Worked examples
The question
Question: write the polar point (4, 60°) in rectangular form (x, y).
(r, θ) = (4, 60°)
1 Plot the point
Draw the polar grid: circles 1, 2, 3 around the center, and lines for the angles.
Turn 60° from the right side of the x-axis, going up and round.
Then go out 4.
2 Use x = r cos θ, y = r sin θ
For any polar point: x = r cos θ and y = r sin θ.
They come from the right triangle under the point.
x = r cos θ y = r sin θ
First x: 4 times cos 60°.
x = 4 × cos 60°
Careful!
x goes with cos, y goes with sin.
Swapped, we would get the point mirrored.
Careful: not x = 4 sin 60° ✗
cos 60° is 1/2.
We know it from the special triangles.
cos 60° = 12
Put it in.
x = 4 × 12
4 times 1/2 is 2.
x = 2
Now y: 4 times sin 60°.
y = 4 × sin 60°
sin 60° is √3/2.
We know it from the special triangles.
sin 60° = √32
Put it in.
y = 4 × √32
4 times √3/2 is 2√3.
y = 2√3
So the point is (2, 2√3).
(x, y) = (2, 2√3)
3 Check
Let's check on the grid.
From the point, go straight to the x-axis
and straight to the y-axis.
The grid shows x about 2 and y about 3.5.
2√3 is about 3.5.
The same.
Right!
on the grid: x = 2, y ≈ 3.5
Let's look back.
First: plot the point.
Then: use x = r cos θ, y = r sin θ.
Last: check.
▶ Watch it solved
The question
Question: write the polar point (2, 120°) in rectangular form (x, y).
(r, θ) = (2, 120°)
1 Plot the point
Draw the polar grid: circles 1, 2, 3 around the center, and lines for the angles.
Turn 120° from the right side of the x-axis, going up and round.
Then go out 2.
2 Use x = r cos θ, y = r sin θ
For any polar point: x = r cos θ and y = r sin θ.
They come from the right triangle under the point.
x = r cos θ y = r sin θ
First x: 2 times cos 120°.
x = 2 × cos 120°
Careful!
x goes with cos, y goes with sin.
Swapped, we would get the point mirrored.
Careful: not x = 2 sin 120° ✗
cos 120° is −1/2: the same size as cos 60°, from the special triangles.
It is negative: the point is left of the y-axis.
cos 120° = −12
Put it in.
x = 2 × (−12)
2 times −1/2 is −1.
x = −1
Now y: 2 times sin 120°.
y = 2 × sin 120°
sin 120° is √3/2: the same size as sin 60°, from the special triangles.
It is positive here.
sin 120° = √32
Put it in.
y = 2 × √32
2 times √3/2 is √3.
y = √3
So the point is (−1, √3).
(x, y) = (−1, √3)
3 Check
Let's check on the grid.
From the point, go straight to the x-axis
and straight to the y-axis.
The grid shows x about −1 and y about 1.7.
√3 is about 1.7.
The same.
Right!
on the grid: x = −1, y ≈ 1.7
Let's look back.
First: plot the point.
Then: use x = r cos θ, y = r sin θ.
Last: check.
▶ Watch it solved
The question
Question: write r = 4 sin θ in x and y.
r = 4 sin θ
1 Write the rules
The rules between polar and x, y: x = r cos θ, y = r sin θ, and r² = x² + y²: Pythagoras.
x = r cos θ, y = r sin θ, r2 = x2 + y2
2 Multiply both sides by r
y is r sin θ.
To get r sin θ, we multiply both sides by r.
r × r = 4r sin θ
Careful!
Not only the right side.
The left side is multiplied by r too.
Careful: not r = 4r sin θ ✗
On the left, r times r is r².
r2 = 4r sin θ
3 Use the rules
r² is x² + y².
And r sin θ is y.
So x² + y² = 4y.
x2 + y2 = 4y
4 Complete the square
Move 4y to the left: it becomes − 4y.
Group the y terms.
x2 + (y2 − 4y) = 0
Half of −4 is −2.
Squared: 4.
Add 4 to both sides.
x2 + (y2 − 4y + 4) = 4
−4 ÷ 2 = −2, (−2) × (−2) = 4
The bracket is a full square: (y − 2)².
So x² + (y − 2)² = 4.
x2 + (y − 2)2 = 4
5 Draw it
Draw the polar grid.
A circle: center (0, 2), radius 2.
6 Check
Let's check with one polar point: (4, 90°).
Put θ = 90° in: r = 4 sin 90° = 4.
So the point is (0, 4).
Put it in.
It gives 4, the right side.
Right!
0² + (4 − 2)² = 4
Let's look back.
First: write the rules.
Then: multiply both sides by r.
Then: use the rules.
Then: complete the square.
Then: draw it.
Last: check.
▶ Watch it solved
