We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\sin kx$sinkx and $\cos kx$coskx.
The fact that the sine and cosine functions both have period $2\pi$2π is expressed by the statements
$\sin(x+2n\pi)=\sin x$sin(x+2nπ)=sinx and $\cos(x+2n\pi)=\cos x$cos(x+2nπ)=cosx
for all integers $n$n. So it must be true that since $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx, then $\tan(x+2n\pi)=\tan x$tan(x+2nπ)=tanx. However, $2\pi$2π is not the smallest interval at which the tangent function repeats.
A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $\pi$π. That is, $\tan(x+n\pi)=\tan x$tan(x+nπ)=tanx for all integers $n$n.
The fact that the tangent function repeats at intervals of $\pi$π can be verified by considering the unit circle diagram. If $\pi$π is added to an angle $\alpha$α, then the diagram shows that $\sin(\alpha+\pi)$sin(α+π) has the same magnitude as $\sin\alpha$sinα but opposite sign. The same relation holds between $\cos(\alpha+\pi)$cos(α+π) and $\cos\alpha$cosα.
We make use of the definition: $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$tanα=sinαcosα
$\tan(\alpha+\pi)$tan(α+π) | $=$= | $\frac{\sin(\alpha+\pi)}{\cos(\alpha+\pi)}$sin(α+π)cos(α+π) |
$=$= | $\frac{-\sin\alpha}{-\cos\alpha}$−sinα−cosα | |
$=$= | $\tan\alpha$tanα |
We are now in a position to determine the period of the function $\tan kx$tankx where $k$k is any number.
We can define a new variable $x'=kx$x′=kx so that $\tan kx=\tan x'$tankx=tanx′. But, we have seen that $\tan x'=\tan(x'+\pi)$tanx′=tan(x′+π). So,
$\tan x'$tanx′ | $=$= | $\tan(kx+\pi)$tan(kx+π) |
$=$= | $\tan\left(kx+\frac{k\pi}{k}\right)$tan(kx+kπk) | |
$=$= | $\tan k\left(x+\frac{\pi}{k}\right)$tank(x+πk) |
This says, we need to advance $x$x by an amount $\frac{\pi}{k}$πk in order to reach the same function value as $\tan kx$tankx. We conclude that the coefficient $k$k in $\tan kx$tankx changes the period by the factor $\frac{1}{k}$1k compared with the period of $\tan x$tanx.
What is the period of the function $\tan\frac{3x}{2}$tan3x2?
In this example, $k=\frac{3}{2}$k=32. So the period of the function is $\frac{1}{k}$1k times the period of $\tan x$tanx. Thus, $\tan\frac{3x}{2}$tan3x2 has period $\frac{2}{3}\times\pi$23×π, or$\frac{2\pi}{3}$2π3.
You are given the graph of a function that looks like the graph of $\tan x$tanx except that it repeats at intervals of $3\pi$3π rather than at intervals of $\pi$π. Assuming the function has the form $\tan kx$tankx, what is $k$k?
The period of $\tan kx$tankx is $\frac{\pi}{k}$πk and we know this to be $3\pi$3π.
So, $\frac{\pi}{k}=3\pi$πk=3π and hence, $k=\frac{1}{3}$k=13.
Consider the equation $y=\tan7x$y=tan7x.
Complete the table of values for $y=\tan7x$y=tan7x.
$x$x | $-\frac{\pi}{28}$−π28 | $0$0 | $\frac{\pi}{28}$π28 | $\frac{3\pi}{28}$3π28 | $\frac{\pi}{7}$π7 | $\frac{5\pi}{28}$5π28 |
---|---|---|---|---|---|---|
$y$y | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Graph the equation.
The graph of $f\left(x\right)=\tan x$f(x)=tanx is plotted below. On the same set of axes, plot $g\left(x\right)=\tan4x$g(x)=tan4x.
The function $f\left(x\right)=\tan6x$f(x)=tan6x is to be plotted on the interval $\left[\frac{\pi}{12},\frac{5\pi}{12}\right]$[π12,5π12].
First, plot the interval on the number line below.
State the period of the function $f\left(x\right)=\tan6x$f(x)=tan6x.
Find the asymptotes of the function that occur on this interval. Write your answers separated by a comma.
Find the $x$x-intercepts of the function that occur on this interval. Write your answers separated by a comma.
The graph of $y=\tan x$y=tanx has been plotted below, where the interval $\left[\frac{\pi}{12},\frac{5\pi}{12}\right]$[π12,5π12] has also been highlighted. By moving the points, plot the function $f\left(x\right)=\tan6x$f(x)=tan6x.
Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions