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3.13 - Trigonometry and Polar Coordinates

Overview

Every point in the plane has rectangular coordinates (x,y)(x, y): how far right and how far up. Polar coordinates describe the same point with an angle and a distance from the origin. Trigonometry converts between the two systems, and the same idea gives complex numbers a polar form.

The polar coordinate system

The polar grid is built from circles centered at the origin and lines through the origin. The origin is called the pole, and the positive x-axis is called the polar axis. Angles are measured from the polar axis, counterclockwise for positive angles.
A point's polar coordinates are an ordered pair (r,θ)(r, \theta):
  • θ\theta is the measure of an angle in standard position whose terminal ray lies on a line through the point.
  • rr is the radial displacement: the signed distance from the origin along that line.
To plot (r,θ)(r, \theta), rotate to the angle θ\theta, then move rr units out along the terminal ray. In the figure, A=(3,π6)A = \left(3, \frac{\pi}{6}\right) is 3 units out along the π6\frac{\pi}{6} ray, and B=(2,3π4)B = \left(2, \frac{3\pi}{4}\right) is 2 units out along the 3π4\frac{3\pi}{4} ray.
3.13 - Trigonometry and Polar Coordinates

Negative values of rr

When rr is negative, move along the line containing the terminal ray, but in the opposite direction. To plot C=(−2,π3)C = \left(-2, \frac{\pi}{3}\right), face the π3\frac{\pi}{3} ray, then go 2 units backward through the origin (the red dashed line). CC lands in Quadrant III, on the 4π3\frac{4\pi}{3} ray.
So (−2,π3)\left(-2, \frac{\pi}{3}\right) and (2,4π3)\left(2, \frac{4\pi}{3}\right) name the same point. In general, (r,θ)(r, \theta) and (−r,θ+π)(-r, \theta + \pi) are always the same point.

One point, many names

Adding or subtracting 2π2\pi lands on the same ray, so (r,θ)(r, \theta) and (r,θ+2π)(r, \theta + 2\pi) are the same point. Combined with negative values of rr, this gives every point infinitely many polar representations. The origin is (0,θ)(0, \theta) for any θ\theta.

Example 1: Other names for a point

Give three other polar representations of (3,π6)\left(3, \frac{\pi}{6}\right) with −2π≤θ≤2π-2\pi \le \theta \le 2\pi.
  • Subtract 2π2\pi: (3,π6−2π)=(3,−11π6)\left(3, \frac{\pi}{6} - 2\pi\right) = \left(3, -\frac{11\pi}{6}\right).
  • Use −r-r and add π\pi: (−3,π6+π)=(−3,7π6)\left(-3, \frac{\pi}{6} + \pi\right) = \left(-3, \frac{7\pi}{6}\right).
  • Use −r-r and subtract π\pi: (−3,π6−π)=(−3,−5π6)\left(-3, \frac{\pi}{6} - \pi\right) = \left(-3, -\frac{5\pi}{6}\right).
All of these, along with the original, plot at point AA in the figure.

Converting polar to rectangular

Picture a point PP at distance rr from the origin, at angle θ\theta. Dropping a perpendicular to the x-axis makes a right triangle with hypotenuse rr, horizontal leg xx, and vertical leg yy.
3.13 - Trigonometry and Polar Coordinates
From the triangle, cos⁡θ=xr\cos\theta = \frac{x}{r} and sin⁡θ=yr\sin\theta = \frac{y}{r}. Multiplying both by rr:
x=rcos⁡θy=rsin⁡θx = r\cos\theta \qquad y = r\sin\theta
These formulas work for every point, including negative values of rr and angles in any quadrant. The signs take care of themselves.

Example 2: Polar to rectangular

Convert (4,5π6)\left(4, \frac{5\pi}{6}\right) and (−2,π3)\left(-2, \frac{\pi}{3}\right) to rectangular coordinates.
(a)
x=4cos⁡5π6=4(−32)=−23y=4sin⁡5π6=4(12)=2x = 4\cos\frac{5\pi}{6} = 4\left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} \qquad y = 4\sin\frac{5\pi}{6} = 4\left(\frac{1}{2}\right) = 2
The point is (−23,2)\left(-2\sqrt{3}, 2\right), in Quadrant II, as expected for an angle of 5π6\frac{5\pi}{6} with positive rr.
(b)
x=−2cos⁡π3=−2(12)=−1y=−2sin⁡π3=−2(32)=−3x = -2\cos\frac{\pi}{3} = -2\left(\frac{1}{2}\right) = -1 \qquad y = -2\sin\frac{\pi}{3} = -2\left(\frac{\sqrt{3}}{2}\right) = -\sqrt{3}
The point is (−1,−3)\left(-1, -\sqrt{3}\right). That's point CC in the figure: the negative rr put it in Quadrant III.

Converting rectangular to polar

Going the other way, the Pythagorean Theorem gives rr and the tangent ratio gives θ\theta:
r=x2+y2r = \sqrt{x^2 + y^2}θ=arctan⁡(yx) for x>0θ=arctan⁡(yx)+π for x<0\theta = \arctan\left(\frac{y}{x}\right) \text{ for } x > 0 \qquad\qquad \theta = \arctan\left(\frac{y}{x}\right) + \pi \text{ for } x < 0
Why add π\pi? Arctangent only outputs angles between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}, which point into Quadrants I and IV, the right half of the plane. If the point has x<0x < 0, it's in the left half, so adding π\pi turns the angle onto the correct ray.
If x=0x = 0, the point is on the y-axis: use θ=π2\theta = \frac{\pi}{2} when y>0y > 0 and θ=3π2\theta = \frac{3\pi}{2} (or −π2-\frac{\pi}{2}) when y<0y < 0. These formulas give the name with r≥0r \ge 0. Any of the other names for the point is also correct.

Example 3: Rectangular to polar

Convert (−3,3)(-3, 3) and (2,−23)\left(2, -2\sqrt{3}\right) to polar coordinates.
(a) r=(−3)2+32=18=32r = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}. Since x<0x < 0, add π\pi:
θ=arctan⁡(3−3)+π=−π4+π=3π4\theta = \arctan\left(\frac{3}{-3}\right) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}
The point is (32,3π4)\left(3\sqrt{2}, \frac{3\pi}{4}\right). Without the +π+\pi you'd get −π4-\frac{\pi}{4}, which points into Quadrant IV, the wrong side of the origin.
(b) r=22+(−23)2=16=4r = \sqrt{2^2 + \left(-2\sqrt{3}\right)^2} = \sqrt{16} = 4. Since x>0x > 0:
θ=arctan⁡(−232)=arctan⁡(−3)=−π3\theta = \arctan\left(\frac{-2\sqrt{3}}{2}\right) = \arctan\left(-\sqrt{3}\right) = -\frac{\pi}{3}
The point is (4,−π3)\left(4, -\frac{\pi}{3}\right), or equivalently (4,5π3)\left(4, \frac{5\pi}{3}\right).

Example 4: With a calculator (AP-style)

Point PP has rectangular coordinates (−5,−12)(-5, -12). Find two polar representations of PP with 0≤θ<2π0 \le \theta < 2\pi: one with r>0r > 0 and one with r<0r < 0.
Step 1: r=(−5)2+(−12)2=169=13r = \sqrt{(-5)^2 + (-12)^2} = \sqrt{169} = 13.
Step 2: x<0x < 0, so add π\pi. With the calculator in radian mode:
θ=arctan⁡(−12−5)+π=arctan⁡(2.4)+π≈1.176+3.142=4.318\theta = \arctan\left(\frac{-12}{-5}\right) + \pi = \arctan(2.4) + \pi \approx 1.176 + 3.142 = 4.318
Step 3: For a negative rr, use the opposite ray: θ−π≈1.176\theta - \pi \approx 1.176.
Answer: (13,4.318)(13, 4.318) and (−13,1.176)(-13, 1.176). Check the second one: x=−13cos⁡(1.176)≈−13(0.385)=−5.0x = -13\cos(1.176) \approx -13(0.385) = -5.0 and y=−13sin⁡(1.176)≈−13(0.923)=−12.0y = -13\sin(1.176) \approx -13(0.923) = -12.0.

Complex numbers in polar form

A complex number a+bia + bi can be plotted as the point (a,b)(a, b) in the complex plane, where the horizontal axis is the real axis and the vertical axis is the imaginary axis. For example, −1+3 i-1 + \sqrt{3}\,i is the point (−1,3)\left(-1, \sqrt{3}\right).
3.13 - Trigonometry and Polar Coordinates
Like any point, it also has polar coordinates (r,θ)(r, \theta). Substituting a=rcos⁡θa = r\cos\theta and b=rsin⁡θb = r\sin\theta gives the polar form of the complex number:
a+bi=(rcos⁡θ)+i(rsin⁡θ)a + bi = (r\cos\theta) + i(r\sin\theta)
Here r=a2+b2r = \sqrt{a^2 + b^2} is the number's distance from 0 in the complex plane, and θ\theta is found exactly as for any other point.

Example 5: Writing a complex number in polar form

Write z=−1+3 iz = -1 + \sqrt{3}\,i in polar form.
Step 1: a=−1a = -1 and b=3b = \sqrt{3}, so r=1+3=2r = \sqrt{1 + 3} = 2.
Step 2: a<0a < 0, so add π\pi:
θ=arctan⁡(3−1)+π=−π3+π=2π3\theta = \arctan\left(\frac{\sqrt{3}}{-1}\right) + \pi = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}
Answer: z=2cos⁡2π3+i(2sin⁡2π3)z = 2\cos\frac{2\pi}{3} + i\left(2\sin\frac{2\pi}{3}\right). Check: 2cos⁡2π3=−12\cos\frac{2\pi}{3} = -1 and 2sin⁡2π3=32\sin\frac{2\pi}{3} = \sqrt{3}.

Example 6: Polar form back to a+bia + bi

Write 6cos⁡7π6+i(6sin⁡7π6)6\cos\frac{7\pi}{6} + i\left(6\sin\frac{7\pi}{6}\right) in the form a+bia + bi.
a=6cos⁡7π6=6(−32)=−33b=6sin⁡7π6=6(−12)=−3a = 6\cos\frac{7\pi}{6} = 6\left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} \qquad b = 6\sin\frac{7\pi}{6} = 6\left(-\frac{1}{2}\right) = -3
Answer: −33−3i-3\sqrt{3} - 3i.

Common mistakes

  • Forgetting the +π+\pi when x<0x < 0. Arctangent alone always points into Quadrant I or IV.
  • Plotting a negative rr along the terminal ray. A negative rr goes in the opposite direction, through the origin.
  • Swapping the formulas. xx goes with cosine and yy goes with sine.
  • Using degree mode. An angle like 1.176 only makes sense in radians.
  • Thinking a point has only one name. (r,θ)(r, \theta), (r,θ+2π)(r, \theta + 2\pi), and (−r,θ+π)(-r, \theta + \pi) are all the same point.

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