<s>[INST] maximum of $f(x)=\sin{(\pi Ax)}\left(\,\csc{(\pi x)}+\csc{(\pi (\frac{1}{A}-x))}\,\right)$ Let a function $f(x)$ be 
$f(x)=\sin{(\pi Ax)}\left(\,\csc{(\pi x)}+\csc{(\pi (\frac{1}{A}-x))}\,\right)$
where $A\geq 2$ is a positive integer and $\csc{(x)}=\frac{1}{\sin{(x)}}$.
I want to prove that on the interval $x\in[0, \frac{1}{2A}]$, $f(x)$ has its maximum at $x=\frac{1}{2A}$.
To prove that, I think we have two approaches.
First, prove it directly.
Second, prove that $f(x)$ is an increasing function  on the interval $x\in[0, \frac{1}{2A}]$.
When $A=2$, we are able to take the second approach thanks to a user named egreg.
Prove that $\sin{(\pi 2x)}\left(\,\csc{(\pi x)}+\csc{(\pi (0.5-x))}\,\right)$ is an increasing function
Now could anybody prove it generally?
I am not sure for any $A$, $f(x)$ is increasing on the interval $x\in[0, \frac{1}{2A}]$ (maybe true?), but strongly believe the maximum is $f(\frac{1}{2A})$.
The followings are graphs of $f(x)$ when $A=2, 3, 4$, respectively, where $\frac{1}{2A}=0.25, 0.1667, 0.125$.



 [/INST] It took me a good chunk of the day and a bunch of headaches, but I was able to write an informal proof that that $f(x)=\sin{(\pi Ax)}\left(\,\csc{(\pi x)}+\csc{(\pi (\frac{1}{A}-x))}\,\right)$ has a maximum at $x=\frac{1}{2A}$ for any $A$ not only for integers >= 2, but for all reals $A < -1$ or $A > 1$.
The trick to proving this is we have to prove f(x)' is 0 and f(x)'' is negative. 
Let $A$ be any number for the reals excluding 0 and 
$f(x) = \sin( \pi A x)\csc(\pi x) + \sin(x \pi A)\csc( \pi(\frac{1}{A} - x))$. 
Now we are going to derive f(x). 
By deriving f(x), we get that:
$f'(x) =  A \pi \cos( \pi A x) \csc( \pi x) 
    - \pi\sin(\pi A x)\csc( \pi x)\cot(\pi x) 
    + \pi A\cos(\pi A x)\csc(\pi (\frac{1}{A}-x))
    +  \pi\sin(\pi Ax)\csc(\pi(\frac{1}{A}-x))\cot( \pi(\frac{1}{A}-x))$
Which reduces to:
$f'(x) =  \pi A \cos(\pi Ax)(\csc(\pi x) + \csc(\pi(\frac{1}{A} - x)) + \pi\sin(\pi Ax)(\csc(\pi(\frac{1}{A}-x))\cot(\pi(\frac{1}{A}-x) - \csc(\pi x)\cot(\pi x))$ 
Now we will substitute in $x = \frac{1}{2A}$ and get:
$f'(x) = \pi A \cos(\pi A(\frac{1}{2A}))(\csc(\pi(\frac{1}{2A})) + \csc(\pi(1/A - (\frac{1}{2A})) + \pi \sin(\pi A(\frac{1}{2A}))(\csc(\pi(1/A-(\frac{1}{2A})))\cot(\pi(1/A-(\frac{1}{2A})) - \csc(\pi(\frac{1}{2A}))\cot(\pi(\frac{1}{2A}))$
Which simplified looks like:
$f'(x) =  \pi A\cos(\frac{\pi}{2})(\csc(\frac{\pi}{2}) + \csc(\frac{\pi}{2A})) 
+ \pi \sin(\frac{\pi}{2})(\csc(\frac{\pi}{2A})\cot(\frac{\pi}{2A}) - \csc(\frac{\pi}{2A})\cot(\frac{\pi}{2A}))$
Since $cos({\pi}{2})) = 0$ and $\csc(\frac{\pi}{2A})\cot(\frac{\pi}{2A}) - \csc(\frac{\pi}{2A})\cot(\frac{\pi}{2A}) = 0$, we can further reduce the problem to:
$f'(x) =  \pi A(0)(\csc(\frac{\pi}{2}) + \csc(\frac{\pi}{2A})) + \pi \sin(\frac{\pi}{2})(0)$
So $f'(x) = 0$ for any real A that does not equal 0 so by definition, $\frac{1}{2A}$ is a critical point.
Now to prove that $x = \frac{1}{2A}$ is a maximum bigger headache.
In order to prove $x =\frac{1}{2A}$ is a maximum, we must prove $f(x)'' < 0$. 
We now have to take the derivative of f(x)' and after much simplification, substituting  $x = \frac{1}{2A}$, and further simplifying, we get:
$f(x)'' = 2\pi^2\sin(\frac{\pi}{2})(\csc(\frac{\pi}{2A})(-A^2+2\cot(\frac{\pi}{2A})^2+1)$
(I have all the work, it's just a lot to post.)
Now $2\pi^2\sin(\frac{\pi}{2})(\csc(\frac{\pi}{2A})> 0$ for all reals except $A=0$,  so when $-A^2+2\cot(\frac{\pi}{2A})^2+1 < 0$ then $\frac{1}{2A}$ is a maximum.
Plugging in $-A^2+2\cot(\frac{\pi}{2A})^2+1$ into Wolfram Alpha, when $-1<=A<=1$ then $-A^2+2\cot(\frac{\pi}{2A})^2+1 > 0$ and when  $A < -1$ or $A > 1$ then $-A^2+2\cot(\frac{\pi}{2A})^2+1 < 0$.
So since f(x)'' is negative and f(x)' is 0, by definition $x=\frac{1}{2A}$ is a maximum for
all reals except for -1 <= A <= 1.
QED
(Thanks to Maple and Wolfram Alpha for helping derive and simplify the derivatives and checking the work.)
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