Question

Find the values of x for which the trigonometric function $f(x)=\sqrt{\sin x-\cos x}$ is defined, $x\in \left[ 0,2\pi \right]$.

Answer 1

Hint: In this question, first work out on the value of x that means to work on the domain of the given trigonometric function. Second work out on the domain of the function $\sqrt{f(x)}$ is $f(x)\ge 0$.

Complete step-by-step answer:

The given trigonometric function is

$f(x)=\sqrt{\sin x-\cos x}$

We know that, the domain of the function $\sqrt{f(x)}$ is $f(x)\ge 0$

$\sin x-\cos x\ge 0$

Rearranging the term, we get

$\sin x\ge \cos x$

For what values of x, the sine trigonometric function is greater than or equal to cosine trigonometric function. This will be shown in the below figure.

The figure shows that the sine trigonometric function is greater than or equal to cosine trigonometric function in the interval $\left[ \dfrac{\pi }{4},\dfrac{5\pi }{4} \right]$ .

Hence the values of x for the given trigonometric function lie in the interval $\left[ \dfrac{\pi }{4},\dfrac{5\pi }{4} \right]$

Note: You might get confused about the difference between the domain of a function and the range of a function. Domain is the independent variable and range is the dependent variable. On the other hand, range is defined as a set of all probable output values. Domain is what is put into a function, whereas range is what is the result of the function with the domain value.