Trigonometric Graphs
Symmetrical and periodic nature of trig functions
Intro to sin(x), cos(x) and tan(x)
Key features of sine and cosine curves
Amplitude of sine and cosine
Period changes for sine and cosine
Phase shifts for sine and cosine
Transformations of sine and cosine curves and equations
Domain and range of sine and cosine curves
Graphing sine curves
Sine Waves and Music (Investigation)
Graphing cosine curves
Finding equations of sine and cosine curves
Graphical solution of trigonometric equations involving sine and cosine
Applications of sine and cosine functions
Graph sums of sine and cosine (rad)
Key features of tangent curves
Dilation of tangent curves
Period changes for tangents
Phase shifts for tangents
Transformations of tangent curves and equations
Domain and range of tangent curves
Graphing tangent curves
Find the equation of a tangent curve
Intro to sec(x), cosec(x) and cot(x)
Key features of cot, sec and cosec curves
Transformations of cot, sec and cosec curves and equations
Domain and range of cot, sec and cosec curves
Graphing cot, sec and cosec curves
Find the equation of a cot, sec and cosec curve
Solve harmonic motion problems
This material is very similar to that in another chapter in which phase shift is explained for the sine and cosine functions.
Phase shift for trigonometric functions means moving the graph of the function to the right or to the left. This transformation occurs when a constant is added to (or subtracted from) the angle or number to which the function is applied.
For example, the following functions include a phase shift transformation.
$f\left(\theta\right)=\sin\left(\theta+\frac{\pi}{4}\right)$f(θ)=sin(θ+π4​)
$g\left(x\right)=\cos(x-0.5)$g(x)=cos(x−0.5)
$h\left(\alpha\right)=\tan\left(\alpha+\frac{22}{7}\right)$h(α)=tan(α+227​)
A more general idea
What is called phase shift in trigonometric functions is the same as the transformation of any real number function that moves the graph to the left or to the right. Thus, there is no need to think of this transformation as something peculiar to the trigonometric functions. It applies similarly to all kinds of functions.
Suppose $f$f is a function with its domain in the real numbers. Let $g$g be another function such that $g(x)=f(x+h)$g(x)=f(x+h) for all values of the domain variable $x$x, and where $h$h is a fixed positive number. Equivalently, $g(v-h)=f(v)$g(v−h)=f(v).
Consider a function value such that $f(v)=k$f(v)=k and $g(u)=k$g(u)=k. Then, $f(v)=g(u)$f(v)=g(u). But $f(v)=g(v-h)$f(v)=g(v−h). Hence, the function value attaind by $f$f at the point $v$v is attained by $g$g a distance $h$h to the left of $v$v.
Thus, all the function values attained by $f$f at points $x$x are attained by $g$g at points $x-h$x−h, which are a distance $h$h to the left of $x$x.
The conclusion to be drawn from this discussion is that the graph of $f(x+h)$f(x+h) should look the same as the graph of $f(x)$f(x) shifted $h$h units to the left.
Similarly, $f(x-h)$f(x−h) represents a shift to the right by $h$h units.
Since this applies to all real functions, it certainly applies to the trigonometric functions and therefore to the tangent function.
Sketch the graph of $f(x)=\tan\left(x-\frac{\pi}{4}\right)$f(x)=tan(x−π4​).
At what points $x$x is the function $f$f undefined?
With graphing software, we obtain the following representation.
We see that the function $\tan x$tanx has been moved to the right a distance $\frac{\pi}{4}$π4​. Since $\tan x$tanx is undefined at $x=\frac{\pi}{2}+n\pi$x=π2​+nπ for all integers $n$n, the undefined points for $\tan\left(x-\frac{\pi}{4}\right)$tan(x−π4​) must be $x=\frac{3\pi}{4}+n\pi$x=3π4​+nπ.
Given a graph diagram that looks identical to the graph of $\tan x$tanx except that it has been shifted to the left by $\frac{\pi}{6}$π6​, find the formula for the new function.
The shift is to the left. So, the function must be given by $\tan\left(x+\frac{\pi}{6}\right).$tan(x+π6​).
We want to identify how the coordinates of key points on the graph of $f\left(x\right)=\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\left(x\right)=\tan\left(x-\frac{\pi}{3}\right)$g(x)=tan(x−π3​).
The point $A$A on the graph of $f\left(x\right)$f(x) has the coordinates $\left(0,0\right)$(0,0).
What are the coordinates of the corresponding point on the graph of $g\left(x\right)$g(x)? Give your answer in the form $\left(\editable{},\editable{}\right)$(
The point $B$B on the graph of $f\left(x\right)$f(x) has the coordinates $\left(\frac{\pi}{4},1\right)$(π4​,1).
The graph of $f\left(x\right)$f(x) has an asymptote passing through point $C$C with coordinates $\left(\frac{\pi}{2},0\right)$(π2​,0).
Using the answers from the previous parts, apply a phase shift to the graph of $f\left(x\right)=\tan x$f(x)=tanx to draw the graph of $g\left(x\right)=\tan\left(x-\frac{\pi}{3}\right)$g(x)=tan(x−π3​).
Reveal Solution
The graph of $f\left(x\right)=\tan x$f(x)=tanx has been drawn below. On the same set of axes draw the graph of $g\left(x\right)=\tan\left(x+\frac{\pi}{2}\right)$g(x)=tan(x+π2​).
The graph of a function in the form $f\left(x\right)=\tan\left(x-h\right)$f(x)=tan(x−h), where $0\le h<\pi$0≤h<π, is drawn below.
Compared to the graph of $y=\tan x$y=tanx, state the phase shift of the graph of $f\left(x\right)$f(x).
Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions