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Math Help - Trigonometric Inequality

  1. #1
    MHF Contributor Drexel28's Avatar
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    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 0\le x\le\sqrt{\frac{\pi}{2}}. Prove that \sin^2(x)\le\sin\left(x^2\right)
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    MHF Contributor Drexel28's Avatar
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    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 \forall x\in\left[0,1\right] it is true that x\le x^2. Furthermore, since \sin(x) is increasing on this interval it can readily be seen that \sin^2(x)\le\sin(x)\le\sin\left(x^2\right). So this reduces to proving that \forall x\in\left[0,1\right]\quad \sin^2(x)\le\sin\left(x^2\right).
    Last edited by Drexel28; November 6th 2009 at 01:40 PM.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    \forall x\in\left[0,\sqrt{\frac{\pi}{2}}\right] it is true that x\le x^2.
    This is not true ( x=1/2). did you mean x \geq 1?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jose27 View Post
    This is not true ( x=1/2). did you mean x \geq 1?
    Oops. Of course. Thank you
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    MHF Contributor Bruno J.'s Avatar
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    It can be made slightly easier by considering that \forall x\in\left[0,1\right] it is true that x\le x^2.
    Still false! 1/2 is still in there.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Still false! 1/2 is still in there.
    Something's wrong with me haha

    \color{red}\left[1,\sqrt{\frac{\pi}{2}}\right] haha.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    A couple people have asked me about my solution to this problem. Well here it is.

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

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

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

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

    Using this lemma and the previous deductinos we see that \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 \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 \frac{1}{2}\sin^2(z)\le\frac{1}{2}\sin\left(z^2\ri  ght)\quad 0<z<1 and the result follows.
    Last edited by Drexel28; November 10th 2009 at 12:41 PM. Reason: Added Spoiler
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