Inequality quick question context cauchy fresnel integral

[binbagsss]

Homework Statement

please see attached, I am stuck on the second inequality.

Homework Equations

attached

The Attempt at a Solution

I have no idea where the $2/\pi$ has come from, I'm guessing it is a bound on $sin \theta$ for $\theta$ between $\pi/4$ and $0$ ?

I know $sin \theta \approx \theta$ for small $\theta$ so we could bound it to $\pi/4$ since $sin$ increases in this range $0, \pi /4$, however that would be a stronger approximation , loosing the actual integration over $\theta$ which is clearly not what has been done.

Many thanks in advance.

 

[Charles Link]

I think I managed to prove it. Let $x=\frac{\pi}{4}-x'$. Starting with $sin(\frac{\pi}{4}-x') \geq \frac{\frac{\pi}{4}-x'}{\frac{\pi}{2}}$ ==>> (This must be proven to be the case), This leads to $cos(x')-sin(x')-\sqrt{2}(\frac{1}{2}-(\frac{2}{\pi})x') \geq 0$ for $0<x'< \frac{\pi}{4}$. The function (on the left side of the second inequality) can be shown to be monotonically decreasing from $1 -\frac{\sqrt{2}}{2}$ to $0$ as $x'$ goes from $0$ to $\frac{\pi}{4}$ by looking at the first derivative. You can try to work through what I did and see if you agree. $\\$ Editing: I found a sign error when I took the derivative=scratch the proof=it is incomplete at present... $\\$ Additional editing: A little algebra on the derivative shows it indeed is monotonically decreasing=the proof appears to work. The proof of the first derivative being less than zero on the interval $0<x'< \frac{\pi}{4}$ has one step to it that may not be so obvious, but I believe it works. I would be happy to show you a couple of the steps if you get stuck.

 

[binbagsss]

The function (on the left side of the second inequality) can be shown to be monotonically decreasing from $1 -\frac{\sqrt{2}}{2}$ to $0$ as $x'$ goes from $0$ to $\frac{\pi}{4}$ by looking at the first derivative.

mmm okay thanks.

I am stuck on making the final conclusion from what has been done,

 

[Charles Link]

The first derivative is constantly negative over the interval and at the very right ($x'=\frac{\pi}{4}$) the function has the value zero. The function is thereby positive for the interval $0<x'< \frac{\pi}{4}$. The first derivative is $f'(x')=-sin(x')-cos(x')+2 \sqrt{2}/\pi$. To show this is less than zero over the interval $0<x'< \frac{\pi}{4}$, it helps to see that $sin(x')+cos(x')=\sqrt{2} cos(x'-\frac{\pi}{4})$. Then the point $x'=0$ is the point of most concern, because that is where $cos(x'-\frac{\pi}{4})$ has its minimum value on the interval $0<x'<\frac{\pi}{4}$..