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  1. MHB Apprentice

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    #1
    $ \displaystyle \sin{(\pi x)}>\cos{(\pi \sqrt{x})} $

    I don't know how to solve this. I would really appreciate some help.
    I tried to do something, but didn't get anything.

    If I move cos to the left side, I can't apply formulas for sum. Since arguments of sin and cos have $ \displaystyle \pi $, I think there is no way I can somehow make it simpler by using addition formulas. If I could somehow get rid of that square root, but how?! I know that $ \displaystyle x=(\sqrt{x})^2 $, but what's use of that when I don't see how to get rid of that power of 2. I tried squaring everything and doing something, but I didn't get anything from that. I don't know how to proceed. I don't see there are any formulas which I could use to make this simpler.

    Must solve this somehow, would appreciate your help.

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    #2
    Quote Originally Posted by karseme View Post
    $ \displaystyle \sin{(\pi x)}>\cos{(\pi \sqrt{x})} $

    I don't know how to solve this. I would really appreciate some help.
    I tried to do something, but didn't get anything.

    If I move cos to the left side, I can't apply formulas for sum. Since arguments of sin and cos have $ \displaystyle \pi $, I think there is no way I can somehow make it simpler by using addition formulas. If I could somehow get rid of that square root, but how?! I know that $ \displaystyle x=(\sqrt{x})^2 $, but what's use of that when I don't see how to get rid of that power of 2. I tried squaring everything and doing something, but I didn't get anything from that. I don't know how to proceed. I don't see there are any formulas which I could use to make this simpler.

    Must solve this somehow, would appreciate your help.
    Hi karseme!

    Suppose we assume that both angles are in the first quadrant.
    That is, $0\le \pi x \le \frac\pi 2$ and $0\le \pi \sqrt x \le \frac\pi 2$.
    So $0\le x \le \frac 14$.

    Then:
    $$\sin{(\pi x)}>\sin{(\frac\pi 2 - \pi \sqrt{x})}$$
    Since the sine is increasing in the first quadrant, we get:
    $$\pi x>\frac\pi 2 - \pi \sqrt{x}$$
    It follows that:
    $$\sqrt x > \frac 12 (\sqrt 3 - 1)$$
    And since $\sqrt{}$ is increasing, we find:
    $$\frac 12 (2-\sqrt 3) < x \le \frac 14$$

    We can continue with considering cases if the angles are in other quadrants.

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    #3 Thread Author
    Thank you very much. It was very helpful.

  4. MHB Apprentice

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    #4 Thread Author
    So, if we assume that both angles are in the second quadrant, then the following must be true:
    $ \displaystyle \dfrac{\pi}{2} \leq \pi x \leq \pi \qquad \land \qquad \dfrac{\pi}{2} \leq \pi \sqrt{x} \leq \pi \, \implies \dfrac{1}{2} \leq x \leq 1 $

    Since $ \displaystyle \pi \sqrt{x} $ is in the second quadrant we have:

    $ \displaystyle cos{(\pi \sqrt{x})}=-\cos{(\pi-\pi \sqrt{x})}=-\sin{(\dfrac{\pi}{2} + \pi - \pi \sqrt{x})}=-\sin{(\dfrac{3\pi}{2}- \pi \sqrt{x})}=\sin{(\pi \sqrt{x}-\dfrac{3\pi}{2})} $

    So,

    $ \displaystyle \sin{(\pi x)} > \sin{(\pi \sqrt{x}-\dfrac{3\pi}{2})} $

    Since sine is decreasing in the second quadrant then we have:

    $ \displaystyle \pi x < \pi \sqrt{x}-\dfrac{3\pi}{2} $

    This quadratic inequality does not have solution(for t=\sqrt{x}).

    Is this good? But, when I solve the given inequality for $ \displaystyle \dfrac{1}{2} \leq x \leq 1 $ it is true.

    So, those are the solutions...is it any good?

    But what about periodicity of sine? then those can't be the only solutions?

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    #5
    Quote Originally Posted by karseme View Post
    So, if we assume that both angles are in the second quadrant, then the following must be true:
    $ \displaystyle \dfrac{\pi}{2} \leq \pi x \leq \pi \qquad \land \qquad \dfrac{\pi}{2} \leq \pi \sqrt{x} \leq \pi \, \implies \dfrac{1}{2} \leq x \leq 1 $

    Since $ \displaystyle \pi \sqrt{x} $ is in the second quadrant we have:

    $ \displaystyle cos{(\pi \sqrt{x})}=-\cos{(\pi-\pi \sqrt{x})}=-\sin{(\dfrac{\pi}{2} + \pi - \pi \sqrt{x})}=-\sin{(\dfrac{3\pi}{2}- \pi \sqrt{x})}=\sin{(\pi \sqrt{x}-\dfrac{3\pi}{2})} $

    So,

    $ \displaystyle \sin{(\pi x)} > \sin{(\pi \sqrt{x}-\dfrac{3\pi}{2})} $

    Since sine is decreasing in the second quadrant then we have:

    $ \displaystyle \pi x < \pi \sqrt{x}-\dfrac{3\pi}{2} $

    This quadratic inequality does not have solution(for t=\sqrt{x}).

    Is this good? But, when I solve the given inequality for $ \displaystyle \dfrac{1}{2} \leq x \leq 1 $ it is true.

    So, those are the solutions...is it any good?

    But what about periodicity of sine? then those can't be the only solutions?
    I'm afraid we have to consider $\pi x$ to be in any quadrant and $\pi\sqrt x$ to be in the same or any other quadrant.
    Furthermore, we have to consider that they may have an additional $2k\pi$ added to them that can be different for each of them.

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