Is there a way to prove that $x > \sin^2(x) , x >0$

Question

The inequality of $$ x > \sin^2(x), \forall x \in ( 0 , + \infty) $$ can be visualized, and easily, we can prove that it's true (via the graph).

However, how can we prove that it's true algebraically?

Answer 1

If $f(x)=\sin^2(x)$, then $f'(x)=2\sin(x)\cos(x)=\sin(2x)<1$, unless $x$ is of the form $\frac\pi4+k\pi$ ($k\in\mathbb Z$). So,$$x=\int_0^x1\,\mathrm dt>\int_0^xf'(t)\,\mathrm dt=f(x)=\sin^2(x).$$

Answer 2

Using $|\sin x | \leqslant 1 $ and the mean value theorem, for $x > 0$

$$\sin^2 x \leqslant |\sin x| = x\left|\frac{\sin x - \sin 0}{x} \right| = x|\cos \xi| \leqslant x$$

Answer 3

Let

$$f(x) = x - \sin^2(x) \tag{1}\label{eq1A}$$

Then you have

$$f'(x) = 1 - 2\sin(x)\cos(x) = 1 - \sin(2x) \ge 0 \tag{2}\label{eq2A}$$

For all $x \neq \frac{\pi}{4} + n\pi$, for some $n \in \mathbb{Z}$, the derivative is always positive. Note $f(0) = 0$, so $f(x)$ will be increasing for $x \gt 0$ up to $x = \frac{\pi}{4}$. However, since $x \gt 1 \ge \sin^2(x)$ at that point and $x$ is continuing to increase, you get that $f(x) \gt 0$ for $x \gt 0$. This means that

$$x \gt \sin^2(x), \; x \gt 0 \tag{3}\label{eq3A}$$