<s>[INST] Evaluate position of first secondary maximum of $\frac{\sin N (x/2)}{\sin (x/2)}$ The function
$$f(x) = \displaystyle \left | \frac{\sin \left( N \displaystyle \frac{x}{2} \right)}{\sin \left( \displaystyle \frac{x}{2} \right)} \right |$$
when evaluated for $x > 0$, has its first zero in
$$x = \frac{2 \pi}{N}$$
But the first secondary maximum is in
$$x = \frac{3 \pi}{N}$$
and not in
$$x = \frac{\pi}{N}$$
because here the function is decreasing from the maximum reached for $x = 0$ (where $f(x) = N$).
My questions are: why? And how can it be seen?
I tried to compute the derivative of $f(x)$:
$$f'(x) = \displaystyle \frac{ \displaystyle \frac{N}{2} \cos \left( \frac{N x}{2} \right) \sin \left( \frac{x}{2} \right) - \frac{1}{2} \sin \left( \frac{N x}{2} \right) \cos \left( \frac{x}{2} \right)}{\sin^2 \left( \displaystyle \frac{x}{2} \right)}$$
for $0 \leq x < \displaystyle \frac{2 \pi}{N}$. It is not immediately visible that it is always negative in that range. I tried to rearrange the numerator with the Werner formulas, but I did not obtain a more clear expression.
Are there any other methods or assumptions that can be done in order to prove that between $x = 0$ and $x = \displaystyle \frac{2 \pi}{N}$ the function $f(x)$ is monothonically decreasing?
 [/INST] By using your formula for $f'(x)$ we have to show that
$$N \cos\left(\frac{Nx}{2}\right) \sin\left(\frac{x}{2}\right) \le \sin\left(\frac{Nx}{2}\right) \cos\left(\frac{x}{2}\right)$$
when $0 \le x < \frac{2\pi}{N}$. This clearly holds if $N = 1$. Assume thus that $N \ge 2$ and write $y = x/2$.
If $\frac{\pi}{2} \le Ny < \pi$, then $\cos(Ny) \le 0$, and the inequality is trivial. 
Assume therefore that $0 \le Ny < \frac{\pi}{2}$. The inequality is then equivalent to
$$N\tan(y) \le \tan(N y).$$
This is true because $\tan(y)$ is a superadditive function in that range. (It is enough to check that it is convex and $0$ at $0$.)
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