Question

1.draw the graph of sin inverse x and cos inverse x using the graph of sinx and cos x
2.discuss the concept of mirror image of the two graphs, in each case, about the line y=x.
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Answer 1

Answer:

1) Graph of

${sin}^{ - 1}x$

is symmetric about the line y=x

2) Graph of

${cos}^{ - 1} x$

is not symmetric about the line y=x.

Step-by-step explanation:

We know that sin x is periodic function with time period 2π.So sin function can not be invertible in its complete domain.

If a function is Bijective than only it can be invertible.

So,to defined

${sin}^{ - 1}x$

we have to first adjust the domain of the sin function,in which it can be one-one and onto(i.e. Bijective)

So,[-π/2,π/2] is the suitable part,which is near to origin.

Now,

${sin}^{ - 1}x$

has domain [-π/2,π/2] and range [-1,1],because range and domain are interchanged while a function is being inverse.

Hence the graph of

${sin}^{ - 1} x$

is shifted to Y-axis.(graph1)

Graph 2 is graph of sin x.

Graph 3 is both sin and sin inverse functions.

Graph 4 shown symmetry along line y= x.

By the same way graph of cos inverse can be drawn as shown in graph 5

Hope it helps you.