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That is, how can one solve the equation

$\cos(\pi t)\sin(t) = \cos[\pi(t+T)]\sin[t+T]\text{ for all } t\in\mathbb{R}^+$

for $T$ assuming that $T$ is also a positive real? I have tried using a whole load of trig identities and tricks but frankly, it's just a mess and I don't want to regurgitate that nasty stuff here. Upon plotting the function I can see that $T\approx 50$ (that it indeed has regular periodicity on a seemingly large-ish interval) but I have no idea how to determine this analytically. Any help is appreciated.

Ern
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  • Hint: there is no such period $T$. Related: Period of the sum/product of two functions. – dxiv Sep 19 '16 at 23:52
  • I'm supposed to find the Fourier series for this thing. Does this mean that there is no such series? Have I not looked closely enough at my plot such that I have been deceived?! If there's no period then it's really close. Anyway thanks for the link. I will check it out. – Ern Sep 19 '16 at 23:55
  • Technically not, but that depends on the context. See Can a non-periodic function have a Fourier series?. – dxiv Sep 20 '16 at 00:01
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    wow, you're very helpful off the cuff. Thanks very much! Seems I have much to learn here – Ern Sep 20 '16 at 00:02
  • If there's no period then it's really close. Note that $cos(r t) cos(t)$ is periodic for any rational $r \in \mathbb{Q}$. For example, replacing $\pi$ with its well known approximation $\frac{22}{7}$ gives the function $cos(\frac{22}{7} t) cos(t)$ with the period $14 \pi \approx 44$ which is in the ballpark range of $50$. – dxiv Sep 20 '16 at 00:27

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