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.
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°.
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.
Consider the equation $y=\tan9x$y=tan9x.
Complete the table of values for $y=\tan9x$y=tan9x.
$x$x | $\left(-5\right)^\circ$(−5)° | $0^\circ$0° | $5^\circ$5° | $15^\circ$15° | $20^\circ$20° | $25^\circ$25° |
---|---|---|---|---|---|---|
$y$y | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Graph the equation.
The function $f\left(x\right)$f(x) on a restricted domain has the form $f\left(x\right)=\tan bx$f(x)=tanbx. Two neighbouring asymptotes of this function are known to be at $x=30^\circ$x=30° and $x=90^\circ$x=90°.
Find the $x$x-intercept on the same domain. Give your answer in exact form.
The graph of a function in the form $y=\tan bx$y=tanbx is plotted below.
State the period of the function.
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
Apply graphical methods in solving problems