Unit 3 Topic 3.8

The Tangent Function

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4 min readfreeNot practiced

Sine and cosine are the coordinates of a point on the unit circle. The third function you met in Topic 3.2, tangent, is their quotient. It behaves very differently: it has no maximum, its graph breaks into separate branches, and it repeats twice as often. This page builds the graph of tangent from sine and cosine, then reads its period, zeros and asymptotes, and transforms it.

Tangent as a slope and as a height

Put an angle θ\theta in standard position. Its terminal ray meets the unit circle at (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). The slope of that ray is rise over run:

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

Extend the ray until it hits the vertical line x=1x = 1. Because the ray goes through the origin with slope tan⁡θ\tan\theta, it reaches that line at height tan⁡θ\tan\theta (Figure 1). As θ\theta grows toward π2\frac{\pi}{2}, the ray gets steeper and the height shoots up. At θ=π2\theta = \frac{\pi}{2} the ray is vertical, parallel to the line, and never meets it. That is a vertical asymptote.

A unit circle with an angle theta in standard position. Its terminal ray passes through (cos theta, sin theta) and continues to the dashed line x = 1, which it meets at (1, tan theta). The vertical segment from the x-axis to that point has length tan theta.
Figure 1. The terminal ray meets x=1x = 1 at height tan⁡θ\tan\theta. LevlPrep original diagram.

Zeros, asymptotes and period

A quotient is 00 where its top is 00 and is undefined where its bottom is 00. So:

  • Zeros: tan⁡x=0\tan x = 0 where sin⁡x=0\sin x = 0, at x=kπx = k\pi for every integer kk.
  • Vertical asymptotes: where cos⁡x=0\cos x = 0, at x=π2+kπx = \frac{\pi}{2} + k\pi. Near each one the outputs grow without bound. Write both sides in limit notation: limx→π/2−tan⁡x\lim_{x \to \pi/2^-} \tan x=∞= \infty and limx→π/2+tan⁡x\lim_{x \to \pi/2^+} \tan x=−∞= -\infty.
  • Period π\pi: adding π\pi to the angle points the ray in the opposite direction. Both sin⁡\sin and cos⁡\cos change sign, so their quotient does not change: tan⁡(x+π)=tan⁡x\tan(x + \pi) = \tan x.
  • Range: every real number, (−∞,∞)(-\infty, \infty).

On each branch, such as (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), tangent increases. Left of the zero the graph is concave down, and right of it the graph is concave up: on (0,π2)\left(0, \frac{\pi}{2}\right) the rate of change of tan⁡x\tan x is increasing, so equal steps in xx give bigger and bigger rises.

FunctionPeriodRangeZerosVertical asymptotes
sin⁡x\sin x2π2\pi[−1,1][-1, 1]x=kπx = k\pinone
cos⁡x\cos x2π2\pi[−1,1][-1, 1]x=π2+kπx = \frac{\pi}{2} + k\pinone
tan⁡x\tan xπ\pi(−∞,∞)(-\infty, \infty)x=kπx = k\pix=π2+kπx = \frac{\pi}{2} + k\pi

Transforming tangent

The same four changes you used for sinusoids work here. In f(x)f(x)=atan⁡(b(x−c))+d= a\tan(b(x - c)) + d:

  • aa stretches the graph vertically (and reflects it if a<0a < 0). There is no amplitude, because tangent has no maximum.
  • bb changes the period to π|b|\frac{\pi}{|b|}. Notice the π\pi, not 2π2\pi.
  • cc shifts the graph right cc units, so every zero and asymptote moves right cc.
  • dd shifts the graph up dd units. The "center" points of the branches move to height dd.

Worked example. Find the period, the vertical asymptotes and one zero of f(x)=3tan⁡(2(x−π4))f(x) = 3\tan\left(2\left(x - \frac{\pi}{4}\right)\right).

Period. b=2b = 2, so the period is π2\frac{\pi}{2}.

Asymptotes. Set the input of tangent equal to π2+kπ\frac{\pi}{2} + k\pi: 2(x−π4)=π2+kπ2\left(x - \frac{\pi}{4}\right) = \frac{\pi}{2} + k\pi. Divide by 22: x−π4=π4+kπ2x - \frac{\pi}{4} = \frac{\pi}{4} + \frac{k\pi}{2}. Add π4\frac{\pi}{4}: x=π2+kπ2x = \frac{\pi}{2} + \frac{k\pi}{2}. The asymptotes are π2\frac{\pi}{2} apart, one period, as they should be.

A zero. Set the input equal to 00: 2(x−π4)=02\left(x - \frac{\pi}{4}\right) = 0 gives x=π4x = \frac{\pi}{4}. Check: f(π4)=3tan⁡0=0f\left(\frac{\pi}{4}\right) = 3\tan 0 = 0.

Rates of change on a branch

Average rates of change on tangent behave like those of any concave-up or concave-down graph. Take g(x)=tan⁡xg(x) = \tan x on [0,π3]\left[0, \frac{\pi}{3}\right] with two equal intervals.

Worked example. The average rate of change of gg on [0,π6]\left[0, \frac{\pi}{6}\right] is tan⁡(π/6)−tan⁡0π/6\frac{\tan(\pi/6) - \tan 0}{\pi/6}=3/3π/6= \frac{\sqrt{3}/3}{\pi/6}=23π= \frac{2\sqrt{3}}{\pi}, the slope of the secant line through (0,0)(0, 0) and (π6,33)\left(\frac{\pi}{6}, \frac{\sqrt{3}}{3}\right). On [π6,π3]\left[\frac{\pi}{6}, \frac{\pi}{3}\right] it is 3−3/3π/6\frac{\sqrt{3} - \sqrt{3}/3}{\pi/6}=43π= \frac{4\sqrt{3}}{\pi}. The rates increase over equal intervals, so gg is concave up on [0,π3]\left[0, \frac{\pi}{3}\right], and the rate of change of gg is increasing there.

Tangent is not proportional to its input. Doubling the angle from π6\frac{\pi}{6} to π3\frac{\pi}{3} triples the output, from 33\frac{\sqrt{3}}{3} to 3\sqrt{3}. Always compute the values instead of scaling them.

Common slips

  • Using 2πb\frac{2\pi}{b} for the period. That is the sine rule; tangent uses πb\frac{\pi}{b}.
  • Putting asymptotes where sin⁡x=0\sin x = 0. Those are the zeros; asymptotes come from cos⁡x=0\cos x = 0.
  • Writing only one side of an asymptote. Give both one-sided limits, each with its sign.

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