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Question

Question: Sketch the graph for \(y={{\sec }^{-1}}\sec x\)....

Sketch the graph for y=sec1secxy={{\sec }^{-1}}\sec x.

Explanation

Solution

Hint: In this question, we will first write secant function in terms of cos then understand the relation of cosx\cos x and cos1x{{\cos }^{-1}}x. We will observe the graph of cosx\cos x , and how they change signs and use it to plot a graph of y=sec1secxy={{\sec }^{-1}}\sec x.
Complete step-by-step answer:
In given question, we have,
y=sec1secxy={{\sec }^{-1}}\sec x
This can be written as
secy=secx\sec y=\sec x.
We know that secx=1cosx\sec x=\dfrac{1}{\cos x}, therefore, we can write above equation as,
1cosy=1cosx\dfrac{1}{\cos y}=\dfrac{1}{\cos x}
Cross multiplying this, we get,
cosx=cosy\cos x=\cos y
This can be written as,
y=cos1cosxy={{\cos }^{-1}}\cos x.
So, the graph of y=sec1secxy={{\sec }^{-1}}\sec x is the same as the graph of y=cos1cosxy={{\cos }^{-1}}\cos x.
Now, cosx\cos x is a periodic function with period 2π2\pi , which means its values repeat in the same pattern after 2π2\pi increases in xx. That is, cosx=cos(2π+x)\cos x=\cos \left( 2\pi +x \right).
Since, cos x is periodic with period 2π2\pi . Therefore, y=cos1cosxy={{\cos }^{-1}}\cos x is also period with period 2π2\pi .
Also, the domain here is a set of those values of xx for which cos1cosx{{\cos }^{-1}}\cos x is defined. And, range is the set of values where cos1cosx{{\cos }^{-1}}\cos x lies.
Now, for all real values of xx, cosx\cos x lies between -1 and 1. And, between -1 and 1, the inverse function of cosine is defined. Therefore, cos1cosx{{\cos }^{-1}}\cos x is defined for all real values ofxx.
We know, graph of y=cosxy=\cos x is:

We see that, in the interval [π,π]\left[ -\pi ,\pi \right] , for two different values of xx , we have the same value of yy.
Also, from definition of cosine inverse, in this graph, we get,
cos1y=x{{\cos }^{-1}}y=x
If we substitute y=cosxy=\cos x here, we get,
cos1cosx=x{{\cos }^{-1}}\cos x=x.
Now, in graph of cos1cosx{{\cos }^{-1}}\cos x, we have,
y=cos1cosxy={{\cos }^{-1}}\cos x
y=x\Rightarrow y=x
But, in the interval [π,π]\left[ -\pi ,\pi \right] , for two different values of xx , we have the same value of yy .
Let those two different values be represented by y1,y2{{y}_{1}},{{y}_{2}}.
Now, as xx increases from π-\pi to 0, cosx\cos x increases from -1 to 1, and hence, cos1cosx{{\cos }^{-1}}\cos x decreases from π\pi to 0. Therefore, here we will have, y1=x{{y}_{1}}=-x.
And as it increases from 0 to π\pi , cosx\cos x decreases from 1 to -1, and hence, cos1cosx{{\cos }^{-1}}\cos x increases from 0 to π\pi . Therefore, here we will have, y2=x{{y}_{2}}=x .
Also, from π-\pi toπ\pi , the length of interval is 2π2\pi and cos1cosx{{\cos }^{-1}}\cos x periodic with period 2π2\pi . Therefore, the rest of the graph will repeat the same as in interval [π,π]\left[ -\pi ,\pi \right].
Hence, the graph of cos1cosx{{\cos }^{-1}}\cos x is given by:

Hence for the graph of y=sec1secxy={{\sec }^{-1}}\sec x is plotted above.

Note: While plotting the graph, keep in mind that for two different values of xx , cos1cosx{{\cos }^{-1}}\cos x will have the same value in interval of length 2π2\pi . So, looking at y=xy=x, do not directly plot a graph of an infinite straight line.