Skills Beta

Trigonometry and the Unit Circle

4 min read · freeNot practiced

Trigonometry returns in nearly every unit of calculus: the rates of change of sine and cosine in Unit 2, oscillating motion in Unit 4, areas under waves in Unit 6. You do not need every identity from precalculus. You need radians, the unit circle, the exact values at a handful of angles, the graphs, and three identities. All of it grows from one picture.

Radians

A radian measures an angle by the length of the arc it cuts on a circle of radius 1. Going all the way around a unit circle covers its circumference, 2π2\pi, so a full turn is 2π2\pi radians. A half turn is π\pi radians, which is 180∘180^{\circ}. To convert:

  • degrees to radians: multiply by π180\frac{\pi}{180}, so 135∘135^{\circ}=135π180= \frac{135\pi}{180}=3π4= \frac{3\pi}{4};
  • radians to degrees: multiply by 180π\frac{180}{\pi}, so π6=30∘\frac{\pi}{6} = 30^{\circ}.

Why does calculus insist on radians? Because its formulas are simplest there. For example, in Topic 1.8 you will meet limx→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}=1= 1; that limit is 1 only when xx is in radians. In degrees, it would be the awkward number π180\frac{\pi}{180}. So every formula in this course assumes radians, and so must your calculator. Checking the mode is worth a point on the exam.

The unit circle

The unit circle is x2+y2=1x^2 + y^2 = 1. Start at (1,0)(1, 0) and walk counterclockwise through an angle θ\theta. The point you reach defines cosine and sine: (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) Cosine is the xx-coordinate and sine is the yy-coordinate. Negative angles go clockwise; adding 2π2\pi brings you back to the same point.

The unit circle with points at the angles 0, pi/6, pi/4, pi/3 and pi/2 and their coordinates (1, 0), (square root 3 over 2, 1/2), (square root 2 over 2, square root 2 over 2), (1/2, square root 3 over 2) and (0, 1); also (-1, 0) at pi and (0, -1) at 3 pi/2. Each quadrant is labeled with the signs of sine and cosine.
Figure 1. The first-quadrant special angles and the signs of sine and cosine in each quadrant. LevlPrep original diagram.
θ\theta00π6\frac{\pi}{6}π4\frac{\pi}{4}π3\frac{\pi}{3}π2\frac{\pi}{2}
sin⁡θ\sin\theta0012\frac{1}{2}22\frac{\sqrt{2}}{2}32\frac{\sqrt{3}}{2}11
cos⁡θ\cos\theta1132\frac{\sqrt{3}}{2}22\frac{\sqrt{2}}{2}12\frac{1}{2}00
tan⁡θ\tan\theta0033\frac{\sqrt{3}}{3}113\sqrt{3}undefined

A memory aid: the sines go 02,\frac{\sqrt{0}}{2},12,\frac{\sqrt{1}}{2},22,\frac{\sqrt{2}}{2},32,\frac{\sqrt{3}}{2},42\frac{\sqrt{4}}{2}, and the cosines are the same list backward.

Any special angle: reference angle plus sign

Every other special angle is a mirror image of one in the first quadrant. Find the reference angle (the acute angle to the xx-axis), take its value from the table, then attach the sign for the quadrant. Sine is positive in quadrants I and II (upper half); cosine is positive in I and IV (right half).

Worked example 1. Find sin⁡5π6\sin\frac{5\pi}{6}, cos⁡4π3\cos\frac{4\pi}{3} and tan⁡7π4\tan\frac{7\pi}{4}.

5π6\frac{5\pi}{6}: reference angle π6\frac{\pi}{6}, quadrant II, sine positive. sin⁡5π6=12\sin\frac{5\pi}{6} = \frac{1}{2}.

4π3=π+π3\frac{4\pi}{3} = \pi + \frac{\pi}{3}: reference angle π3\frac{\pi}{3}, quadrant III, cosine negative. cos⁡4π3=−12\cos\frac{4\pi}{3} = -\frac{1}{2}.

7π4\frac{7\pi}{4}: reference angle π4\frac{\pi}{4}, quadrant IV, sine negative and cosine positive. tan⁡7π4\tan\frac{7\pi}{4}=−2/22/2= \frac{-\sqrt{2}/2}{\sqrt{2}/2}=−1= -1.

The six functions

Tangent and the three reciprocal functions are all built from sine and cosine: tan⁡x=sin⁡xcos⁡x,sec⁡x=1cos⁡x,csc⁡x=1sin⁡x,cot⁡x=cos⁡xsin⁡x\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x} Each is undefined where its denominator is 0. For example, tan⁡x\tan x and sec⁡x\sec x are undefined at x=π2x = \frac{\pi}{2}, where cos⁡x=0\cos x = 0, and their graphs have vertical asymptotes there. One notation to know cold: sin2⁡x\sin^2 x means (sin⁡x)2(\sin x)^2, not sin⁡(x2)\sin(x^2).

The graphs

The graphs of y = sin x (solid) and y = cos x (dashed) from x = -pi/2 to 2 pi: waves between -1 and 1 with period 2 pi. Sine passes through (0, 0); cosine starts at (0, 1).
Figure 2. Sine and cosine: waves of height 1 that repeat every 2π2\pi. LevlPrep original diagram.

Both y=sin⁡xy = \sin x and y=cos⁡xy = \cos x are waves between −1-1 and 11 that repeat every 2π2\pi (the period). Sine starts at 0 and rises; cosine starts at its peak, 1. Cosine is sine shifted left by π2\frac{\pi}{2}. Sine is 0 at multiples of π\pi; cosine is 0 at odd multiples of π2\frac{\pi}{2}. The graph of tan⁡x\tan x repeats every π\pi and has vertical asymptotes wherever cos⁡x=0\cos x = 0.

Identities you will use

A trig identity is an equation true for every angle. Three cover almost everything in AB:

  • The Pythagorean identity sin2⁡x+cos2⁡x=1\sin^2 x + \cos^2 x = 1. It is just x2+y2=1x^2 + y^2 = 1 for the point on the unit circle. Dividing by cos2⁡x\cos^2 x gives tan2⁡x+1=sec2⁡x\tan^2 x + 1 = \sec^2 x.
  • Double angle: sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x.
  • Double angle: cos⁡2x\cos 2x=cos2⁡x−sin2⁡x= \cos^2 x - \sin^2 x=1−2sin2⁡x= 1 - 2\sin^2 x=2cos2⁡x−1= 2\cos^2 x - 1.

Worked example 2. θ\theta is in quadrant II and sin⁡θ=35\sin\theta = \frac{3}{5}. Find cos⁡θ\cos\theta and tan⁡θ\tan\theta.

cos2⁡θ\cos^2\theta=1−925= 1 - \frac{9}{25}=1625= \frac{16}{25}, so cos⁡θ=±45\cos\theta = \pm\frac{4}{5}. In quadrant II cosine is negative: cos⁡θ=−45\cos\theta = -\frac{4}{5}.

tan⁡θ\tan\theta=3/5−4/5= \frac{3/5}{-4/5}=−34= -\frac{3}{4}.

Solving a trig equation

Worked example 3. Solve 2sin⁡x=12\sin x = 1 for 0≤x<2π0 \le x < 2\pi.

sin⁡x=12\sin x = \frac{1}{2}. The reference angle is π6\frac{\pi}{6}. Sine is positive in quadrants I and II.

So x=π6x = \frac{\pi}{6} or x=π−π6=5π6x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.

Always look for the second solution: on [0,2π)[0, 2\pi), sin⁡x=k\sin x = k has two solutions for −1<k<1-1 < k < 1.

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