A friend of mine thought about this problem, and I got very interested in it, but I couldn't develop it. Could some help me, because I got stuck trying to evaluate the expression using $n=2k+1$ and analyzing it for $n=2k$ too?

Prove that $\frac{\sin(nx)}{\sin^n(x)}$, for natural $n$, only has minimum values for n odd.

Thanks for any help :)


For $n$ even, the function is odd and as it goes to infinity in absolute value when $x$ goes to zero, it obviously goes to minus infinity for $x$ going to zero from the left.

For $n > 1$ odd, the function is even, periodic with period $\pi$ and if we restrict it say on $[-\pi/2, \pi/2]$ which is a full period, the only singularity is at zero, where it goes to plus infinity from both sides, so the minimum is attained by continuity if we restrict it to the compact set $[-\pi/2, \pi/2] - (-\epsilon, \epsilon)$ for some small positive $\epsilon$ chosen such that the function is big (greater than zero will do actually but we can say greater than 100 for example) on $[-\epsilon, \epsilon]$ which is possible since the limit at zero is plus infinity.


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