N sin^2(x) versus sin^2(Nx)

Hi guys, I'm working through a problem on violations of Bell's inequality using polarised photons passing through filters. I've managed to get right through to the end but for my work to be correct the following statement must be true:

$\displaystyle \forall N \geqslant 2 $ $\displaystyle \exists \theta $ such that $\displaystyle N \sin^2{\theta} - \sin^2{(N \theta)} < 0$

I know this is true, and I'd probably get away with just stating this as fact, but I'm not satisfied with that. It should also be really easy to prove and I'm probably missing something obvious, but any help would be appreciated.

Here's what I've tried so far.

Let $\displaystyle f(\theta) = N \sin^2{\theta} - \sin^2{(N \theta)}$

Then extrema of f occur at $\displaystyle \frac{df}{d \theta} = 0$ $\displaystyle \Rightarrow$ $\displaystyle 2 N \sin{\theta} \cos{\theta} - 2 N \sin{(N \theta)} \cos{(N \theta)} = 0$

$\displaystyle N \sin{2 \theta} - N \sin{(2 N \theta)} = 0$

So extrema occur at $\displaystyle \theta = \frac{k \pi}{N+1} $ and $\displaystyle \frac{(\frac{1}{2} + k) \pi}{N-1}$

My idea was to show that a minimum occurs under the theta axis, but I can't seem to be able to show this (Doh)

pomp.