Polar Coordinates Introduction

In Cartesian coordinates every point in the plane is fixed by an ordered pair $(x,\,y)$(x,y) — a horizontal displacement and a vertical displacement. Polar coordinates describe the same plane in a completely different language: a distance and a direction. A point is written $(r,\,\theta)$(r,θ), where $r$r is how far it lies from a fixed origin and $\theta$θ is the angle the line to it makes with a fixed reference ray. This is the natural language for anything built around rotation — circles, spirals, the petals of a rose curve, planetary orbits — and it is the gateway to one of the most elegant results in the AQA Further Mathematics pure content: the polar area integral.

---

1 Spec mapping

This lesson opens the polar coordinates strand of the compulsory pure content examined on Paper 1 and Paper 2 of AQA Further Mathematics (7367) — each paper 2 hours, 100 marks, $33\tfrac{1}{3}\%$3331​% of the qualification. Setting up the coordinate system and the conversion identities is the foundation for the two assessed skills that follow: sketching polar curves and computing the area $\tfrac12\int r^2\,\mathrm{d}\theta$21​∫r2dθ. The content here is overwhelmingly AO1 (selecting and carrying out a routine procedure — converting between systems), with the quadrant-fixing and "interpret the locus" parts reaching into AO2 (reasoning and communicating). Polar work appears in the compulsory papers, so every Further Maths candidate meets it regardless of the Paper 3 options they choose.

---

2 Core theory — the polar coordinate system

Fix a point $O$O, the pole (it plays the role of the origin), and a fixed ray from $O$O, the initial line (conventionally the positive $x$x-axis). A point $P$P is then located by:

$r = |OP| \quad\text{(the radial coordinate)}, \qquad \theta = \text{angle from the initial line to } OP \ \text{(the angular coordinate).}$r=∣OP∣(the radial coordinate),θ=angle from the initial line to OP (the angular coordinate).

The angle $\theta$θ is measured anticlockwise-positive and almost always in radians in Further Maths. We write the pair as $(r,\,\theta)$(r,θ).

Element	Symbol	Convention
Pole (origin)	$O$O	fixed reference point
Initial line	—	usually the positive $x$x-axis
Radial coordinate	$r$r	distance $
Angular coordinate	$\theta$θ	radians, anticlockwise positive

The conversion identities

Drop a perpendicular from $P$P to the initial line, meeting it at the foot $N$N. In the right-angled triangle $ONP$ONP the hypotenuse is $OP = r$OP=r and the angle at $O$O is $\theta$θ; the horizontal side is $ON = x$ON=x and the vertical side is $NP = y$NP=y. Elementary trigonometry — $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{x}{r}$cosθ=hypotenuseadjacent​=rx​ and $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{y}{r}$sinθ=hypotenuseopposite​=ry​ — rearranges instantly to the polar-to-Cartesian formulae:

$\boxed{\,x = r\cos\theta, \qquad y = r\sin\theta\,}$x=rcosθ,y=rsinθ​

These two equations are the workhorses of the whole topic; commit them to memory, because they are also the parametrisation used for tangents and for arc length later in the course. Squaring and adding, and using the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$cos2θ+sin2θ=1:

$x^2 + y^2 = r^2\cos^2\theta + r^2\sin^2\theta = r^2\big(\cos^2\theta + \sin^2\theta\big) = r^2.$x2+y2=r2cos2θ+r2sin2θ=r2(cos2θ+sin2θ)=r2.

So $r^2 = x^2 + y^2$r2=x2+y2, the single most-used identity in the whole topic. Dividing the two conversion formulae gives the angle:

$\boxed{\,r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}\,}$r=x2+y2​,tanθ=xy​​

Quadrant warning. $\theta = \arctan\!\left(\tfrac{y}{x}\right)$θ=arctan(xy​) returns a value in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$(−2π​,2π​), so it is only directly correct when $x>0$x>0. If $x<0$x<0 add $\pi$π (the point is in the second or third quadrant); on the $y$y-axis use $\theta = \pm\tfrac{\pi}{2}$θ=±2π​. Always sketch the point first and read off the quadrant — never trust the calculator's $\arctan$arctan blindly.

---

3 Worked examples with mark scheme

Worked Example 1 — polar $\to$→ Cartesian

Convert the polar point $\left(4,\,\tfrac{\pi}{3}\right)$(4,3π​) to Cartesian coordinates.

$\begin{aligned} x &= r\cos\theta = 4\cos\tfrac{\pi}{3} = 4\cdot\tfrac12 = 2, \\[2pt] y &= r\sin\theta = 4\sin\tfrac{\pi}{3} = 4\cdot\tfrac{\sqrt3}{2} = 2\sqrt3. \end{aligned}$xy​=rcosθ=4cos3π​=4⋅21​=2,=rsinθ=4sin3π​=4⋅23​​=23​.​

$\therefore\ (x,y) = \big(2,\ 2\sqrt3\big).$∴ (x,y)=(2, 23​).

Mark scheme: M1 apply $x=r\cos\theta,\ y=r\sin\theta$x=rcosθ, y=rsinθ with the given $r,\theta$r,θ; A1 $x=2$x=2; A1 $y=2\sqrt3$y=23​ (exact surd, not $3.46$3.46).

Worked Example 2 — Cartesian $\to$→ polar (quadrant care)

Convert the Cartesian point $(-3,\,3)$(−3,3) to polar coordinates with $r>0,\ 0\le\theta<2\pi$r>0, 0≤θ<2π.

$r = \sqrt{(-3)^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2.$r=(−3)2+32​=9+9​=18​=32​.

The naive calculator value is $\arctan\!\left(\tfrac{3}{-3}\right) = \arctan(-1) = -\tfrac{\pi}{4}$arctan(−33​)=arctan(−1)=−4π​. But $(-3,3)$(−3,3) lies in the second quadrant ($x<0,\ y>0$x<0, y>0), so add $\pi$π:

$\theta = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}. \qquad \therefore\ (r,\theta) = \left(3\sqrt2,\ \tfrac{3\pi}{4}\right).$θ=π−4π​=43π​.∴ (r,θ)=(32​, 43π​).

Mark scheme: M1 $r=\sqrt{x^2+y^2}$r=x2+y2​; A1 $r=3\sqrt2$r=32​; M1 evaluate $\tan\theta=\tfrac{y}{x}$tanθ=xy​ and adjust for the correct quadrant; A1 $\theta=\tfrac{3\pi}{4}$θ=43π​. The quadrant-adjustment mark is the one most candidates drop.

Worked Example 3 — converting a Cartesian equation to a polar locus

Express $x^2 + y^2 = 4x$x2+y2=4x in polar form and identify the curve.

Substitute $x^2+y^2=r^2$x2+y2=r2 and $x = r\cos\theta$x=rcosθ:

$\begin{aligned} r^2 &= 4r\cos\theta \\ r^2 - 4r\cos\theta &= 0 \\ r\,(r - 4\cos\theta) &= 0 \ \implies\ r = 0 \ \text{ or }\ r = 4\cos\theta. \end{aligned}$r2r2−4rcosθr(r−4cosθ)​=4rcosθ=0=0 ⟹ r=0  or  r=4cosθ.​

The factor $r=0$r=0 is just the pole, which already lies on $r=4\cos\theta$r=4cosθ (at $\theta=\tfrac{\pi}{2}$θ=2π​), so the curve is

$\boxed{\,r = 4\cos\theta\,}.$r=4cosθ​.

Completing the square in the Cartesian form, $(x-2)^2 + y^2 = 4$(x−2)2+y2=4: a circle of radius $2$2, centre $(2,0)$(2,0) — i.e. diameter $4$4 sitting on the initial line and passing through the pole.

Mark scheme: M1 substitute $r^2$r2 and $r\cos\theta$rcosθ; M1 factor out $r$r (do not just "divide by $r$r" and silently lose the pole); A1 $r=4\cos\theta$r=4cosθ; B1 identify circle, centre $(2,0)$(2,0), radius $2$2.

Worked Example 4 — a polar equation back to Cartesian

Express the polar equation $r = 6\sin\theta$r=6sinθ in Cartesian form, and hence describe the curve.

The trick whenever the equation contains $r\sin\theta$rsinθ or $r\cos\theta$rcosθ is to multiply through by $r$r so that every term becomes one of the recognisable groups $r^2,\ r\cos\theta,\ r\sin\theta$r2, rcosθ, rsinθ:

$\begin{aligned} r &= 6\sin\theta \\ r^2 &= 6\,r\sin\theta \qquad (\times\, r) \\ x^2 + y^2 &= 6y \qquad (r^2 = x^2+y^2,\ r\sin\theta = y). \end{aligned}$rr2x2+y2​=6sinθ=6rsinθ(×r)=6y(r2=x2+y2, rsinθ=y).​

Now complete the square in $y$y:

$x^2 + (y-3)^2 = 9.$x2+(y−3)2=9.

A circle of radius $3$3, centre $(0,3)$(0,3), passing through the pole.

Mark scheme: M1 multiply by $r$r; M1 replace $r^2$r2 and $r\sin\theta$rsinθ with their Cartesian groups; A1 $x^2+y^2=6y$x2+y2=6y; A1 complete the square and identify the circle. Notice that multiplying by $r$r here is the inverse operation to the dangerous "divide by $r$r" — it can introduce the pole $r=0$r=0 as a spurious solution, but since the pole already lies on the circle nothing is gained or lost.

---

4 Specimen-style exam question

(Specimen-style.) The point $P$P has Cartesian coordinates $(-2,\,-2\sqrt3)$(−2,−23​).

(a) Find polar coordinates of $P$P with $r>0$r>0 and $-\pi<\theta\le\pi$−π<θ≤π. (3 marks) (b) Hence write down the polar coordinates of the reflection of $P$P in the initial line. (2 marks)

Model solution.

(a) $r = \sqrt{(-2)^2 + (-2\sqrt3)^2} = \sqrt{4+12} = \sqrt{16} = 4.$r=(−2)2+(−23​)2​=4+12​=16​=4. The point is in the third quadrant ($x<0,\ y<0$x<0, y<0). The reference angle is $\arctan\!\left(\tfrac{2\sqrt3}{2}\right) = \arctan\sqrt3 = \tfrac{\pi}{3}$arctan(223​​)=arctan3​=3π​; in the third quadrant with $-\pi<\theta\le\pi$−π<θ≤π this is $\theta = -\left(\pi - \tfrac{\pi}{3}\right) = -\tfrac{2\pi}{3}.$θ=−(π−3π​)=−32π​.

$\therefore\ P = \left(4,\ -\tfrac{2\pi}{3}\right).$∴ P=(4, −32π​).

(b) Reflection in the initial line (the $x$x-axis) sends $\theta\mapsto-\theta$θ↦−θ and leaves $r$r unchanged:

$P' = \left(4,\ \tfrac{2\pi}{3}\right).$P′=(4, 32π​).

---

5 Synoptic links

• A-Level Maths trigonometry. Every conversion rests on right-angled-triangle trig and the identity $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1; the quadrant rule is exactly the CAST/unit-circle reasoning from AS.
• Circles (A-Level Maths). $r=2a\cos\theta$r=2acosθ is the polar form of $(x-a)^2+y^2=a^2$(x−a)2+y2=a2 — completing the square links the two pictures.
• Parametric equations. $x=r(\theta)\cos\theta,\ y=r(\theta)\sin\theta$x=r(θ)cosθ, y=r(θ)sinθ is a parametrisation by angle; this is the bridge used in the tangents lesson where $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm d\theta}{\mathrm dx/\mathrm d\theta}$dxdy​=dx/dθdy/dθ​.
• Complex numbers (Further Maths). Modulus–argument form $z=r(\cos\theta+\mathrm i\sin\theta)$z=r(cosθ+isinθ) is the polar coordinate of the Argand point $(x,y)$(x,y): $r=|z|,\ \theta=\arg z$r=∣z∣, θ=argz. Polar coordinates and complex numbers are two views of one idea, and the quadrant rule for $\theta$θ is identical to the rule for choosing the principal argument $\arg z \in (-\pi,\pi]$argz∈(−π,π]. Mastering the quadrant fix here pays off directly when you compute arguments in the complex-numbers module.
• Areas (Further Maths, next lessons). The conversion identities are the launch-pad for the polar area formula $A=\tfrac12\int_\alpha^\beta r^2\,\mathrm d\theta$A=21​∫αβ​r2dθ: the factor of $r^2$r2 is precisely $x^2+y^2$x2+y2 reappearing, and getting the tracing interval $[\alpha,\beta]$[α,β] right depends on the multiple-representations idea from §10.

---

6 Mark-scheme literacy

Polar conversion questions are short, but the marks are unusually easy to drop because so much of the credit is for communicating the method, not just landing the number. The command words tell you exactly what the examiner is buying.

• "Convert" demands the conversion identities shown explicitly — a bare answer with no $x=r\cos\theta$x=rcosθ line risks losing the method mark even when the number is right.
• The quadrant adjustment is its own mark on Cartesian$\to$→polar questions. Examiners write "correct quadrant" against it; $-\tfrac{\pi}{4}$−4π​ where $\tfrac{3\pi}{4}$43π​ was wanted scores the $r$r mark only.
• Exact surds. Leave $2\sqrt3$23​, $3\sqrt2$32​; a rounded decimal in a "find the exact value" context loses the accuracy mark.
• "Identify the curve" wants the name plus its data — "circle, centre $(2,0)$(2,0), radius $2$2", not just "a circle".
• Follow-through. A wrong $r$r carried correctly into part (b) usually keeps the (b) method marks — so always continue.

---

7 Grade-band model answers

Question. Convert $x^2+y^2-2y=0$x2+y2−2y=0 into polar form, simplifying fully, and describe the curve. (4 marks)

Mid-band.

$r^2 - 2y = 0$r2−2y=0, and $y=r\sin\theta$y=rsinθ, so $r^2 = 2r\sin\theta$r2=2rsinθ, giving $r = 2\sin\theta$r=2sinθ. It is a circle.

Examiner-style commentary. The substitutions are right and $r=2\sin\theta$r=2sinθ is correct, so this earns the method and answer marks. It loses the final mark for the vague "it is a circle" — no centre, no radius — and it divided by $r$r without commenting on the lost $r=0$r=0 factor.

Stronger.

Using $x^2+y^2=r^2$x2+y2=r2 and $y=r\sin\theta$y=rsinθ: $r^2 = 2r\sin\theta$r2=2rsinθ, so $r(r-2\sin\theta)=0$r(r−2sinθ)=0. Hence $r=0$r=0 (the pole) or $r=2\sin\theta$r=2sinθ. The pole already lies on $r=2\sin\theta$r=2sinθ ($\theta=0$θ=0), so the curve is $r=2\sin\theta$r=2sinθ, a circle.

Examiner-style commentary. Factoring rather than cancelling shows control of the $r=0$r=0 subtlety — exactly the AO2 reasoning examiners reward. Still slightly short on the description.

Top-band.

From $x^2+y^2=r^2$x2+y2=r2 and $y=r\sin\theta$y=rsinθ, the equation $x^2+y^2-2y=0$x2+y2−2y=0 becomes $r^2 = 2r\sin\theta$r2=2rsinθ, i.e. $r(r-2\sin\theta)=0$r(r−2sinθ)=0. The factor $r=0$r=0 is the pole, which is also reached by $r=2\sin\theta$r=2sinθ at $\theta=0$θ=0, so no points are lost and the polar equation is $r = 2\sin\theta.$r=2sinθ. Completing the square on the Cartesian form, $x^2+(y-1)^2=1$x2+(y−1)2=1: a circle of radius $1$1, centre $(0,1)$(0,1), passing through the pole, traced once as $\theta$θ runs over $[0,\pi]$[0,π].

Examiner-style commentary. Full marks. Every identity is stated, the $r=0$r=0 factor is handled honestly, and the description is complete and cross-checked against the Cartesian circle — including the $[0,\pi]$[0,π] tracing interval, which pre-empts the double-counting trap that bites in area questions later.

---

8 Common misconceptions

1. "$r$r and $\theta$θ are interchangeable like $x$x and $y$y." They are not symmetric: $r$r is a length and $\theta$θ is an angle, two entirely different kinds of quantity, so the order in the ordered pair is fixed. The point $\left(4,\tfrac{\pi}{3}\right)$(4,3π​) (distance $4$4, angle $\tfrac{\pi}{3}$3π​) is nothing like $\left(\tfrac{\pi}{3},4\right)$(3π​,4) (distance $\tfrac{\pi}{3}\approx1.05$3π​≈1.05, angle $4$4 radians). Read the bracket as (how far, which way).
2. "$\theta=\arctan(y/x)$θ=arctan(y/x) always." The arctangent only returns angles between $-\tfrac{\pi}{2}$−2π​ and $\tfrac{\pi}{2}$2π​, so it can never name a second- or third-quadrant direction on its own. Whenever $x<0$x<0 you must add $\pi$π. Forgetting this is comfortably the single most frequent polar error, and it is worth its own mark on the mark scheme.
3. "A point has one set of polar coordinates." Because a full turn returns you to the same direction, $(r,\theta)$(r,θ) and $(r,\theta+2n\pi)$(r,θ+2nπ) are the same point for every integer $n$n, and signed-$r$r (§10) adds yet more names. Cartesian coordinates are unique; polar coordinates are deliberately not. This non-uniqueness is exactly what makes finding the intersection of two polar curves subtle.
4. "Dividing an equation by $r$r is harmless." Cancelling an $r$r quietly throws away the solution $r=0$r=0, which is the pole. Always factor instead — write $r(r-f(\theta))=0$r(r−f(θ))=0 — and then ask explicitly whether the pole already lies on the remaining curve. Usually it does, but stating it is what earns the reasoning mark.
5. "$r$r can never be negative." Treated as a raw distance it is $\ge 0$≥0, but the Further Maths convention permits a signed $r$r, where a negative value means "measure backwards along the ray", i.e. $(r,\theta)=(-r,\theta+\pi)$(r,θ)=(−r,θ+π). This convention is what allows a single tidy equation such as $r=2\cos 3\theta$r=2cos3θ to sweep out an entire rose.
6. "Degrees and radians are interchangeable here." Polar angles in Further Maths are in radians throughout, so it is $\sin\tfrac{\pi}{3}$sin3π​, never $\sin 60^\circ$sin60∘. A calculator left in degree mode quietly corrupts every coordinate that follows; check the mode before you start.
7. "The pole is at $\theta=0$θ=0." The pole is the single point $r=0$r=0; it has no well-defined angle at all, because every ray passes through it. That is why curves can meet at the pole with completely different $\theta$θ-values, a fact that matters enormously when locating intersections.

---

9 Common errors

• Reading $\arctan$arctan in degrees with the calculator in the wrong mode.
• Squaring carelessly: $(-3)^2=9$(−3)2=9, not $-9$−9; $(2\sqrt3)^2 = 4\cdot3 = 12$(23​)2=4⋅3=12, not $6$6.
• "Simplifying" $\sqrt{18}$18​ to $9\sqrt2$92​ instead of $3\sqrt2$32​ ($\sqrt{18}=\sqrt{9}\,\sqrt2$18​=9​2​).
• Cancelling $r$r and losing the pole (see misconception 4).
• Giving a decimal when an exact surd or exact multiple of $\pi$π is required.
• Putting the angle outside the requested range (e.g. returning $\tfrac{5\pi}{4}$45π​ when $-\pi<\theta\le\pi$−π<θ≤π was demanded).

---

10 Negative $r$r and multiple representations

A signed radial coordinate is read as follows: for $r<0$r<0, the point $(r,\theta)$(r,θ) is plotted a distance $|r|$∣r∣ along the ray pointing the opposite way — that is, you face in the direction $\theta$θ and then step backwards. Algebraically this means

$(-r,\ \theta) = (r,\ \theta+\pi), \qquad (r,\ \theta) = (r,\ \theta + 2n\pi)\ \ (n\in\mathbb Z).$(−r, θ)=(r, θ+π),(r, θ)=(r, θ+2nπ)  (n∈Z).

For example $(-2,\,0)$(−2,0) is the same point as $(2,\,\pi)$(2,π), which is $(-2,0)$(−2,0) in Cartesian: facing along the positive $x$x-axis but stepping back two units lands you on the negative $x$x-axis. This convention is what lets a single equation such as $r=2\cos 3\theta$r=2cos3θ (a three-petal rose) trace a complete petal across the ranges where $r$r dips negative, rather than the curve simply vanishing there — essential reading for the curve-sketching and intersection work that follows. When a question demands $r>0$r>0 you must rewrite any negative-$r$r representation using the first identity above before quoting your final answer.

---

11 A standard-curves reference

The real pay-off of polar coordinates is that families of curves which are awkward in Cartesian form become one-line equations. A circle through the origin, a heart-shaped cardioid, a multi-petalled rose and an ever-widening spiral are all just $r=f(\theta)$r=f(θ) for a simple $f$f. The table below collects the standard forms you should recognise on sight; the next lesson sketches each of them carefully, but it pays to start memorising the shape each equation produces now.

Polar equation	Curve	Key feature
$r=a$r=a	circle, centre $O$O	radius $a$a
$\theta=\alpha$θ=α	line through $O$O	direction $\alpha$α
$r=2a\cos\theta$r=2acosθ	circle	centre $(a,0)$(a,0), through $O$O
$r=2a\sin\theta$r=2asinθ	circle	centre $(0,a)$(0,a), through $O$O
$r=a(1+\cos\theta)$r=a(1+cosθ)	cardioid	cusp at $O$O ($\theta=\pi$θ=π)
$r=a\theta$r=aθ	Archimedean spiral	turns spaced $2\pi a$2πa
$r^2=a^2\cos 2\theta$r2=a2cos2θ	lemniscate	figure-eight

---

12 Going further (stretch)

The conversion identities are a single complex equation in disguise. Writing $z=x+\mathrm iy$z=x+iy and using Euler's formula,

$z = x+\mathrm iy = r\cos\theta + \mathrm i\,r\sin\theta = r\,e^{\mathrm i\theta},$z=x+iy=rcosθ+irsinθ=reiθ,

so a polar coordinate $(r,\theta)$(r,θ) is literally the modulus and argument of the Argand point. The "multiply by $r$r then substitute $r^2=x^2+y^2$r2=x2+y2" trick is then just clearing $|z|^2 = z\bar z$∣z∣2=zzˉ.

A genuine STEP-flavoured extension: not all conics are pole-centred, and the cleanest way to handle a conic with a focus at the origin is in polar form. The focus-directrix polar equation of a conic is

$r = \frac{\ell}{1 + e\cos\theta},$r=1+ecosθℓ​,

with the pole at a focus, semi-latus-rectum $\ell$ℓ and eccentricity $e$e: $e=0$e=0 gives a circle, $0<e<1$0<e<1 an ellipse, $e=1$e=1 a parabola, and $e>1$e>1 a hyperbola. The single parameter $e$e sweeps through every conic section — a unification that is far less obvious in Cartesian form, where each conic looks like a different equation. This is exactly the relation Newton derived for planetary orbits: a single elegant polar formula that encodes Kepler's first law (planets move on ellipses with the Sun at a focus).

To see why $e=1$e=1 is a parabola, set $e=1$e=1, clear the fraction to get $r + r\cos\theta = \ell$r+rcosθ=ℓ, then substitute $r=\sqrt{x^2+y^2}$r=x2+y2​ and $r\cos\theta=x$rcosθ=x:

$\sqrt{x^2+y^2} = \ell - x \ \implies\ x^2+y^2 = \ell^2 - 2\ell x + x^2 \ \implies\ y^2 = \ell^2 - 2\ell x,$x2+y2​=ℓ−x ⟹ x2+y2=ℓ2−2ℓx+x2 ⟹ y2=ℓ2−2ℓx,

which is indeed a parabola opening in the negative-$x$x direction. Carrying out the same elimination for general $e$e and recognising the resulting Cartesian conic is a superb piece of algebra practice, and it shows concretely why polar coordinates — not Cartesian — are the natural language for orbital mechanics and any central-force problem.

---

13 Additional A* practice

P1. Convert $\left(6,\,\tfrac{5\pi}{6}\right)$(6,65π​) to Cartesian coordinates. P2. Convert $(1,\,-\sqrt3)$(1,−3​) to polar form with $r>0,\ -\pi<\theta\le\pi$r>0, −π<θ≤π. P3. Express $y = x\sqrt3$y=x3​ ($x\ge0$x≥0) in polar form. P4. Convert $x^2 + y^2 = 6x - 8y$x2+y2=6x−8y to polar form and identify the curve. P5. A point has polar coordinates $\left(-3,\,\tfrac{\pi}{4}\right)$(−3,4π​). Give an equivalent representation with $r>0$r>0 and $0\le\theta<2\pi$0≤θ<2π, then its Cartesian coordinates.

Worked answers.

P1. $x = 6\cos\tfrac{5\pi}{6} = 6\!\left(-\tfrac{\sqrt3}{2}\right) = -3\sqrt3$x=6cos65π​=6(−23​​)=−33​; $y = 6\sin\tfrac{5\pi}{6} = 6\cdot\tfrac12 = 3$y=6sin65π​=6⋅21​=3. So $(-3\sqrt3,\ 3)$(−33​, 3).

P2. $r=\sqrt{1+3}=2$r=1+3​=2; fourth quadrant ($x>0,\ y<0$x>0, y<0) with reference angle $\arctan\sqrt3=\tfrac{\pi}{3}$arctan3​=3π​, so $\theta=-\tfrac{\pi}{3}$θ=−3π​. Answer $\left(2,\,-\tfrac{\pi}{3}\right)$(2,−3π​).

P3. $r\sin\theta = \sqrt3\,r\cos\theta \Rightarrow \tan\theta=\sqrt3 \Rightarrow \theta=\tfrac{\pi}{3}$rsinθ=3​rcosθ⇒tanθ=3​⇒θ=3π​ (the $x\ge0$x≥0 branch). A half-line through the pole at $\tfrac{\pi}{3}$3π​.

P4. $r^2 = 6r\cos\theta - 8r\sin\theta \Rightarrow r = 6\cos\theta - 8\sin\theta$r2=6rcosθ−8rsinθ⇒r=6cosθ−8sinθ. Completing the square on $(x-3)^2+(y+4)^2 = 25$(x−3)2+(y+4)2=25: a circle, centre $(3,-4)$(3,−4), radius $5$5, through the pole.

P5. $\left(-3,\tfrac{\pi}{4}\right) = \left(3,\tfrac{\pi}{4}+\pi\right) = \left(3,\tfrac{5\pi}{4}\right)$(−3,4π​)=(3,4π​+π)=(3,45π​). Cartesian: $x=3\cos\tfrac{5\pi}{4} = -\tfrac{3\sqrt2}{2}$x=3cos45π​=−232​​, $y=3\sin\tfrac{5\pi}{4} = -\tfrac{3\sqrt2}{2}$y=3sin45π​=−232​​, i.e. $\left(-\tfrac{3\sqrt2}{2},\,-\tfrac{3\sqrt2}{2}\right)$(−232​​,−232​​).

---

14 Board alignment

Aligned to AQA Further Mathematics 7367, compulsory pure (Papers 1 & 2). The same polar-coordinate content and conversion identities appear in Edexcel Further Pure (FP1/Core Pure) and OCR (MEI) Further Maths; conventions are identical, so this lesson transfers directly across boards.

---

15 Visual summary

The diagram shows a point $P$P with both descriptions and the right-angled triangle that links them.

$\boxed{\ x=r\cos\theta,\quad y=r\sin\theta,\quad r^2=x^2+y^2,\quad \tan\theta=\tfrac{y}{x}\ \text{(fix the quadrant)}\ }$ x=rcosθ,y=rsinθ,r2=x2+y2,tanθ=xy​ (fix the quadrant) ​

AQA exam tip. Show the conversion identities explicitly, sketch the point before fixing $\theta$θ, keep answers exact, and factor (never cancel) an $r$r so you don't lose the pole.