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Question: How do you find the domain and range of ln (inverse \[\sin x\] )?...

How do you find the domain and range of ln (inverse sinx\sin x )?

Explanation

Solution

Hint : The domain of a function is the set of all acceptable input values (x-values). The range of a function is the set of all output values (y-values) and to find the domain and range of a given trigonometric function we must know the range and domain of the sine function, hence by applying the range and domain of trigonometric functions we can solve the given function.

Complete step-by-step answer :
Let us write the given function:
ln (inverse sinx\sin x )
Let,
f(x)=ln(sin1x)f\left( x \right) = \ln \left( {{{\sin }^{ - 1}}x} \right)
Now as lny\ln y is defined only if y>0y > 0 , therefore for ln(sin1x)\ln \left( {{{\sin }^{ - 1}}x} \right) to be defined we must have
sin1x>0{\sin ^{ - 1}}x > 0 ,
We know that sin1x{\sin ^{ - 1}}x is defined in the interval [1,1]\left[ { - 1,1} \right] and sin1x>0{\sin ^{ - 1}}x > 0 for x(0,1]x \in \left( {0,1} \right] .
Therefore, the domain of f(x)f\left( x \right) is x(0,1]x \in \left( {0,1} \right] .
And as sin1y{\sin ^{ - 1}}y and lny\ln y are increasing functions, the range of f(x)f\left( x \right) will be
(,ln(π2)]\left( { - \infty ,\ln \left( {\dfrac{\pi }{2}} \right)} \right] .
Hence,
Domain: (0,1]\left( {0,1} \right]
Range: (,ln(π2)]\left( { - \infty ,\ln \left( {\dfrac{\pi }{2}} \right)} \right]

Note : The domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. They may also have been called the input and output of the function. The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.