shown as a ladder of expressions, one row per legal move
\[ (x, y) = \left(2\sqrt{3}, -2\right) \]
Verify: check the quadrant
Why: Five sixths of pi points into quadrant two, and the negative r sends the point to quadrant four. The answer has x positive and y negative, which is quadrant four. Consistent.
Trap
\[ \text{radar at } (3, 1), \; \text{contact at } (5, 40^\circ) \;\Longrightarrow\; (x, y) = (5\cos 40^\circ, 5\sin 40^\circ) \]
Apply the conversion formulas directly
Why: The formulas are correct and the arithmetic is right.
But the pole here is the radar at 3 comma 1, not the origin. The formulas give the contact's position relative to the radar, which is not its position in the map's coordinates.
\[ (x, y) = \left(3 + 5\cos 40^\circ, \; 1 + 5\sin 40^\circ\right) \]
Convert relative to the pole, then translate by the pole's position
Why: The conversion and the translation are separate steps.
The theorem is stated for the pole at the origin, and every application inherits that assumption. Whenever the pole is elsewhere, convert first and translate second — and say which you have done, because the two answers look equally plausible.
Prediction
A point has polar coordinates with r equal to 7.
Predict first
What can you say about x squared plus y squared?
Correct: It equals 49.
Why: The theorem says x squared plus y squared equals r squared, and that holds whatever the angle is. Geometrically, all points at distance 7 from the origin lie on a circle of radius 7, and the sum of the squares is the square of the distance by the Pythagorean Theorem. The angle only determines where on that circle the point sits.
Sorting
Each task uses a different part of the theorem.
Sort into buckets
Sort each task by the relation it starts from.
Socratic
The theorem states the tangent relation only when x is not zero.
Discussion prompt
What goes wrong when x is zero, and how are such points handled?
Hint: What is the tangent of a right angle?
Answer:
If x is zero the quotient y over x is undefined, and correspondingly the tangent is undefined at pi over 2 and 3 pi over 2. So the relation says nothing at all there.
Those are exactly the points on the vertical axis, and they are handled by inspection instead: a point at 0 comma negative 3 is plainly 3 units from the pole along the negative y-axis, so theta is 3 pi over 2 with r equal to 3.
This is worth noticing as a pattern rather than an exception. A formula that divides will always have a locus where it fails, and the locus is usually the geometrically special one — here, the axis the tangent's asymptotes correspond to. Reading the exclusion tells you where the special cases are before you meet them.
Section
Section 2
Concept
The sum of squares gives r up to a sign and the tangent gives the reference angle, but two opposite quadrants share the same tangent. The only reliable resolution is to plot the point first.
The book opens its worked example by recommending exactly this, noting that plotting first avoids many common mistakes. It is not a beginner's crutch; it is the method.
Figure (svg): Two points in opposite quadrants sharing the same value of y over x, showing that the tangent alone cannot distinguish them
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-926
Picture it
Two points, opposite quadrants, identical quotient.
Figure (svg): Two points in opposite quadrants sharing the same value of y over x, showing that the tangent alone cannot distinguish them
The arctangent returns values only in the fourth and first quadrants, so for a point in the second or third it always reports the wrong one of the two candidates.
Worked example
Example 11.4.2, part 1, with r non-negative and theta from 0 up to 2 pi.
\[ \text{Convert } P\left(2, -2\sqrt{3}\right) \text{ to polar form.} \]
Plot it
Why: Positive x and negative y puts it in quadrant four.
Find r
Why: Four plus 12 is 16.
\[ r = 4 \]
Find the reference angle
Why: The tangent is negative root three, so the reference angle is pi over 3.
\[ r e f = \frac{\pi}{3} \]
Place it in quadrant four
Why: Two pi minus pi over 3.
\[ \theta = 5 \pi / 3 \]
Figure (svg): The solution to Worked example a quadrant four point shown as a ladder of expressions, one row per legal move
\[ (r, \theta) = \left(4, \tfrac{5\pi}{3}\right) \]
Verify: convert back
Why: Four cosine of 5 pi over 3 is 4 times one half, which is 2. Four sine of 5 pi over 3 is 4 times negative root three over two, which is negative 2 root 3. Both match the original point.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 925-925
Sorting
The arctangent gives the reference angle; the plot gives the quadrant.
Sort into buckets
Sort each rectangular point by the adjustment its angle needs.
Worked example
Example 11.4.2, part 4. The tangent is not one of the common values.
\[ \text{Convert } S(-3, 4) \text{ to polar form.} \]
Plot it
Why: Negative x and positive y puts it in quadrant two.
Find r
Why: Nine plus 16 is 25.
\[ r = 5 \]
Find the reference angle
Why: The tangent's magnitude is four thirds, not a common value.
\[ \arctan(\frac{4}{3}) \]
Place it in quadrant two
Why: Pi minus the reference angle.
\[ \theta = \pi - \arctan(\frac{4}{3}) \]
Figure (svg): The solution to Worked example an angle that is not standard shown as a ladder of expressions, one row per legal move
\[ (r, \theta) = \left(5, \;\pi - \arctan\left(\tfrac{4}{3}\right)\right) \approx (5, 2.21) \]
Verify: check the decimal against the quadrant
Why: The arctangent of four thirds is about 0.927, so theta is about 2.21 radians, which is about 127 degrees. That is in quadrant two, matching the plot. Had we written the arctangent alone we would have got about negative 0.927, in quadrant four, which is wrong.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 926-926
Error analysis
A student converts the point (-3, -3) to polar form.
Annotate
On: \( \tan\theta = \frac{-3}{-3} = 1 \;\Longrightarrow\; \theta = \arctan(1) = \tfrac{\pi}{4}, \quad r = 3\sqrt{2} \)
The reference angle was correct; the quadrant was not. Plotting the point first would have made the error impossible, which is the whole reason the book insists on it.
Faded example
Convert (-2, 2 root 3) to polar form with r at least 0 and theta from 0 up to 2 pi.
Fill in the blanks
r = \sqrt4 = 2, \quad \text___ = \tfrac______, \quad \theta = \pi - \tfrac______ = \tfrac___\pi}___
Why: The sum of squares is 16, so r is 4. The tangent's magnitude is two root three over two, namely root three, giving a reference angle of pi over 3. Negative x with positive y is quadrant two, so the angle is pi minus pi over 3, which is two thirds of pi.
Prediction
A point lies on the negative y-axis, at (0, -6).
Predict first
What goes wrong if you use the tangent relation?
Correct: The quotient is undefined, so it gives nothing.
Why: The relation requires x not to be zero, and here it is. Dividing negative 6 by 0 is undefined, so the relation simply does not apply. Such points are handled by inspection instead: the point is 6 units from the pole straight down, so r is 6 and theta is 3 pi over 2. The book does exactly this for its third example rather than forcing the formula.
Edge cases
The conversion into polar form asks you to choose a quadrant.
Discussion prompt
What happens at the origin, where there is no quadrant to choose?
Hint: What does the sum of squares give, and what does the tangent give?
Answer:
The sum of squares gives r equal to zero, correctly. The tangent gives zero over zero, which is not merely undefined but indeterminate — it carries no information at all.
That is the analytic shadow of the geometric fact from the previous lesson: the pole has no direction, so no angle can be preferred. Every pair 0 comma theta names it.
The conventional answer is 0 comma 0, chosen for tidiness rather than derived. It is worth being clear that this is a choice: unlike every other point, the origin's polar name is a convention agreed on rather than a value computed.
Section
Section 3
Concept
Replace every x with r cosine theta and every y with r sine theta, then simplify using the Pythagorean Identity and factoring. This direction always works.
The redundancy check matters. If the other factor already produces r equal to zero for some angle, the separate factor adds no points and can be dropped without loss.
Figure (svg): Two columns contrasting the straightforward direction of equation conversion with the awkward one
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 927-928
Picture it
Whether polar helps depends entirely on the shape.
Figure (svg): Four equations shown in both coordinate systems, illustrating that a simple equation in one system may be complicated in the other
Circles through the pole and lines through the pole become very simple. A parabola, which is organised around a directrix and an axis rather than a centre, becomes worse.
Worked example
Example 11.4.3, part 1a. A rectangular circle becomes strikingly simple.
\[ \text{Convert } (x-3)^2 + y^2 = 9 \text{ to polar form.} \]
Substitute
Why: Every x becomes r cosine and every y becomes r sine.
\[ (r \cos - 3) ^{2} + (r \sin) ^{2} = 9 \]
Expand
Why: The square of the binomial gives three terms.
\[ r ^{2} \cos ^{2} - 6 r \cos + 9 + r ^{2} \sin ^{2} = 9 \]
Collapse with the identity
Why: Subtract 9 and group the two squared terms.
\[ r ^{2} - 6 r \cos = 0 \]
Factor
Why: One factor is r; the other gives the circle.
\[ r(r - 6 \cos) = 0 \]
Figure (svg): The solution to Worked example a circle through the pole shown as a ladder of expressions, one row per legal move
\[ r = 6\cos(\theta) \]
Verify: check the discarded factor
Why: Setting theta equal to pi over 2 in r equals 6 cosine theta gives r equal to zero, which is the pole. So the pole is already on the curve and the separate factor r equals zero adds nothing. Discarding it loses no points.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 927-927
Faded example
Convert x squared plus y squared equal to 25 into polar form.
Fill in the blanks
(r\cos\theta)^2 + (r\sin\theta)^2 = 25 \;\Longrightarrow\; r^2\left(\cos^2 + \sin^2\right) = r^2 \cdot 1 = 25 \;\Longrightarrow\; r = 5
Why: The Pythagorean Identity collapses the bracket to 1, leaving r squared equal to 25 and so r equal to 5, taking the positive root. This is the clearest case of polar form being simpler: a circle centred at the pole is just a constant radius.
Worked example
Example 11.4.3, part 1c. The same method, a much uglier result.
\[ \text{Convert } y = x^2 \text{ to polar form.} \]
Substitute
Why: The right side is squared, so the r is too.
\[ r \sin = r ^{2} \cos ^{2} \]
Move everything to one side and factor
Why: Factor out an r.
\[ r(r \cos ^{2} - \sin) = 0 \]
Solve the second factor for r
Why: Divide by the squared cosine.
\[ r = \sin / \cos ^{2} \]
Rewrite with reciprocal functions
Why: Split into a secant and a tangent.
\[ r = \sec \theta \tan \theta \]
Figure (svg): The solution to Worked example a parabola gets worse shown as a ladder of expressions, one row per legal move
\[ r = \frac{\sin(\theta)}{\cos^2(\theta)} = \sec(\theta)\tan(\theta) \]
Verify: check the division was safe
Why: Dividing by squared cosine is only legitimate where the cosine is not zero. If the cosine were zero, the equation r squared cosine squared equals r sine would force the sine to be zero as well, and no angle has both zero. So nothing was lost, and the book makes exactly this argument.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 927-928
Trap
\[ r\cos^2(\theta) = \sin(\theta) \;\Longrightarrow\; r = \frac{\sin(\theta)}{\cos^2(\theta)} \quad \text{(no comment)} \]
Divide both sides by the squared cosine
Why: It is the obvious move and it produces the desired form.
The move happens to be safe here, but nothing in the working says so. As a general rule, dividing by an expression that may be zero deletes solutions silently.
State why the divisor cannot vanish, then divide. Here: if the cosine were zero the equation would force the sine to be zero too, and no angle has both.
Where the divisor genuinely can vanish, split into cases — the case where it is zero and the case where it is not — and solve each.
This is the habit that separates a correct answer from a correct-looking one. Dividing is the only algebraic move in this lesson that can lose solutions, and it deserves a sentence every time.
Matching
The substitution is mechanical; the simplification is where the character shows.
Match the pairs
Why: The circle centred at the pole and the line through the pole both become as simple as an equation can be. The two axis-parallel lines become reciprocal functions, because a vertical line is at constant x but at steadily changing distance from the pole. That pattern is general: shapes centred on the pole simplify, and shapes aligned with the axes do not.
Prediction
You convert a rectangular equation and obtain the factored form r times some expression equals zero.
Predict first
When may the factor r equals 0 be discarded?
Correct: When the other factor already yields r equal to 0 for some angle.
Why: The factor r equals 0 contributes exactly one point, the pole. If the other factor also passes through the pole for some angle, that point is already included and the factor is redundant. If it does not, discarding the factor would delete a genuine point of the curve. So the check is specific rather than automatic, and it takes one substitution.
Explain it to yourself
The rectangular-to-polar direction is described as mechanical.
Discussion prompt
Explain what makes it mechanical, and why the reverse direction is not.
Hint: In each direction, is there a single substitution available?
Answer:
Forwards, there is one substitution that always applies: every x becomes r cosine theta and every y becomes r sine theta. Nothing needs to be decided, and the rest is algebra.
Backwards, there is no such substitution. Solving the relations for r and theta gives a plus-or-minus square root and an arctangent with an added multiple of pi, neither of which is usable in an equation.
So the reverse direction becomes a search: rearrange until one of r squared, r cosine theta, r sine theta, or y over x happens to appear. That is a strategy rather than a procedure, and it may require multiplying by r or squaring, both of which change the equation. Being mechanical is precisely the difference between having a substitution and having to hunt for one.
Section
Section 4
Concept
There is no single substitution. Instead, manipulate the polar equation until one of four expressions shows up, then replace it with its rectangular equivalent.
Both of those moves can change the set of solutions, so each obliges a check afterwards. In polar coordinates the check often succeeds anyway, because an apparently new branch turns out to be the same curve under a different name.
Figure (svg): The three algebraic moves used when converting a polar equation, each labelled with the risk it carries to the solution set
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 928-929
Picture it
Three moves, three different consequences for the solution set.
Figure (svg): The three algebraic moves used when converting a polar equation, each labelled with the risk it carries to the solution set
Multiplying and squaring add points, and dividing removes them. Adding a point is easier to check for than losing one, which is why dividing deserves the most care.
Worked example
Example 11.4.3, part 2a. A negative constant radius.
\[ \text{Convert } r = -3 \text{ to rectangular form.} \]
Square both sides
Why: This makes r squared appear.
\[ r ^{2} = 9 \]
Substitute
Why: The sum of squares replaces r squared.
\[ x ^{2} + y ^{2} = 9 \]
Ask what squaring added
Why: The squared equation also allows r equal to positive 3.
\[ r = +- 3 \]
Check that nothing new was gained
Why: Every point with r equal to 3 has a name with r equal to negative 3.
Figure (svg): The solution to Worked example squaring both sides shown as a ladder of expressions, one row per legal move
\[ x^2 + y^2 = 9 \]
Verify: justify the extra branch
Why: A point named 3 comma theta is also named negative 3 comma theta plus pi, so it satisfies r equals negative 3 under a different name. The branch squaring introduced describes the same set of points in the plane, so the conversion is sound.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 928-928
Faded example
Convert r equal to 4 cosine theta into rectangular form.
Fill in the blanks
r^2 = 4r\cos\theta \;\Longrightarrow\; x^2 + y^2 = 4x \;\Longrightarrow\; (x-2)^2 + y^2 = 4
Why: Multiplying by r creates an r squared and an r cosine, both convertible. Completing the square in x gives a circle of radius 2 centred at 2 comma 0 — a circle through the origin, which is exactly the family that becomes simple in polar form.
Worked example
Example 11.4.3, part 2c. The cardioid, and the price of converting it.
\[ \text{Convert } r = 1 - \cos(\theta) \text{ to rectangular form.} \]
Multiply through by r
Why: This creates an r squared and an r cosine.
\[ r ^{2} = r - r \cos \]
Rearrange to isolate the lone r
Why: Move the r cosine across.
\[ r = r ^{2} + r \cos \]
Square both sides
Why: Now every term is convertible.
\[ r ^{2} = (r ^{2} + r \cos) ^{2} \]
Substitute
Why: Sum of squares for r squared, and x for r cosine.
\[ x ^{2} + y ^{2} = (x ^{2} + y ^{2} + x) ^{2} \]
Figure (svg): The solution to Worked example multiplying by r shown as a ladder of expressions, one row per legal move
\[ x^2 + y^2 = \left(x^2 + y^2 + x\right)^2 \]
Verify: check the multiplication by r
Why: Multiplying by r makes r equal to zero a solution. But the original equation already gives r equal to zero when theta is zero, so the pole was on the curve to begin with and nothing new was introduced. The squaring is justified by the same name-swapping argument as the previous example.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 928-929
Error analysis
A student converts the polar equation theta equal to 4 pi over 3.
Annotate
On: \( \theta = \tfrac{4\pi}{3} \;\Longrightarrow\; \tan\theta = \sqrt{3} \;\Longrightarrow\; y = x\sqrt{3} \quad \text{(done)} \)
The equation theta equals a constant leaves r free, so it describes the entire line through the pole in that direction, not a ray. That is exactly the set y equals x root three, so the conversion is right — but only the geometric argument establishes it.
Sorting
Each polar equation needs a different first step.
Sort into buckets
Sort each equation by the move it needs.
Prediction
You multiply a polar equation through by r during a conversion.
Predict first
What is the one point you must check afterwards?
Correct: The pole.
Why: Multiplying by r makes r equal to zero satisfy the new equation whatever theta is, so the pole is added to the solution set. If the original curve already passed through the pole for some angle, nothing changed; if it did not, the conversion has introduced a spurious point. One substitution settles it, and it is the only check the move requires.
Counterexample
A student proposes: squaring both sides of an equation always changes the set of points it describes.
Discussion prompt
Give a case where it does not, and say what makes polar coordinates unusually forgiving here.
Hint: What does a point with a negative r also get called?
Answer:
Take r equals negative 3. Squaring gives r squared equal to 9, which allows both negative 3 and positive 3. On the face of it that is a strictly larger solution set.
But every point named 3 comma theta is also named negative 3 comma theta plus pi, so it already satisfied the original equation under a different name. The set of points in the plane is unchanged, even though the set of ordered pairs is not.
What makes polar coordinates forgiving is exactly the non-uniqueness from the previous lesson. A branch that looks new is often the same curve wearing a different label, and the check is whether every point of the new branch has a name satisfying the original equation. In rectangular coordinates, where names are unique, no such rescue is available and squaring genuinely does enlarge the solution set.
Section
Section 5
Concept
A shape has no intrinsic complexity; it has a complexity relative to a coordinate system. Choosing the system that matches the shape's symmetry is a real technique, not an aesthetic preference.
The book's closing remark on the section is exactly this point, and it is the argument for the next lesson: rather than convert an awkward polar equation, learn to graph it where it stands.
Figure (svg): Four equations shown in both coordinate systems, illustrating that a simple equation in one system may be complicated in the other
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 929-929
Picture it
One is a procedure and the other is a search.
Figure (svg): Two columns contrasting the straightforward direction of equation conversion with the awkward one
That asymmetry is a reason to develop tools that work directly in polar coordinates rather than always translating back to familiar ground.
Worked example
Example 11.4.3, part 1b. The simplest possible polar equation.
\[ \text{Convert } y = -x \text{ to polar form.} \]
Substitute
Why: Both sides pick up an r.
\[ r \sin = -r \cos \]
Collect and factor
Why: Move everything to one side.
\[ r(\cos + \sin) = 0 \]
Solve the second factor
Why: The cosine and sine are opposite when the angle is negative pi over 4 plus a multiple of pi.
\[ \theta = -\frac{\pi}{4} + \pi k \]
Argue geometrically
Why: With r free, one such angle traces the whole line.
\[ \theta = -\frac{\pi}{4} \]
Figure (svg): The solution to Worked example a line through the pole shown as a ladder of expressions, one row per legal move
\[ \theta = -\tfrac{\pi}{4} \]
Verify: confirm the whole line is covered
Why: In theta equal to negative pi over 4, the variable r is free and may be positive, negative or zero. Positive r gives the fourth-quadrant half of the line, negative r gives the second-quadrant half, and r equal to zero gives the origin. So the single equation traces the entire line.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 927-927
Sorting
Look at what the shape is organised around.
Sort into buckets
Sort each curve.
Worked example
Deciding which system a problem belongs in, before doing any work.
\[ \text{Which system suits } x^2 + y^2 = 4x \text{ better, and why?} \]
Identify the shape
Why: Complete the square in x.
\[ (x - 2) ^{2} + y ^{2} = 4 \]
Note where it sits
Why: A circle of radius 2 centred at 2 comma 0.
Predict the polar form
Why: A circle through the pole should be very simple.
\[ \text{expect } r = a \cos \]
Convert and confirm
Why: Substituting and dividing by r gives it.
\[ r = 4 \cos \theta \]
Figure (svg): The solution to Worked example choosing before converting shown as a ladder of expressions, one row per legal move
\[ x^2 + y^2 = 4x \;\Longleftrightarrow\; r = 4\cos(\theta) \]
Verify: compare the two forms
Why: The rectangular form needs completing the square to be recognisable; the polar form is a single product. Any question about this curve — where it crosses an axis, its maximum distance from the origin — is easier in the polar form, and that is the practical payoff of choosing well.
Trap
\[ r = 1 - \cos(\theta) \;\Longrightarrow\; x^2 + y^2 = \left(x^2+y^2+x\right)^2 \]
Convert to rectangular form because it is more familiar
Why: Rectangular coordinates are where all previous graphing was done.
But the result is a fourth-degree equation that cannot be solved for y, cannot be sketched by hand, and reveals nothing. The original was a one-line description of a heart-shaped curve.
Leave it in polar form and develop the tools to graph it there. That is what the next lesson does.
The book's own conclusion: nice things in rectangular coordinates can turn ugly in polar coordinates, and the reverse. Conversion is a tool, not an obligation.
The practical rule: convert when the target system is simpler for the question you are asking, and not otherwise. Familiarity is not the same as simplicity.
Prediction
You convert the polar equation r equal to a constant into rectangular form, and then convert the result back.
Predict first
Do you necessarily get the equation you started with?
Correct: No, you may get the version with the opposite sign.
Why: Starting from r equal to negative 3, squaring gives the circle of radius 3, and converting that back naturally gives r equal to 3. The two equations describe the same set of points but are different equations, because squaring destroyed the sign information. Round-tripping through a non-reversible step does not always return the original expression, though it does return the original curve.
Matching
Symmetry is what decides.
Match the pairs
Why: The first two are the two things a polar pair can hold constant, so they are the two shapes polar coordinates describe most naturally. Everything else requires a genuine relationship between r and theta, and how ugly that relationship is depends on how far the shape is from being centred on the pole.
Real world
A directional microphone's sensitivity is specified by a polar pattern: how strongly it picks up sound arriving from each direction. The cardioid pattern, r equal to 1 minus cosine theta, is the most common.
Discussion prompt
Explain why the specification is given in polar form, and what the rectangular form would cost.
Hint: What does the manufacturer actually want to communicate?
Answer:
The quantity being described is sensitivity as a function of direction, which is exactly a radius as a function of an angle. Polar form is not a convenient choice here — it is the literal shape of the data.
The rectangular form of the cardioid is x squared plus y squared equal to the square of x squared plus y squared plus x: a fourth-degree implicit equation that cannot be solved for y, cannot be read at a glance, and hides the one fact that matters, which is the null at the rear.
In the polar form that fact is immediate: at theta equal to zero the radius is zero, so the microphone is deaf directly behind it. That is the entire reason the cardioid is chosen for stage use, and it is a one-line consequence of the polar equation and effectively invisible in the rectangular one.
The general point is worth carrying: the right coordinate system makes the important property obvious, and a specification is written in whichever system does that.
Comparison
Fill the blanks from memory. The asymmetry between the columns is the point.
Comparison matrix
| Rectangular into polar | Polar into rectangular | |
|---|---|---|
| for points | sum of squares and tangent, then choose the quadrant | substitute into x = r cos and y = r sin |
| difficulty for points | needs a sketch to fix the quadrant | direct, no decisions |
| for equations | substitute for x and y everywhere | rearrange until a standard form appears |
| difficulty for equations | mechanical | a search, and may need squaring or multiplying by r |
| what to check | whether a factor of r is redundant | what multiplying and squaring added |
The easy direction for points is the hard direction for equations, and the reverse. Noticing which of the four boxes you are in tells you what to expect before you begin.
Pattern
Five moves, covering both objects and both directions.
Dividing by an expression is the only move that can lose solutions. Say why the divisor cannot be zero, every time.
OpenStax Algebra and Trigonometry 2e, §10.3 Polar Coordinates §10.3
Check
A point, into polar form.
Check your understanding
Convert (-1, root 3) to polar form with r at least 0 and theta from 0 up to 2 pi.
Answer: B
Why: The sum of squares is 1 plus 3, so r is 2. The tangent's magnitude is root three, giving a reference angle of pi over 3. Negative x with positive y is quadrant two, so the angle is pi minus pi over 3, namely two thirds of pi.
Check
An equation, into polar form.
Check your understanding
What is the polar form of x squared plus y squared equal to 6y?
Answer: B
Why: The left side is r squared and the right is 6 times r sine theta. Dividing by r, which is legitimate since the pole is recovered at theta equal to zero, gives r equal to 6 sine theta. This is a circle of radius 3 centred at 0 comma 3, passing through the origin.
Check
An equation, into rectangular form.
Check your understanding
What is the rectangular form of r cosine theta equal to negative 2?
Answer: B
Why: The expression r cosine theta is one of the four standard forms and converts directly to x. So the equation is x equal to negative 2, a vertical line two units to the left of the origin. No multiplying or squaring is needed.
Real world
A robot vacuum builds a map with a spinning laser rangefinder. The sensor produces, several thousand times a second, a pair: an angle of the spinning head and a distance to whatever it hit. The mapping software must assemble these into a floor plan and merge it with the map built from an earlier position in the room.
Discussion prompt
Explain which coordinate system each stage of that pipeline wants, and why the conversion cannot be skipped.
Hint: What does the sensor produce, and what does merging two maps require?
Answer:
The sensor is inherently polar: it measures a bearing from its own spin angle and a range from the time of flight. There is no conversion at the source, and imposing rectangular coordinates on the raw stream would only lose information.
Merging two maps is inherently rectangular. The two scans were taken from different poles, so their polar pairs are not comparable at all — the same wall has different ranges and bearings from each position. Only after both are converted into a common rectangular frame, and translated by the robot's movement, can the two describe the same wall.
So the conversion is unavoidable, and it runs in one direction: polar in, rectangular out, which is the easy direction — a pair of multiplications per reading, with no quadrant decision and no ambiguity. That is not a coincidence. Sensors measure polar because rotation is what a sensor does; maps are stored rectangular because translation is what merging needs; and the conversion between them is cheap precisely in the direction the pipeline needs it.
The reverse direction, rectangular into polar, appears only when the software must predict what the sensor should see from a hypothetical position — and there it needs the quadrant care of this lesson, done correctly for millions of points.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
You convert (-5, -5) to polar form and your calculator's arctangent of y over x returns pi over 4. What should you do?
Correct: Add pi, because the point is in quadrant three.
\[ (-5, -5) \;\Longrightarrow\; r = 5\sqrt{2}, \; \theta = \tfrac{\pi}{4} + \pi = \tfrac{5\pi}{4} \]
Why: The two minus signs cancel in the quotient, so the tangent is 1 and the arctangent returns pi over 4 — a quadrant one angle. But the point has both coordinates negative and lies in quadrant three, reached by adding pi to the reference angle, giving 5 pi over 4. Negating would give quadrant four and subtracting from a full turn would too. Plotting the point first makes this immediate.
Explain it
They have converted three points from rectangular to polar form and got the quadrant wrong on the two that were not in quadrant one. They believe the calculator is broken.
Discussion prompt
In no more than five sentences, explain what the arctangent can and cannot tell them, and give them a reliable method.
Hint: What is the range of the arctangent, and how much of the plane does it cover?
Answer:
A usable answer: the arctangent's range runs only from negative pi over 2 to pi over 2, which covers quadrants one and four and nothing else. A point in quadrant two or three has a tangent identical to one in quadrant four or one, and the calculator has no way to tell them apart from the quotient alone.
So treat what it returns as the reference angle, not the answer. Then plot the point, see which quadrant it is actually in, and adjust: pi minus for quadrant two, pi plus for quadrant three, two pi minus for quadrant four.
The calculator is not broken — it is answering a question with two correct answers, and it has to pick one.
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 quadrant is fixed by plotting before computing and treating the arctangent as a reference angle only. The forward equation conversion is fixed by the single substitution and the Pythagorean Identity. The reverse is fixed by memorising the four standard expressions to hunt for. Choosing a system is fixed by asking whether the shape is centred on a point or aligned with the axes. 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 draw the two coordinate systems overlaid, with the pole at the origin and the polar axis along the positive x-axis, mark one point, and write all four relations of Theorem 11.7 beside it. Underneath, write in one line why the tangent relation excludes the case where x is zero. In the middle of the page draw four points, one in each quadrant, all with the same tangent magnitude, and beside each write the adjustment its reference angle needs. In the bottom left, convert x squared plus y squared equal to 6x into polar form, showing the substitution, the identity, the factoring, and the check on the discarded factor. In the bottom right, convert r equal to 1 minus cosine theta into rectangular form, and note beside each step whether it added points, lost points, or neither. Finally, circle the one algebraic move on the page that can lose solutions.
The circled move is dividing. Multiplying by r and squaring both add points, which is easy to check for; dividing removes them silently, which is not. That asymmetry is why the divisor deserves a sentence every time.
Recap
Five things, and the second is where nearly all the marks in this section are won and lost.
| If the question says | Your first move is |
|---|---|
| Convert this polar point | Substitute into x = r cos and y = r sin |
| Convert this rectangular point | Plot it, then find r and the reference angle |
| Convert this rectangular equation | Substitute for x and y everywhere |
| Convert this polar equation | Hunt for r squared, r cos, r sin or the tangent |
| You multiplied by r or squared | Check what that added, especially the pole |
The cardioid's rectangular form is a fourth-degree equation nobody would choose to graph. Its polar form is one line. The next lesson takes that seriously and develops the tools to graph polar equations where they stand, without converting at all.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.4 Polar Coordinates §11.4, pp. 924-930 — everything on these slides traces back here
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