The domain of
$\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ was given to be all
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ such that
$\text{\hspace{0.17em}}x\ne k\pi \text{\hspace{0.17em}}$ for any integer
$\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$Would the domain of$\text{\hspace{0.17em}}y=A\mathrm{csc}(Bx-C)+D\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}x\ne \frac{C+k\pi}{B}?$
Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input.
Given a function of the form
$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right),\text{\hspace{0.17em}}$ graph one period.
Express the function given in the form
$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
$\text{\hspace{0.17em}}\left|A\right|.$
Identify
$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and determine the period,
$\text{\hspace{0.17em}}P=\frac{2\pi}{\left|B\right|}.$
Draw the graph of
$\text{\hspace{0.17em}}y=A\mathrm{sin}\left(Bx\right).$
Use the reciprocal relationship between
$\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to draw the graph of
$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
Sketch the asymptotes.
Plot any two reference points and draw the graph through these points.
Graphing a variation of the cosecant function
Graph one period of
$\text{\hspace{0.17em}}f(x)=\mathrm{-3}\mathrm{csc}(4x).$
Step 1. The given function is already written in the general form,
$\text{\hspace{0.17em}}y=A\mathrm{csc}\left(Bx\right).$
Step 2.$\text{\hspace{0.17em}}\left|A\right|=\left|-3\right|=3,$ so the stretching factor is 3.
Step 3.$\text{\hspace{0.17em}}B=4,$ so
$\text{\hspace{0.17em}}P=\frac{2\pi}{4}=\frac{\pi}{2}.\text{\hspace{0.17em}}$ The period is
$\text{\hspace{0.17em}}\frac{\pi}{2}\text{\hspace{0.17em}}$ units.
Step 4. Sketch the graph of the function
$\text{\hspace{0.17em}}g(x)=\mathrm{-3}\mathrm{sin}(4x).$
Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the
cosecant function .
Steps 6–7. Sketch three asymptotes at
$\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}x=\frac{\pi}{4},\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}x=\frac{\pi}{2}.\text{\hspace{0.17em}}$ We can use two reference points, the local maximum at
$\text{\hspace{0.17em}}\left(\frac{\pi}{8},\mathrm{-3}\right)\text{\hspace{0.17em}}$ and the local minimum at
$\text{\hspace{0.17em}}\left(\frac{3\pi}{8},3\right).$[link] shows the graph.
Graph one period of
$\text{\hspace{0.17em}}f(x)=0.5\mathrm{csc}(2x).$
Given a function of the form
$\text{\hspace{0.17em}}f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D,\text{\hspace{0.17em}}$ graph one period.
Express the function given in the form
$\text{\hspace{0.17em}}y=A\mathrm{csc}(Bx-C)+D.$
Identify the stretching/compressing factor,
$\text{\hspace{0.17em}}\left|A\right|.$
Identify
$\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and determine the period,
$\text{\hspace{0.17em}}\frac{2\pi}{\left|B\right|}.$
Identify
$\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ and determine the phase shift,
$\text{\hspace{0.17em}}\frac{C}{B}.$
Draw the graph of
$\text{\hspace{0.17em}}y=A\mathrm{csc}(Bx)\text{\hspace{0.17em}}$ but shift it to the right by and up by
$\text{\hspace{0.17em}}D.$
Sketch the vertical asymptotes, which occur at
$\text{\hspace{0.17em}}x=\frac{C}{B}+\frac{\pi}{\left|B\right|}k,$ where
$\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is an integer.
Graphing a vertically stretched, horizontally compressed, and vertically shifted cosecant
Sketch a graph of
$\text{\hspace{0.17em}}y=2\mathrm{csc}\left(\frac{\pi}{2}x\right)+1.\text{\hspace{0.17em}}$ What are the domain and range of this function?
Step 1. Express the function given in the form
$\text{\hspace{0.17em}}y=2\mathrm{csc}\left(\frac{\pi}{2}x\right)+1.$
Step 2. Identify the stretching/compressing factor,
$\text{\hspace{0.17em}}\left|A\right|=2.$
Step 3. The period is
$\text{\hspace{0.17em}}\frac{2\pi}{\left|B\right|}=\frac{2\pi}{\frac{\pi}{2}}=\frac{2\pi}{1}\cdot \frac{2}{\pi}=4.$
Step 4. The phase shift is
$\text{\hspace{0.17em}}\frac{0}{\frac{\pi}{2}}=0.$
Step 5. Draw the graph of
$\text{\hspace{0.17em}}y=A\mathrm{csc}(Bx)\text{\hspace{0.17em}}$ but shift it up
$\text{\hspace{0.17em}}D=1.$
Step 6. Sketch the vertical asymptotes, which occur at
$\text{\hspace{0.17em}}x=0,x=2,x=4.$
Given the graph of
$\text{\hspace{0.17em}}f(x)=2\mathrm{cos}\left(\frac{\pi}{2}x\right)+1\text{\hspace{0.17em}}$ shown in
[link] , sketch the graph of
$\text{\hspace{0.17em}}g(x)=2\mathrm{sec}\left(\frac{\pi}{2}x\right)+1\text{\hspace{0.17em}}$ on the same axes.
Analyzing the graph of
y = cot
x
The last trigonometric function we need to explore is
cotangent . The cotangent is defined by the
reciprocal identity$\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x=\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}x}.\text{\hspace{0.17em}}$ Notice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph at
$\text{\hspace{0.17em}}0,\pi ,\text{\hspace{0.17em}}$ etc. Since the output of the tangent function is all real numbers, the output of the
cotangent function is also all real numbers.
We can graph
$\text{\hspace{0.17em}}y=\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ by observing the graph of the tangent function because these two functions are reciprocals of one another. See
[link] . Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases.
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