Unit 6, lesson 45

Unit-Circle Coordinates and Exact Values

How an angle picks out a point on the unit circle, why its coordinates are the cosine and sine, and how two special triangles, reference angles and quadrant signs give exact values.

  • Theory part 89
  • Practice part 90
  • Angles and the Unit Circle

Stand at the center of a round table with radius one meter and point at the edge. Where your finger lands depends only on the angle you turn through, and the two numbers that locate that spot are the cosine and the sine of the angle. Calculus asks for these numbers constantly, and it wants them exactly, so it pays to read them straight off a circle and two small triangles.

Key terms

Circle equation

Definition The circle with center (h,k)(h, k) and radius r>0r > 0 is the set of all points (x,y)(x, y) in the plane whose distance from (h,k)(h, k) is rr. By the distance formula a point lies on it exactly when

(x−h)2+(y−k)2=r2.(x - h)^2 + (y - k)^2 = r^2.

In plain words A circle is every point at one fixed distance from the center, and the equation is the distance formula squared.

Example The circle with center (2,−1)(2, -1) and radius 33 has equation

(x−2)2+(y+1)2=9.(x - 2)^2 + (y + 1)^2 = 9.

The point (2,2)(2, 2) is on it, since 02+32=90^2 + 3^2 = 9. The point (4,1)(4, 1) is not, since 22+22=82^2 + 2^2 = 8.

Unit circle

Definition The unit circle is the circle of radius 11 centered at the origin. Its equation is

x2+y2=1.x^2 + y^2 = 1.

Every point (x,y)(x, y) on it satisfies −1≤x≤1-1 \le x \le 1 and −1≤y≤1-1 \le y \le 1, because each square is at most 11.

In plain words It is the simplest circle there is, with the center at the origin and the radius equal to one. With radius 11 the coordinates of a point on it need no rescaling, which is what lets them serve directly as cosine and sine.

Example The point (35,45)\left(\frac{3}{5}, \frac{4}{5}\right) lies on the unit circle because

925+1625=1.\frac{9}{25} + \frac{16}{25} = 1.

The point (0.6,0.7)(0.6, 0.7) does not, since 0.36+0.49=0.850.36 + 0.49 = 0.85, so it sits inside the circle.

Cosine and sine as coordinates

Definition Put an angle θ\theta in standard position, with its vertex at the origin and its initial side along the positive xx-axis, measured in radians, counterclockwise for positive θ\theta and clockwise for negative θ\theta. Its terminal side meets the unit circle at exactly one point P(θ)P(\theta). The cosine and sine of θ\theta are the coordinates of that point,

P(θ)=(cos⁡θ,sin⁡θ).P(\theta) = (\cos \theta, \sin \theta).

So cos⁡θ\cos \theta is the horizontal coordinate and sin⁡θ\sin \theta is the vertical coordinate. Since the point is on the unit circle, for every real θ\theta

cos⁡2θ+sin⁡2θ=1,−1≤cos⁡θ≤1,−1≤sin⁡θ≤1.\begin{gathered} \cos^2 \theta + \sin^2 \theta = 1, \\[0.4em] -1 \le \cos \theta \le 1, \\[0.4em] -1 \le \sin \theta \le 1. \end{gathered}

Here cos⁡2θ\cos^2 \theta means (cos⁡θ)2(\cos \theta)^2.

In plain words Turn through the angle, land on the circle, and read the point. How far right or left you are is the cosine, and how far up or down you are is the sine.

Example A quarter turn, θ=π2\theta = \frac{\pi}{2}, lands at the top point (0,1)(0, 1), so cos⁡π2=0\cos \frac{\pi}{2} = 0 and sin⁡π2=1\sin \frac{\pi}{2} = 1. A half turn, θ=π\theta = \pi, lands at (−1,0)(-1, 0), so cos⁡π=−1\cos \pi = -1 and sin⁡π=0\sin \pi = 0. If an angle θ\theta lands at (35,45)\left(\frac{3}{5}, \frac{4}{5}\right), then cos⁡θ=35\cos \theta = \frac{3}{5} and sin⁡θ=45\sin \theta = \frac{4}{5}.

Coterminal angles

Definition Two angles in standard position are coterminal when they have the same terminal side. In radians, α\alpha and β\beta are coterminal exactly when β=α+2πk\beta = \alpha + 2\pi k for some integer kk. Coterminal angles give the same point on the unit circle, so for every real θ\theta and every integer kk

cos⁡(θ+2πk)=cos⁡θ,sin⁡(θ+2πk)=sin⁡θ.\begin{gathered} \cos(\theta + 2\pi k) = \cos \theta, \\[0.4em] \sin(\theta + 2\pi k) = \sin \theta. \end{gathered}

In plain words Adding or removing full turns brings you back to the same spot, so the coordinates do not change.

Example The angle 13π3\frac{13\pi}{3} equals 4π+π34\pi + \frac{\pi}{3}, two full turns plus π3\frac{\pi}{3}. The angle −5π3-\frac{5\pi}{3} equals −2π+π3-2\pi + \frac{\pi}{3}. Both are coterminal with π3\frac{\pi}{3} and land at the same point as π3\frac{\pi}{3}.

Special right triangles

Definition Two right triangles have side lengths that can be found exactly. Scaled so that the hypotenuse is 11,

  • the isosceles right triangle has angles π4\frac{\pi}{4}, π4\frac{\pi}{4}, π2\frac{\pi}{2} and two legs of length 22\frac{\sqrt{2}}{2},
  • the triangle with angles π6\frac{\pi}{6}, π3\frac{\pi}{3}, π2\frac{\pi}{2} has short leg 12\frac{1}{2} opposite the angle π6\frac{\pi}{6} and long leg 32\frac{\sqrt{3}}{2} opposite the angle π3\frac{\pi}{3}.

Both follow from the Pythagorean theorem. Equal legs aa with 2a2=12a^2 = 1 give a=12=22a = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}. Cutting an equilateral triangle of side 11 in half gives a short leg of 12\frac{1}{2}, and then the long leg is

1−14=32.\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}.

In plain words These two triangles are the only ones you need to memorize. Put one inside the unit circle with its hypotenuse as a radius, and its legs are exactly the coordinates of the corner on the circle.

Example Draw the radius at angle π3\frac{\pi}{3} and drop a vertical segment from its end to the xx-axis. That makes a triangle with hypotenuse 11 and angle π3\frac{\pi}{3} at the origin. The side along the axis is adjacent to that angle, so it is the short leg 12\frac{1}{2}, and the vertical side is the long leg 32\frac{\sqrt{3}}{2}. Therefore

P(π3)=(12,32).P\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).

Exact values

Definition An exact value is a number written precisely with integers, fractions, radicals and constants such as π\pi, rather than as a rounded decimal. For example sin⁡π3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} is exact, while sin⁡π3≈0.8660\sin \frac{\pi}{3} \approx 0.8660 is an approximation. The exact values in the first quadrant are

θ\theta00π6\frac{\pi}{6}π4\frac{\pi}{4}π3\frac{\pi}{3}π2\frac{\pi}{2}
cos⁡θ\cos \theta1132\frac{\sqrt{3}}{2}22\frac{\sqrt{2}}{2}12\frac{1}{2}00
sin⁡θ\sin \theta0012\frac{1}{2}22\frac{\sqrt{2}}{2}32\frac{\sqrt{3}}{2}11

In plain words Keep the square roots and fractions instead of pressing a calculator button. Exact values make cancellations visible and keep later answers exact.

Example The exact values make this simplification immediate,

sin⁡2π6+cos⁡π3=14+12=34.\sin^2 \frac{\pi}{6} + \cos \frac{\pi}{3} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}.

Rounded decimals lose this. Squaring cos⁡π6≈0.8660\cos \frac{\pi}{6} \approx 0.8660 gives 0.7499560.749956, close to but not equal to the true value cos⁡2π6=34\cos^2 \frac{\pi}{6} = \frac{3}{4}.

Reference angle

Definition Let θ\theta be an angle in standard position whose terminal side does not lie on an axis. Its reference angle θref\theta_{\text{ref}} is the acute angle, in (0,π2)\left(0, \frac{\pi}{2}\right), between the terminal side and the xx-axis. For θ\theta in [0,2π)[0, 2\pi) it is

θref={θin quadrant I,π−θin quadrant II,θ−πin quadrant III,2π−θin quadrant IV,\theta_{\text{ref}} = \begin{cases} \theta & \text{in quadrant I}, \\ \pi - \theta & \text{in quadrant II}, \\ \theta - \pi & \text{in quadrant III}, \\ 2\pi - \theta & \text{in quadrant IV}, \end{cases}

and any other angle is first replaced by its coterminal angle in [0,2π)[0, 2\pi). The reference angle gives the sizes of the coordinates,

∣cos⁡θ∣=cos⁡θref,∣sin⁡θ∣=sin⁡θref.\begin{gathered} \lvert \cos \theta \rvert = \cos \theta_{\text{ref}}, \\[0.4em] \lvert \sin \theta \rvert = \sin \theta_{\text{ref}}. \end{gathered}

Reflecting across an axis keeps the size of each coordinate and can change only its sign.

In plain words The reference angle is how far the terminal side is from the horizontal axis. It tells you the numbers, and the quadrant tells you the signs.

Example The angle 5π6\frac{5\pi}{6} lies in quadrant II, so its reference angle is

π−5π6=π6.\pi - \frac{5\pi}{6} = \frac{\pi}{6}.

The point for π6\frac{\pi}{6} is (32,12)\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right). Reflecting it across the vertical axis into quadrant II makes the horizontal coordinate negative and leaves the vertical one alone, so

P(5π6)=(−32,12).P\left(\frac{5\pi}{6}\right) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right).

Quadrant signs

Definition The signs of cos⁡θ\cos \theta and sin⁡θ\sin \theta are the signs of the coordinates in the quadrant where the terminal side of θ\theta lies.

Quadrantcos⁡θ\cos \thetasin⁡θ\sin \theta
Ipositivepositive
IInegativepositive
IIInegativenegative
IVpositivenegative

An angle whose terminal side lies on an axis belongs to no quadrant, and one of its coordinates is 00.

In plain words Right of the vertical axis the cosine is positive, and above the horizontal axis the sine is positive. That is all the table says.

Example Take the negative angle −3π4-\frac{3\pi}{4}. Turning clockwise by 3π4\frac{3\pi}{4} passes the bottom of the circle and stops in quadrant III, where both coordinates are negative. The terminal side makes the angle π4\frac{\pi}{4} with the negative xx-axis, so the isosceles right triangle gives legs of 22\frac{\sqrt{2}}{2}, and

P(−3π4)=(−22,−22).P\left(-\frac{3\pi}{4}\right) = \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right).

As a check, the coterminal angle in [0,2π)[0, 2\pi) is −3π4+2π=5π4-\frac{3\pi}{4} + 2\pi = \frac{5\pi}{4}, which is in quadrant III with reference angle π4\frac{\pi}{4}.

Check yourself

  1. Find the exact point on the unit circle at the angle 2π3\frac{2\pi}{3}.
Show the answer

The angle is in quadrant II with reference angle π−2π3=π3\pi - \frac{2\pi}{3} = \frac{\pi}{3}. The point for π3\frac{\pi}{3} is (12,32)\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), and in quadrant II only the horizontal coordinate turns negative, so the point is (−12,32)\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right).

  1. Find cos⁡(−π6)\cos\left(-\frac{\pi}{6}\right) and sin⁡(−π6)\sin\left(-\frac{\pi}{6}\right) exactly.
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A clockwise turn of π6\frac{\pi}{6} ends in quadrant IV with reference angle π6\frac{\pi}{6}. The cosine is positive and the sine is negative there, so cos⁡(−π6)=32\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} and sin⁡(−π6)=−12\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}.

  1. The angle θ\theta has its terminal side in quadrant II and sin⁡θ=35\sin \theta = \frac{3}{5}. Find cos⁡θ\cos \theta.
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From cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1 we get

cos⁡2θ=1−925=1625,\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25},

so cos⁡θ=±45\cos \theta = \pm \frac{4}{5}. In quadrant II the cosine is negative, so cos⁡θ=−45\cos \theta = -\frac{4}{5}.

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