# Thread: Trigonometric inequality with pi

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.

2. Originally Posted by karseme
$\displaystyle \sin{(\pi x)}>\cos{(\pi \sqrt{x})}$

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.

Thank you very much. It was very helpful.

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?

5. Originally Posted by karseme
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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