Unit 6, lesson 48

The Tangent Function

How tangent is the ratio of sine to cosine, why it is undefined where cosine is zero, and how its signs, zeros, period and vertical asymptotes shape a graph made of repeating branches.

  • Theory part 95
  • Practice part 96
  • The Six Trigonometric Functions

A ramp that rises at an angle θ\theta from the ground has a slope, rise over run, and that slope depends only on the angle. On the unit circle the rise is sin⁡θ\sin \theta and the run is cos⁡θ\cos \theta, so the slope is their quotient. That quotient is the tangent. When the run shrinks to zero the ramp stands straight up and the slope stops being a number, which is exactly where the tangent graph breaks.

Key terms

Tangent as a sine-to-cosine ratio

Definition For every real xx with cos⁡x≠0\cos x \ne 0, the tangent of xx is

tan⁡x=sin⁡xcos⁡x.\tan x = \frac{\sin x}{\cos x}.

Since P(x)=(cos⁡x,sin⁡x)P(x) = (\cos x, \sin x) is the point on the unit circle at angle xx radians, tan⁡x\tan x is the slope of the line through the origin and P(x)P(x). Its sign follows from the quadrant signs, positive in quadrants I and III and negative in quadrants II and IV.

In plain words Divide the height of the circle point by its horizontal coordinate. The result is how steep the radius to that point is.

Example At π3\frac{\pi}{3} the point is (12,32)\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), so

tan⁡π3=3/21/2=3.\tan \frac{\pi}{3} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}.

At 3π4\frac{3\pi}{4} the point is (−22,22)\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), and the quotient is −1-1. A radius pointing up and to the left has negative slope.

Undefined inputs

Definition The quotient sin⁡xcos⁡x\frac{\sin x}{\cos x} is undefined exactly when cos⁡x=0\cos x = 0, that is when

x=π2+kπ,k an integer.x = \frac{\pi}{2} + k\pi, \qquad k \text{ an integer}.

The domain of tangent is the set of all real numbers except these. At an excluded input tan⁡x\tan x has no value at all. It is not equal to ∞\infty.

In plain words At the top and bottom of the circle the radius is vertical, the run is zero, and a vertical line has no slope.

Example At x=3π2x = \frac{3\pi}{2} the point is (0,−1)(0, -1), so the formula asks for −10\frac{-1}{0}, and tan⁡3π2\tan \frac{3\pi}{2} is undefined. At x=2πx = 2\pi the point is (1,0)(1, 0), and tan⁡2π=01=0\tan 2\pi = \frac{0}{1} = 0 is perfectly fine.

Zeros

Definition The zeros of tangent are the inputs where tan⁡x=0\tan x = 0. A quotient is zero exactly when its numerator is zero and its denominator is not, so the zeros are the zeros of sine,

x=kπ,k an integer.x = k\pi, \qquad k \text{ an integer}.

At every one of these cos⁡x=±1≠0\cos x = \pm 1 \ne 0, so tangent is defined there.

In plain words The slope is zero when the radius lies flat along the horizontal axis.

Example In (−2π,2π)(-2\pi, 2\pi) the zeros of tangent are −π-\pi, 00 and π\pi. The endpoints ±2π\pm 2\pi are also zeros but are not in the open interval.

Period

Definition For every xx in the domain of tangent and every integer kk, the input x+kπx + k\pi is also in the domain and

tan⁡(x+kπ)=tan⁡x.\tan(x + k\pi) = \tan x.

The reason is that a half turn sends P(x)P(x) to the opposite point −P(x)-P(x), which flips the sign of both coordinates and leaves their quotient unchanged. The least positive period of tangent is π\pi. To see that nothing smaller works, suppose p>0p > 0 is a period. Then tan⁡p=tan⁡(0+p)=tan⁡0=0\tan p = \tan(0 + p) = \tan 0 = 0. The zeros of tangent are exactly the inputs kπk\pi, so p=kπp = k\pi for a positive integer kk, and therefore p≥πp \ge \pi.

In plain words Opposite points on the circle lie on the same line through the origin, so they give the same slope. Tangent repeats every half turn, twice as often as sine and cosine.

Example Since 7π6=π6+π\frac{7\pi}{6} = \frac{\pi}{6} + \pi,

tan⁡7π6=tan⁡π6=1/23/2=33.\tan \frac{7\pi}{6} = \tan \frac{\pi}{6} = \frac{1/2}{\sqrt{3}/2} = \frac{\sqrt{3}}{3}.

Directly, the point at 7π6\frac{7\pi}{6} is (−32,−12)\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right), and the two minus signs cancel in the quotient.

Vertical asymptote

Definition The vertical line x=ax = a is a vertical asymptote of the graph of ff if, from at least one side of aa, the values f(x)f(x) become and stay larger than any number you name, or become and stay smaller than any negative number you name, once xx is close enough to aa on that side. In limit notation, at least one of the one-sided limits of ff at aa is ∞\infty or −∞-\infty. Every excluded input x=π2+kπx = \frac{\pi}{2} + k\pi is a vertical asymptote of tangent. Near such an input the numerator sin⁡x\sin x is close to 11 or −1-1 while the denominator cos⁡x\cos x is close to 00, so

∣tan⁡x∣=∣sin⁡x∣∣cos⁡x∣\lvert \tan x \rvert = \frac{\lvert \sin x \rvert}{\lvert \cos x \rvert}

exceeds any bound once xx is close enough. On each side the sign of tan⁡x\tan x does not change, and it is the sign of sin⁡x\sin x divided by the sign of cos⁡x\cos x there. For every asymptote, tan⁡x\tan x is large and positive just to its left and large and negative just to its right.

In plain words Dividing a number near 11 by a number near 00 gives something enormous. Whether it is enormous and positive or enormous and negative depends on the signs.

Example Approach π2\frac{\pi}{2} from the left, through quadrant I. There sin⁡x\sin x is near 11 and cos⁡x\cos x is small and positive, so tan⁡x\tan x is large and positive. From the right, through quadrant II, cos⁡x\cos x is small and negative, so tan⁡x\tan x is large and negative.

xxπ2−0.01\frac{\pi}{2} - 0.01π2−0.001\frac{\pi}{2} - 0.001π2+0.001\frac{\pi}{2} + 0.001π2+0.01\frac{\pi}{2} + 0.01
tan⁡x\tan x≈99.997\approx 99.997≈1000.0\approx 1000.0≈−1000.0\approx -1000.0≈−99.997\approx -99.997

The table illustrates the behavior. The sign argument is what explains it.

Tangent graph

Definition The graph of tangent is the set of points (x,tan⁡x)(x, \tan x) for xx in its domain. It is made of identical branches, one on each open interval

(−π2+kπ, π2+kπ),k an integer.\begin{gathered} \left(-\frac{\pi}{2} + k\pi, \ \frac{\pi}{2} + k\pi\right), \\[0.4em] k \text{ an integer}. \end{gathered}

On each such interval tangent is increasing, takes every real value once, and is separated from the next branch by a vertical asymptote. The range of tangent is all real numbers. Tangent is odd, with tan⁡(−x)=−tan⁡x\tan(-x) = -\tan x for every xx in its domain, since sine is odd and cosine is even, so the graph is symmetric about the origin. Tangent is increasing on each branch but not on its whole domain. For example tan⁡π4=1\tan \frac{\pi}{4} = 1 while tan⁡3π4=−1\tan \frac{3\pi}{4} = -1. The middle branch passes through

xx−π3-\frac{\pi}{3}−π4-\frac{\pi}{4}−π6-\frac{\pi}{6}00π6\frac{\pi}{6}π4\frac{\pi}{4}π3\frac{\pi}{3}
tan⁡x\tan x−3-\sqrt{3}−1-1−33-\frac{\sqrt{3}}{3}0033\frac{\sqrt{3}}{3}113\sqrt{3}

In plain words Each branch rises from far below to far above between two neighboring asymptotes, crossing zero halfway. Then the pattern starts over.

Example Graph one branch of y=tan⁡(x−π4)y = \tan\left(x - \frac{\pi}{4}\right). The input x−π4x - \frac{\pi}{4} must avoid π2+kπ\frac{\pi}{2} + k\pi, so the excluded inputs and asymptotes are

x=3π4+kπ,k an integer.x = \frac{3\pi}{4} + k\pi, \qquad k \text{ an integer}.

The zeros are x=π4+kπx = \frac{\pi}{4} + k\pi. The branch through (π4,0)\left(\frac{\pi}{4}, 0\right) lives on (−π4,3π4)\left(-\frac{\pi}{4}, \frac{3\pi}{4}\right) and passes through (0,−1)(0, -1) and (π2,1)\left(\frac{\pi}{2}, 1\right). It is the middle branch of tangent moved right by π4\frac{\pi}{4}, asymptotes and all.

Principal branch

Definition The principal branch of tangent is its restriction to the open interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). On this interval tangent is increasing, hence one-to-one, and its range is still all real numbers. This is the branch that is inverted to define the inverse tangent, so arctan⁡y\arctan y is the unique xx in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) with tan⁡x=y\tan x = y.

In plain words Tangent repeats, so many inputs share each output. Keeping only the middle branch picks one input for every output.

Example The equation tan⁡x=1\tan x = 1 has the solutions x=π4+kπx = \frac{\pi}{4} + k\pi, infinitely many. Only one lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), namely π4\frac{\pi}{4}, so arctan⁡1=π4\arctan 1 = \frac{\pi}{4}. The solution 5π4\frac{5\pi}{4} is just as valid for the equation but lies on a different branch.

Check yourself

  1. Find tan⁡5π6\tan \frac{5\pi}{6} and tan⁡(−π3)\tan\left(-\frac{\pi}{3}\right) exactly.
Show the answer

The point at 5π6\frac{5\pi}{6} is (−32,12)\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), so the quotient is −13=−33-\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}. Since tangent is odd, tan⁡(−π3)=−tan⁡π3=−3\tan\left(-\frac{\pi}{3}\right) = -\tan \frac{\pi}{3} = -\sqrt{3}.

  1. Decide whether tan⁡x\tan x is defined at x=−π2x = -\frac{\pi}{2} and at x=5πx = 5\pi, and give the value where it is.
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At −π2-\frac{\pi}{2} the point is (0,−1)(0, -1) and the cosine is 00, so tangent is undefined there. At 5π5\pi the point is (−1,0)(-1, 0), so the value is 0−1=0\frac{0}{-1} = 0.

  1. For y=tan⁡(x+π3)y = \tan\left(x + \frac{\pi}{3}\right), find the excluded inputs, the zeros, and the open interval of the branch that contains x=0x = 0.
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The input x+π3x + \frac{\pi}{3} must avoid π2+kπ\frac{\pi}{2} + k\pi, so the excluded inputs are x=π6+kπx = \frac{\pi}{6} + k\pi. The zeros come from x+π3=kπx + \frac{\pi}{3} = k\pi, so they are x=−π3+kπx = -\frac{\pi}{3} + k\pi. The branch containing 00 lies between the asymptotes −5π6-\frac{5\pi}{6} and π6\frac{\pi}{6}, on the interval (−5π6,π6)\left(-\frac{5\pi}{6}, \frac{\pi}{6}\right).

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