Prove $\sin^2(x)<\sin(x^2)$ for $0<x<\sqrt{\frac{\pi}{2}}$

Question

I'm trying to prove $\sin^2(x)<\sin(x^2)$ for $0<x<\sqrt{\frac{\pi}{2}}$.

Attempt: This is equivalent to showing $f(x)=\sin(x^2)-\sin^2(x) = \sin(x^2)-\left(\frac{1-\cos(2x)}{2}\right)>0$. Since $f(0)=0$, if we show $f$ is increasing we are done. $f'(x) = 2x\cos(x^2)-\sin(2x)$ but I can't see why it's greater than zero.

Answer 1

Finishing ellya's proof: For $x>1$ we have $$\sin^2(x)< \sin(x) < \sin(x^2)$$ Since $\sin(x)$ is increasing in our domain. For $x\le1$: $$\cos(x^2)\ge\cos(x), x>\sin(x)\implies 2x\cos(x^2)>2\sin(x)\cos(x)=\sin(2x)$$

Answer 2

My thoughts at the moment are to take a step back from what you have put, and see that on $0\lt x\lt\sqrt{\frac{\pi}{2}}$ , $\sin(x)\lt1$ so here $\sin^2(x)\lt\sin(x)$, now it is a question of showing that $\sin(x)\lt \sin(x^2)$

Answer 3

Here's the brute force way:

$$\sin^2(x)<\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)^2<x^2-\frac{(x^2)^3}{3!}<\sin(x^2)$$ for $0<x<\sqrt\frac{\pi}2$.

But it's not pretty to verify the middle inequality, although it's surely doable.

You can check it on Wolfram|Alpha (solution).

Answer 4

By using the Weierstrass product for the sine function, $$\frac{\sin x}{x}=\prod_{k=1}^{+\infty}\left(1-\frac{x^2}{k^2\pi^2}\right),$$ you have that proving: $$\forall k\geq 1,\forall x\in(0,\pi/2),\qquad \left(1-\frac{x}{k^2\pi^2}\right)^2<\left(1-\frac{x^2}{k^2\pi^2}\right)\tag{1}$$ is enough. The RHS of $(1)$ is a concave function over $I=(0,\pi/2)$, so we have, for instance: $$\forall x\in I,\qquad 1-\frac{x^2}{k^2\pi^2}>1-\frac{2x}{\pi k^2},$$ while the LHS of $(1)$ is a convex function over $I$. Since the values of the LHS and the RHS in $(1)$ agree for $x=0$, in order to prove $(1)$ we only need to prove: $$\forall k\geq 1,\qquad \left(1-\frac{1}{2\pi k^2}\right)^2 < 1-\frac{1}{4k^2},$$ that is equivalent to: $$\left(4-\frac{1}{\pi k^2}\right)>\pi.\tag{2}$$ Now $(2)$ is trivial since for any $k\geq 1$ we have $\frac{1}{\pi k^2}\leq\frac{1}{\pi}$ and $4-\frac{1}{\pi}>\pi$ holds.