Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$

Question

Find Domain of $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$

My try:

First of all $x$ belongs to First quadrant

Also $$\sqrt{\sin x}+\sqrt{\cos x}-1 \ge 0$$ Squaring both sides we get

$$\sin x+\cos x+2\sqrt{\sin x\cos x} \ge 1$$

Any clue here?

Answer 1

By periodicity, we can restrict ourselves to $[0, 2\pi[$. As you've noticed, $x$ must be in $X:=[0, \pi/2]$.

One method to continue is to notice that, for $a$ in $[0,1]$, we have $\sqrt{a} \geq a$, therefore for $x$ in $X$ we have $$\sqrt{\sin(x)} + \sqrt{\cos(x)} \geq \sin(x) + \cos(x)$$

It is then easy to check that $\sin(x) + \cos(x) \geq 1$ for $x$ in $X$, for example by squaring. So, the domain of $f$ is $X$.

Answer 2

Hint: you must solve $$\sin(x)\geq 0$$ and $$\cos(x)\geq 0$$ and $$\sqrt{\sin(x)}+\sqrt{\cos(x)}-1\geq 0$$

Answer 3

For $f(x)=\sqrt{\sqrt{\sin x}+\sqrt{\cos x}-1}$, you first of all have to restrict $x$ to the reals.

Then, since $\sin$ and $\cos$ are periodic with period $2\pi$, you can restrict $x \in [0, 2 \pi)$. Any restrictions there are replicated in $[2k\pi, 2(k+1)\pi)$.

Since you need $\sin(x) \ge 0$, this gives $x \in [0, \pi]$.

Since you need $\cos(x) \ge 0$, this gives $x \in [0, \pi/2]\cup [3\pi/2, 2\pi)$.

Combining these gives $x \in [0, \pi/2]$.

Finally, you want $\sqrt{\sin x}+\sqrt{\cos x}-1 \ge 0$.

If $f(x) = \sqrt{\sin x}+\sqrt{\cos x}-1 $, then $f(x) = f(\pi/2-x)$, so we only need to look at $x \in [0, \pi/4]$.

$f(0) = 0$, $f(\pi/4) =\sqrt{2}-1 \gt 0$.

$\begin{array}\\ f'(x) &=\frac12 \cos(x) \sin^{-1/2}(x)-\frac12 \sin(x) \cos^{-1/2}(x)\\ &=\frac12 \dfrac{\cos^{3/2}(x)- \sin^{3/2}(x)}{\sin^{1/2}(x) \cos^{1/2}(x)}\\ \end{array} $

and since $\cos(x) > \sin(x)$ for $x \in [0, \pi/4)$, $f'(x) > 0$ for $0 \le x \lt \pi/4$ so $f(x) \ge 0$ for $0 \le x \le \pi/2$ with equality only at $x=0$ and $x = \pi/2$.

Answer 4

You need $\sqrt{\sin(x)}+\sqrt{\sin(\dfrac{\pi}{2}-x)}\ge 1$. Looking at the graphics you can see that the domain is $$\left([0,\dfrac{\pi}{2}]\cup[2\pi,2\pi+\dfrac{\pi}{2}\cup\cdots\right)\cup\left([-2\pi,-2\pi+\dfrac{\pi}{2}]\cup[-4\pi,4\pi+\dfrac{\pi}{2}]\cup\cdots\right)$$ Finally the domain is the union for all $k\in\mathbb Z$ of the intervals $$\left[2k\pi,2k\pi+\dfrac{\pi}{2}\right]$$