Question

Find the domain of $f\left( x \right)=\sqrt{\cos \left( \sin x \right)}$.

Answer 1

Hint:We will be using the concept of trigonometric and algebraic function to solve the problem. We will first use the fact that for $f\left( x \right)=\sqrt{g\left( x \right)}$ , $g\left( x \right)$ must always be positive. So, we will have only those values in the domain which satisfies it.

Complete step-by-step answer:
Now, we have been given the function $f\left( x \right)=\sqrt{\cos \left( \sin x \right)}$.
Now, we know that for the domain of a function of type,
$f\left( x \right)=\sqrt{g\left( x \right)}$
We have to find the value of x for which $g\left( x \right)>0$ as there cannot be a negative number inside the square root. So, we have,
$\cos \left( \sin x \right)>0..........\left( 1 \right)$
Now, we know that the range of $\sin x$ is,
$-1\le \sin x\le 1\ \ \forall x\in R.........\left( 2 \right)$
So, $\sin x$ always lies between -1 and 1 for all the values of $x\in R$.
Now, from (1) we have that $\cos \left( \sin x \right)>0$ and we have to find the value of x for which this inequality is true.
Now, from (1) we have that $\cos \left( \sin x \right)>0$.
Now, we let $\sin x=\phi$. So, we have $\cos \left( \phi \right)>0$.
We know that the solution for $\cos \left( \phi \right)>0$ is $\left[ -\dfrac{\pi }{2}+2\pi k,\dfrac{\pi }{2}+2\pi k \right]$ where k is any integer. So, we have some intervals which satisfy (2) as for k = - 1, 0, 1 respectively as,
$\left[ \dfrac{-5\pi }{2},\dfrac{-3\pi }{2} \right],\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right],\left[ \dfrac{3\pi }{2},\dfrac{5\pi }{2} \right]$
Now, we have taken $\sin x$ as $\phi$ and we know that $-1\le \sin x\le 1$ for all $x\in R$. So, we can write is as,
$-1\le \phi \le 1$
Now, we can approximate $1\ as\ \dfrac{\pi }{3}\ and\ -1\ as\ -\dfrac{\pi }{3}$ to see in which range the inequality lies. So, we have,
$-\dfrac{\pi }{3}\le \phi \le \dfrac{\pi }{3}............\left( 3 \right)$
Now, since $-1\le \sin x\le 1$. Therefore, we can ignore all values except $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.
Now, we have the only restriction of $\sin x$ is that it lies in $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and from (3).
We can satisfy it for all values of $x\in R$. Hence, the domain of $\sqrt{\cos \left( \sin x \right)}$ is $x\in R$.

Note: To solve these type of question it is important to note that we have used the fact that the solution of $\cos \left( \phi \right)>0$ is $\left[ -\dfrac{\pi }{2}+2\pi k,\dfrac{\pi }{2}+2\pi k \right]$ for all $k\in \mathcal{Z}$. Also, it has to be noted that we have used the fact that for $\sqrt{g\left( x \right)}$ , $g\left( x \right)$ should be greater than 0.