Given:
color(red)(y = f(x) = 4 csc(2x)y=f(x)=4csc(2x)
How to draw a graph for this trigonometric function?
Note that color(green)(y = f(x) = csc(x)y=f(x)=csc(x) is the base function.
Observe that color(blue)(csc(x) = 1/sin(x)csc(x)=1sin(x)
Analyze the graph below:
![enter image source here]()
Note that the function y=f(x)=sin(x)y=f(x)=sin(x) has zeroszeros at x=kpix=kπ, where kk is an integer.
The function y=f(x)=csc(x)y=f(x)=csc(x) has color(blue)"No "No color(blue)(zeroszeros.
For both the functions sin(x) and csc(x)sin(x)andcsc(x), Period = 2pi=2π.
Graph of the function csc(x)csc(x) does not have a maximum or a minimum value, there is color(blue)"No "No color(blue)(amplitudeamplitude.
If values of sin(x)sin(x) is available, one can figure out point by point what the values of csc(x)csc(x) are.
The function goes to infinity periodically and is symmetric with the origin.
At values of xx for which sin(x) = 0sin(x)=0, the function csc(x)csc(x) is undefined.
The x-intercept of y=sin(x)y=sin(x) and the asymptotes of y=csc(x)y=csc(x) are the same.
Next, consider the given trigonometric function:
color(blue)(y = f(x) = 4 csc(2x)y=f(x)=4csc(2x)
Use the form:
A Csc(BX - C) + D.
The variables used gives us the Amplitude and Period.
A=4; B=2; C=0 and D=0A=4;B=2;C=0andD=0 (using the given trigonometric function).
Amplitude = None
Period = (2pi)/B=(2pi)/|2| = pi=2πB=2π|2|=π
Vertical Shift = D = 0
Frequency =1/(Period) = 1/pi=1Period=1π
To draw the graph, we can select a few points as shown below:
![enter image source here]()
csc(x)csc(x) has only Vertical Asymptotes.
Vertical Asymptote = x=(pi n)/2x=πn2, where n is an integer.
Graph of color(blue)(y = f(x) = 4 csc(2x)y=f(x)=4csc(2x)
![enter image source here]()
x-intercepts and y-intercepts = None
