Trigonometric Graphs

Lesson

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 to which the function is applied.

For example, the following functions include a phase shift transformation.

- $f\left(\theta\right)=\sin\left(\theta+45^\circ\right)$
`f`(`θ`)=`s``i``n`(`θ`+45°) - $g\left(x\right)=\cos(x-37.5^\circ)$
`g`(`x`)=`c``o``s`(`x`−37.5°) - $h\left(\alpha\right)=\tan\left(\alpha+180^\circ\right)$
`h`(`α`)=`t``a``n`(`α`+180°)

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 $f(\theta)=\tan(\theta-45^\circ)$`f`(`θ`)=`t``a``n`(`θ`−45°).

At what points $\theta$`θ` is the function $f$`f` undefined?

With graphing software, we obtain the following representation.

We see that the function $\tan\theta$`t``a``n``θ` has been moved to the right a distance $45^\circ$45°. Since $\tan\theta$`t``a``n``θ` is undefined at $\theta=90^\circ+180^\circ n$`θ`=90°+180°`n` for all integers $n$`n`, the undefined points for $\tan(\theta-45^\circ)$`t``a``n`(`θ`−45°) must be $\theta=135^\circ+180^\circ n$`θ`=135°+180°`n`.

Given a graph that looks identical to the graph of $\tan\alpha$`t``a``n``α` except that it has been shifted to the left by $30^\circ$30°, find the formula for the new function.

The shift is to the left. So, the function must be given by $\tan\left(\alpha+30^\circ\right).$`t``a``n`(`α`+30°).

We want to identify how the coordinates of key points on the graph of $f\left(x\right)=\tan x$`f`(`x`)=`t``a``n``x` change as we apply a phase shift to produce the graph of $g\left(x\right)=\tan\left(x-60^\circ\right)$`g`(`x`)=`t``a``n`(`x`−60°).

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The point $A$

`A`on the graph of $f\left(x\right)$`f`(`x`) has the coordinates $\left(0^\circ,0\right)$(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 $\left(45^\circ,1\right)$(45°,1).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 graph of $f\left(x\right)$

`f`(`x`) has an asymptote passing through point $C$`C`with coordinates $\left(90^\circ,0\right)$(90°,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 $\left(\editable{},\editable{}\right)$(,).Using the answers from the previous parts, apply a phase shift to the graph of $f\left(x\right)=\tan x$

`f`(`x`)=`t``a``n``x`to draw the graph of $g\left(x\right)=\tan\left(x-60^\circ\right)$`g`(`x`)=`t``a``n`(`x`−60°).Loading Graph...

The graph of $f\left(x\right)=\tan x$`f`(`x`)=`t``a``n``x` has been drawn below. On the same set of axes draw the graph of $g\left(x\right)=\tan\left(x-45^\circ\right)$`g`(`x`)=`t``a``n`(`x`−45°).

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The graph of a function in the form $f\left(x\right)=\tan\left(x+h\right)$`f`(`x`)=`t``a``n`(`x`+`h`), where $0\le h<180^\circ$0≤`h`<180°, is drawn below.

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Compared to the graph of $y=\tan x$

`y`=`t``a``n``x`, state the phase shift of the graph of $f\left(x\right)$`f`(`x`).Given that $x=60^\circ$

`x`=60° is the first positive asymptote of $f\left(x\right)$`f`(`x`), what is the equation of the second positive asymptote?

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