Unit 4, lesson 52

The Unit Circle

How the unit circle turns every real number into a point whose coordinates are its cosine and sine, why these values repeat after every full turn, and how symmetry gives exact values all around the circle.

  • Theory part 103
  • Practice part 104
  • Triangles and the Unit Circle

Picture a mark on the rim of a wheel of radius 1 that starts level with the hub on the right. As the wheel turns counterclockwise, the mark’s horizontal and vertical positions, measured from the hub, swing between −1-1 and 11 and repeat every lap. Those two positions are exactly the cosine and the sine of the angle the wheel has turned through. One circle holds every value and every sign of the trigonometric functions, for every real input.

Key terms

Unit circle

Definition The unit circle is the circle of radius 1 centered at the origin. It consists of the points (x,y)(x, y) with

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

In plain words It is the set of all points exactly one unit from the origin. With radius 1, the rr in every trigonometric ratio becomes 1 and drops out.

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, because 0.36+0.49=0.850.36 + 0.49 = 0.85.

Radian arc length

Definition On a circle of radius rr, a central angle of θ\theta radians with 0<θ≤2π0 < \theta \le 2\pi cuts off an arc of length s=rθs = r\theta, as in arc length and area of a sector. On the unit circle r=1r = 1, so

s=θ.s = \theta.

This lets every real number tt name a point P(t)P(t) on the unit circle. Start at (1,0)(1, 0) and travel a distance ∣t∣\lvert t \rvert along the circle, counterclockwise if t>0t > 0 and clockwise if t<0t < 0. The point where you stop is P(t)P(t), the point where the terminal side of the angle of tt radians in standard position meets the unit circle. For t=0t = 0 you stay at P(0)=(1,0)P(0) = (1, 0).

In plain words On the unit circle, the radian measure of an angle is the distance you travel along the circle. One radian is the angle whose arc is as long as the radius, about 57.3∘57.3^\circ.

Example The whole circle has length 2π2\pi, so a quarter of it has length π2\frac{\pi}{2}. Traveling π2\frac{\pi}{2} counterclockwise from (1,0)(1, 0) ends at P(π2)=(0,1)P\left(\frac{\pi}{2}\right) = (0, 1), traveling π\pi ends at P(π)=(−1,0)P(\pi) = (-1, 0), and traveling π2\frac{\pi}{2} clockwise ends at P(−π2)=(0,−1)P\left(-\frac{\pi}{2}\right) = (0, -1). Traveling a distance of 1 sweeps out 1 radian and stops in Quadrant I, since 1<π2≈1.5711 < \frac{\pi}{2} \approx 1.571.

Cosine and sine as coordinates

Definition For every real number θ\theta, the point P(θ)P(\theta) on the unit circle at angle θ\theta radians has coordinates

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

This matches the definition by a terminal point, because here r=1r = 1, so cos⁡θ=x1\cos \theta = \frac{x}{1} and sin⁡θ=y1\sin \theta = \frac{y}{1}. The other four functions follow,

tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ,sec⁡θ=1cos⁡θ,csc⁡θ=1sin⁡θ,\begin{gathered} \tan \theta = \frac{\sin \theta}{\cos \theta}, \qquad \cot \theta = \frac{\cos \theta}{\sin \theta}, \\[0.8em] \sec \theta = \frac{1}{\cos \theta}, \qquad \csc \theta = \frac{1}{\sin \theta}, \end{gathered}

each wherever its denominator is not zero. Since P(θ)P(\theta) lies 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}

In plain words Cosine is the xx-coordinate and sine is the yy-coordinate of the point you reach after turning through θ\theta radians. Cosine comes first, just as xx comes before yy.

Example The angle 2π3\frac{2\pi}{3} lies in Quadrant II with reference angle π3\frac{\pi}{3}, so

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

So cos⁡2π3=−12\cos \frac{2\pi}{3} = -\frac{1}{2}, sin⁡2π3=32\sin \frac{2\pi}{3} = \frac{\sqrt{3}}{2} and tan⁡2π3=−3\tan \frac{2\pi}{3} = -\sqrt{3}. As a check, 14+34=1\frac{1}{4} + \frac{3}{4} = 1.

Signed projections

Definition Drop perpendiculars from P(θ)P(\theta) to the two axes. The foot on the xx-axis is (cos⁡θ,0)(\cos \theta, 0) and the foot on the yy-axis is (0,sin⁡θ)(0, \sin \theta). So cos⁡θ\cos \theta is the signed horizontal projection of the radius from the origin to P(θ)P(\theta), positive to the right of the origin and negative to the left, and sin⁡θ\sin \theta is its signed vertical projection, positive above the origin and negative below. On a circle of radius r>0r > 0 centered at the origin, the point at angle θ\theta is

(rcos⁡θ,rsin⁡θ),(r \cos \theta, r \sin \theta),

so its horizontal and vertical projections are rcos⁡θr \cos \theta and rsin⁡θr \sin \theta.

In plain words Cosine tells you how far right of the center the point is, with a negative number meaning left. Sine tells you how far above the center it is, with a negative number meaning below.

Example A Ferris wheel has radius 20 meters. A seat starts level with the hub on the right and turns counterclockwise through 7π6\frac{7\pi}{6} radians, which is 210∘210^\circ. Relative to the hub, the seat is now at

(20cos⁡7π6,20sin⁡7π6)=(−103,−10)≈(−17.32,−10).\begin{aligned} &\left(20 \cos \frac{7\pi}{6}, 20 \sin \frac{7\pi}{6}\right) \\[0.8em] &\quad = \left(-10\sqrt{3}, -10\right) \\[0.4em] &\quad \approx (-17.32, -10). \end{aligned}

The seat is about 17.3217.32 meters to the left of the hub and 10 meters below it. Both signs are negative because the seat is in Quadrant III.

Periodicity

Definition A function ff is periodic if there is a number p>0p > 0 such that for every xx in the domain of ff, the number x+px + p is also in the domain and f(x+p)=f(x)f(x + p) = f(x). Each such pp is a period of ff. The smallest one, when there is one, is the fundamental period, and the period of ff usually means this smallest one. Adding 2π2\pi to θ\theta adds one full lap, which returns to the same point, so P(θ+2π)=P(θ)P(\theta + 2\pi) = P(\theta) and

sin⁡(θ+2πk)=sin⁡θandcos⁡(θ+2πk)=cos⁡θfor every integer k.\begin{gathered} \sin(\theta + 2\pi k) = \sin \theta \\[0.4em] \text{and} \\[0.4em] \cos(\theta + 2\pi k) = \cos \theta \\[0.4em] \text{for every integer } k. \end{gathered}

Sine and cosine have period 2π2\pi, and so do secant and cosecant. Tangent and cotangent have period π\pi. Half a lap takes P(θ)P(\theta) to the opposite point P(θ+π)=(−cos⁡θ,−sin⁡θ)P(\theta + \pi) = (-\cos \theta, -\sin \theta), and the two minus signs cancel in the quotient.

In plain words Every full lap brings the point back to where it started, so the values repeat. For a large or negative angle, add or subtract multiples of 2π2\pi until the angle lands between 00 and 2π2\pi.

Example Taking away one lap,

sin⁡17π6=sin⁡(17π6−2π)=sin⁡5π6=12.\begin{aligned} \sin \frac{17\pi}{6} &= \sin\left(\frac{17\pi}{6} - 2\pi\right) \\[0.8em] &= \sin \frac{5\pi}{6} = \frac{1}{2}. \end{aligned}

Adding two laps gives

cos⁡(−7π3)=cos⁡(−7π3+4π)=cos⁡5π3=12.\begin{aligned} \cos\left(-\frac{7\pi}{3}\right) &= \cos\left(-\frac{7\pi}{3} + 4\pi\right) \\[0.8em] &= \cos \frac{5\pi}{3} = \frac{1}{2}. \end{aligned}

For tangent half a lap is enough, so tan⁡5π4=tan⁡π4=1\tan \frac{5\pi}{4} = \tan \frac{\pi}{4} = 1.

Exact values

Definition In the first quadrant the special right triangles give the points at π6\frac{\pi}{6}, π4\frac{\pi}{4} and π3\frac{\pi}{3}, and the points at 00 and π2\frac{\pi}{2} are (1,0)(1, 0) and (0,1)(0, 1) on the axes.

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

The points at the other multiples of π6\frac{\pi}{6} and π4\frac{\pi}{4} follow by reflecting across the axes or through the origin, since for every real θ\theta

P(π−θ)=(−cos⁡θ,sin⁡θ),P(π+θ)=(−cos⁡θ,−sin⁡θ),P(−θ)=(cos⁡θ,−sin⁡θ).\begin{aligned} P(\pi - \theta) &= (-\cos \theta, \sin \theta), \\[0.4em] P(\pi + \theta) &= (-\cos \theta, -\sin \theta), \\[0.4em] P(-\theta) &= (\cos \theta, -\sin \theta). \end{aligned}

The last of these says cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta and sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin \theta, so cosine is an even function and sine is an odd function.

In plain words Learn the first quadrant well. Every other standard point is a mirror image of one of those, with the same numbers and possibly different signs.

Example Each angle below is written as a first quadrant angle plus a reflection or a full lap.

AngleWritten asPoint on the unit circle
3π4\frac{3\pi}{4}π−π4\pi - \frac{\pi}{4}(−22,22)\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)
7π6\frac{7\pi}{6}π+π6\pi + \frac{\pi}{6}(−32,−12)\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)
5π3\frac{5\pi}{3}2π−π32\pi - \frac{\pi}{3}(12,−32)\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)
−π4-\frac{\pi}{4}−π4-\frac{\pi}{4}(22,−22)\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

For 5π3\frac{5\pi}{3}, subtracting a lap gives −π3-\frac{\pi}{3}, and the rule for P(−θ)P(-\theta) does the rest.

Check yourself

  1. Find the point P(11π6)P\left(\frac{11\pi}{6}\right) on the unit circle, and give cos⁡11π6\cos \frac{11\pi}{6}, sin⁡11π6\sin \frac{11\pi}{6} and tan⁡11π6\tan \frac{11\pi}{6}.
Show the answer

The angle 11π6\frac{11\pi}{6} is 2π−π62\pi - \frac{\pi}{6}, so it has the same point as −π6-\frac{\pi}{6}, namely

P(11π6)=(32,−12).P\left(\frac{11\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right).

So the cosine is 32\frac{\sqrt{3}}{2}, the sine is −12-\frac{1}{2}, and the tangent is

tan⁡11π6=−1/23/2=−13=−33.\tan \frac{11\pi}{6} = \frac{-1/2}{\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.
  1. Use periodicity to find the exact values of cos⁡25π4\cos \frac{25\pi}{4} and sin⁡(−13π3)\sin\left(-\frac{13\pi}{3}\right).
Show the answer

Subtracting three laps gives 25π4−6π=π4\frac{25\pi}{4} - 6\pi = \frac{\pi}{4}, so

cos⁡25π4=cos⁡π4=22.\cos \frac{25\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}.

Adding three laps gives −13π3+6π=5π3-\frac{13\pi}{3} + 6\pi = \frac{5\pi}{3}, so

sin⁡(−13π3)=sin⁡5π3=−32.\sin\left(-\frac{13\pi}{3}\right) = \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}.
  1. A point starts at (1,0)(1, 0) and travels counterclockwise around the unit circle along an arc of length 4. In which quadrant does it stop, and what are the signs of cos⁡4\cos 4 and sin⁡4\sin 4?
Show the answer

The arc length equals the angle in radians, so the point is P(4)P(4). Since π≈3.14\pi \approx 3.14 and 3π2≈4.71\frac{3\pi}{2} \approx 4.71, the angle 4 lies between them and the point is in Quadrant III. Both coordinates are negative there, so cos⁡4<0\cos 4 < 0 and sin⁡4<0\sin 4 < 0. A calculator in radian mode agrees, with cos⁡4≈−0.654\cos 4 \approx -0.654 and sin⁡4≈−0.757\sin 4 \approx -0.757.

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