Hello everyone. This problem is not too difficult, but let's see how many distinct solutions we can find for it.

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

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- Nov 5th 2009, 02:08 PMDrexel28Trigonometric Inequality
Hello everyone. This problem is not too difficult, but let's see how many distinct solutions we can find for it.

Let $\displaystyle 0\le x\le\sqrt{\frac{\pi}{2}}$. Prove that $\displaystyle \sin^2(x)\le\sin\left(x^2\right)$**Problem:** - Nov 6th 2009, 08:37 AMDrexel28
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 PMJose27
- Nov 6th 2009, 01:40 PMDrexel28
- Nov 6th 2009, 02:03 PMBruno J.Quote:

It can be made slightly easier by considering that $\displaystyle \forall x\in\left[0,1\right]$ it is true that $\displaystyle x\le x^2$.

- Nov 7th 2009, 12:56 PMDrexel28
- Nov 10th 2009, 06:55 AMDrexel28
A couple people have asked me about my solution to this problem. Well here it is.

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

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