MHB Trigonometric inequality with pi

[karseme]

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

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 $$ \pi $$, 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 $$ x=(\sqrt{x})^2 $$, 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.

 

[I like Serena]

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

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 $$ \pi $$, 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 $$ x=(\sqrt{x})^2 $$, 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.

 

[karseme]

Thank you very much. :) It was very helpful.

 

[karseme]

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

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

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

So,

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

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

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

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

Is this good? But, when I solve the given inequality for $$ \dfrac{1}{2} \leq x \leq 1 $$ 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 like Serena]

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

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

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

So,

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

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

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

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

Is this good? But, when I solve the given inequality for $$ \dfrac{1}{2} \leq x \leq 1 $$ 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. (Sweating)