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Question: Find the domain and range for function \(y = \csc x\)?...

Find the domain and range for function y=cscxy = \csc x?

Explanation

Solution

Here in this question, we need to find the domain and range of the given function. The domain refers to the set of possible values of x for which the function will be defined and the range refers to the possible range of values that the function can attain for those values of x which are in the domain of the function.

Complete step by step solution:
cscx\csc x is a trigonometric function and it is the reciprocal of other trigonometric function, sinx\sin x. The domain and range of cscx\csc x are related to the domain and range of sinx\sin x.
We know that the range of y=sinxy = \sin x is 1y1 - 1 \leqslant y \leqslant 1. So, we can say that the range of y=cscxy = \csc x is y1y \leqslant - 1 or 1y1 \leqslant y, which encompasses the reciprocal of every value in the range of sinx\sin x.
The domain of y=cscxy = \csc x is every value in the domain of sinx\sin x with the exception of where sinx=0\sin x = 0. Since the reciprocal of zero is undefined. So,
sinx=0 x=0+π(n),  where  nZ  \sin x = 0 \\\ x = 0 + \pi \left( n \right),\;where\;n \in \mathbb{Z} \\\
So, we can say that the domain of y=cscxy = \csc x is xR,xπ(n),nZx \in \mathbb{R},x \ne \pi \left( n \right),n\mathbb{Z}.

Therefore, the domain of y=cscxy = \csc x is xR,xπ(n),nZx \in \mathbb{R},x \ne \pi \left( n \right),n\mathbb{Z} and the range of y=cscxy = \csc x is y1y \leqslant - 1 or 1y1 \leqslant y.

Note: An independent set of those values for a given function which on substitution always gives real value of result but set of value of range is depending upon the value of the set of domains as it is necessary that all ranges must have domain is known as domain. When a function is defined, its domain and co-domain are also mentioned. The domain and co-domain need not be different. They may be the same as well.