MHB Trigonometric inequality with pi

karseme
Messages
13
Reaction score
0
$$ \sin{(\pi x)}>\cos{(\pi \sqrt{x})} $$

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.
 
Mathematics news on Phys.org
karseme said:
$$ \sin{(\pi x)}>\cos{(\pi \sqrt{x})} $$

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.
 
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:
$$ \dfrac{\pi}{2} \leq \pi x \leq \pi \qquad \land \qquad \dfrac{\pi}{2} \leq \pi \sqrt{x} \leq \pi \, \implies \dfrac{1}{2} \leq x \leq 1 $$

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?
 
karseme said:
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)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top