Period changes for tangent curves

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.

 

Example 1

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​.

 

Example 2

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​.

 

Further Examples

QUESTION 1

QUESTION 2

QUESTION 3