# Trigonometric Inequality

• Nov 5th 2009, 02:08 PM
Drexel28
Trigonometric Inequality
Hello everyone. This problem is not too difficult, but let's see how many distinct solutions we can find for it.

Problem: Let $\displaystyle 0\le x\le\sqrt{\frac{\pi}{2}}$. Prove that $\displaystyle \sin^2(x)\le\sin\left(x^2\right)$
• Nov 6th 2009, 08:37 AM
Drexel28
Just in case someone wants to try this but they're having a bit of trouble started. It can be made slightly easier by considering that $\displaystyle \forall x\in\left[0,1\right]$ it is true that $\displaystyle x\le x^2$. Furthermore, since $\displaystyle \sin(x)$ is increasing on this interval it can readily be seen that $\displaystyle \sin^2(x)\le\sin(x)\le\sin\left(x^2\right)$. So this reduces to proving that $\displaystyle \forall x\in\left[0,1\right]\quad \sin^2(x)\le\sin\left(x^2\right)$.
• Nov 6th 2009, 12:55 PM
Jose27
Quote:

Originally Posted by Drexel28
$\displaystyle \forall x\in\left[0,\sqrt{\frac{\pi}{2}}\right]$ it is true that $\displaystyle x\le x^2$.

This is not true ($\displaystyle x=1/2$). did you mean $\displaystyle x \geq 1$?
• Nov 6th 2009, 01:40 PM
Drexel28
Quote:

Originally Posted by Jose27
This is not true ($\displaystyle x=1/2$). did you mean $\displaystyle x \geq 1$?

Oops. Of course. Thank you (Giggle)
• Nov 6th 2009, 02:03 PM
Bruno J.
Quote:

It can be made slightly easier by considering that $\displaystyle \forall x\in\left[0,1\right]$ it is true that $\displaystyle x\le x^2$.
Still false! 1/2 is still in there.
• Nov 7th 2009, 12:56 PM
Drexel28
Quote:

Originally Posted by Bruno J.
Still false! 1/2 is still in there.

Something's wrong with me haha

$\displaystyle \color{red}\left[1,\sqrt{\frac{\pi}{2}}\right]$ haha.
• Nov 10th 2009, 06:55 AM
Drexel28
A couple people have asked me about my solution to this problem. Well here it is.

Problem: Prove that $\displaystyle \sin^2(x)\le\sin\left(x^2\right)\quad\forall 0<x\le\sqrt{\frac{\pi}{2}}$

Spoiler:
Proof: Clearly for $\displaystyle 1\le x\le\sqrt{\frac{\pi}{2}}$ it is obvious that $\displaystyle x\le x^2$. And since $\displaystyle \sin(x)$ is increasing on that interval it stands to reason that $\displaystyle \sin(x)\le\sin\left(x^2\right)\quad 1\le x\le\sqrt{\frac{\pi}{2}}$. Lastly noting that $\displaystyle \sin^2(x)\le\sin(x)\quad\forall x\in\mathbb{R}$. We may finally deduce that the proposed inequality is definitely true for $\displaystyle 1\le x\le\sqrt{\frac{\pi}{2}}$/ Therefore let us restrict our attention to $\displaystyle 0<x<1$. On this interval $\displaystyle x^2\le x$ and since $\displaystyle \cos(x)$ is decreasing on this interval we see that $\displaystyle \cos\left(x\right)\le\cos\left(x^2\right)\quad 0<x<1$. We need the following lemma.

Lemma: $\displaystyle \sin(x)<x\quad 0<x<1$

Proof: Let $\displaystyle \phi(x)=x-\sin(x)$. Then $\displaystyle \phi(0)=0$ and $\displaystyle \phi'(x)=1-\cos(x)$ and since $\displaystyle \cos(x)<1\quad 0<x<1$ the conclusion follows. $\displaystyle \blacksquare$

Using this lemma and the previous deductinos we see that $\displaystyle \cos(x)\le\cos\left(x^2\right)\implies \sin(x)\cos(x)\le\sin(x)\cos\left(x^2\right)\le x\cos\left(x^2\right)$. Now evaluating $\displaystyle \int_0^z \sin(x)\cos(x)dx\le\int_0^z x\cos\left(x^2\right)dx\quad 0<z<1$ gives us $\displaystyle \frac{1}{2}\sin^2(z)\le\frac{1}{2}\sin\left(z^2\ri ght)\quad 0<z<1$ and the result follows.