Period changes for tangents

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 $360^\circ$360° is expressed by the statements

$\sin(x+360^\circ n)=\sin x$sin(x+360°n)=sinx and

$\cos(x+360^\circ n)=\cos x$cos(x+360°n)=cosx 

for all integers $n$n. So it must be true that since $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx​, then $\tan(x+360^\circ n)=\tan x$tan(x+360°n)=tanx. However, $360^\circ$360° 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 $180^\circ$180°. That is, $\tan(x+180^\circ n)=\tan x$tan(x+180°n)=tanx for all integers $n$n.

 

The fact that the tangent function repeats at intervals of $180^\circ$180° can be verified by considering the unit circle diagram. If $180^\circ$180° is added to an angle $\alpha$α, then the diagram shows that $\sin(\alpha+180^\circ)$sin(α+180°) has the same magnitude as $\sin\alpha$sinα but opposite sign. The same relation holds between $\cos(\alpha+180^\circ)$cos(α+180°) and $\cos\alpha$cosα.

We make use of the definition: $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$tanα=sinαcosα​

$\tan(\alpha+180^\circ)$tan(α+180°)	$=$=	$\frac{\sin(\alpha+180^\circ)}{\cos(\alpha+180^\circ)}$sin(α+180°)cos(α+180°)​
	$=$=	$\frac{-\sin\alpha}{-\cos\alpha}$−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 $\tan x'=\tan(x'+180^\circ)$tanx′=tan(x′+180°). So, 

$\tan x'$tanx′	$=$=	$\tan(kx+180^\circ)$tan(kx+180°)
	$=$=	$\tan k\left(x+\frac{180^\circ}{k}\right)$tank(x+180°k​)

This says, we  need to advance $x$x by an amount $\frac{180^\circ}{k}$180°k​ 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}\times180^\circ$23​×180°, or$120^\circ$120°.

 

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 $540^\circ$540° rather than at intervals of $180^\circ$180°. Assuming the function has the form $\tan kx$tankx, what is $k$k?

The period of $\tan kx$tankx is $\frac{180^\circ}{k}$180°k​ and we know this to be $540^\circ$540°.

So, $\frac{180^\circ}{k}=540^\circ$180°k​=540° and hence, $k=\frac{1}{3}$k=13​.

 

Further Examples

QUESTION 1

QUESTION 2

QUESTION 3