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This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arise,s to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos tx}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos x - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3x}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5x}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7x}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.

[added a bit later] Come to think of it, multiplying the $t$-th Fourier coefficient by $2^{-n} {n \choose j}$ amounts to convolution with $\cos^n(x)$, which does exactly what we want. Dealing with the coefficients $1/\sin(\pi t/(2n))$ may be harder, but it feels like only one or two more ideas might be needed, at least for the case $2|m$ where the maximum occurs exactly at $x=0$.

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arises, to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos \phantom. tX}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$ and $X=x/n$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos X - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3X}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5X}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7X}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.

[added a bit later] Come to think of it, multiplying the $t$-th Fourier coefficient by $2^{-n} {n \choose j}$ amounts to convolution with $\cos^n(x)$, which does exactly what we want. Dealing with the coefficients $1/\sin(\pi t/(2n))$ may be harder, but it feels like only one or two more ideas might be needed, at least for the case $2|m$ where the maximum occurs exactly at $x=0$.

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arise,s to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos tx}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos x - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3x}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5x}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7x}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arise,s to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos tx}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos x - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3x}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5x}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7x}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.

[added a bit later] Come to think of it, multiplying the $t$-th Fourier coefficient by $2^{-n} {n \choose j}$ amounts to convolution with $\cos^n(x)$, which does exactly what we want. Dealing with the coefficients $1/\sin(\pi t/(2n))$ may be harder, but it feels like only one or two more ideas might be needed, at least for the case $2|m$ where the maximum occurs exactly at $x=0$.

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arise,s to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos ((n-2j)x)}{\sin\bigl(\bigl(\frac12-\frac{j}{n}\bigr)\pi\bigr)} $$ with $t=n-2j$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin\bigl(\bigl(\frac12-\frac{j}{n}\bigr)\pi\bigr)$ behaves like $2n/\pi t$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos x - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3x}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5x}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7x}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin\bigl(\bigl(\frac12-\frac{j}{n}\bigr)\pi\bigr)$, are much too rough for direct estimates to work.

This seems to be a hard problem. What follows is not a proof, but some observations that may eventually lead to a proof.

There seems to be no simple "closed form"; I can numerically confirm that the maximum seems to occur at $\beta=0,\pi$ or $\beta=\pi/2$ according as $m$ is odd or even, but do not have a proof. Numerical observation also indicates that $f(\beta)$, which is an even function of $\beta -\frac\pi2$ and anti-invariant under trranslation by $n\pi$, stays very close to the maximum for nearly half of its period (and thus close the minimum for nearly another half); and in particular that when $m$ is even, $f(\beta)$ comes extremely close to its maximum at $\beta = -\pi/2$ and $\beta = 3\pi/2$. This suggests that proving the inequality is going to be quite hard. It also suggests that there's a good reason for this behavior, and that (as already suggested in the "meta thread") the proposer of the problem please provide the context where this $f(\beta)$ arise,s to help understand this curious behavior.

Let $n=2m+1$ and $g(x) = f(x+\frac\pi2)$, so $g$ is an even function and satisfies $g(x+n\pi) = -g(x)$. Hence $g$ has a Fourier expansion as a linear combination of $\lbrace\cos(tx/n) : t=1,3,5,\ldots,n\rbrace$. Using the complex-exponential formula for the sine (as suggested in a comment by R.May) yields the explicit expansion $$ f(x) = (-1)^m 2^{-n} \sum_{j=0}^n (-1)^j {n\choose j} \frac{\cos tx}{\sin \frac{\pi t}{2n}} $$ with $t=n-2j$. If we try to understand the behavior as $n \rightarrow \infty$, we might imagine (the following is only heuristic!) that the binomial distribution ${n\choose j} / 2^n$ behaves like the matched Gaussian $(\pi n / 2)^{-1/2} \exp(-t^2/(2n))$, while $\sin(\pi t/(2n))$ behaves like $\pi t/(2n)$. Then our sum would be $$ \left(\frac{32n}{\pi^3}\right)^{1/2} \left( \exp\Bigl(\frac{-1^2}{2n}\Bigr) \cos x - \exp\Bigl(\frac{-3^2}{2n}\Bigr) \frac{\cos 3x}{3} + \exp\Bigl(\frac{-5^2}{2n}\Bigr) \frac{\cos 5x}{5} - \exp\Bigl(\frac{-7^2}{2n}\Bigr) \frac{\cos 7x}{7} + - \cdots \right). $$ For large $n$, if we ignore the exponentials we get a square wave of amplitude $$ (32n/\pi^3)^{1/2} (\pi/4) = (2n/\pi)^{1/2}; $$ restoring the exponentials amounts to smoothing that square wave by applying a heat kernel for time proportional to $1/n$, which yields a function that stays very close to the square wave except near the jumps while never exceeding the original amplitude — which indeed looks a lot like what happens for the actual trigonometric sum.

Is there a rigorous way to show that our sort-of-discrete approximation by a trigonometric polynomial to this "heat-smoothed square wave" behaves similarly, and to verify that its maximum occurs at $x=0$ or $x=\pm\pi/2$ according to the parity of $m$? Note that the Gaussian approximation to near-central binomial coefficients, and the linear approximation to $\sin(\pi t/(2n))$, are much too rough for direct estimates to work.