Is there an identity for the following equation:
$\ f(x) = \sin(x.\pi/2)\cdot \sin(x.\pi/3)\cdot \sin(x.\pi/4) $
I am looking for an equation similar to this equation
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There are ways to change a product of sines and cosines into a sum of sines and cosines; see here for example. The resulting expression is nicer in some ways (it makes the Fourier expansion obvious, for instance), but it's not clear you'll be more satisfied with it than the original formula. – Greg Martin Jan 28 '20 at 21:59
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Where do you stop your product ? Only 3 sines ? Surely not... :) – Jean Marie Jan 28 '20 at 22:19
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With a little experience in this domain (https://math.stackexchange.com/q/1693990), it is doubtful that a simple "closed" formula exists for the product of $\sin(x.\pi/k)$ with $n$ terms. – Jean Marie Jan 28 '20 at 22:30
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Express everything in terms of the trigonometric functions of $\frac {\pi x}{12}$ and working a little $$4\sin \left(\frac{\pi x}{2}\right) \sin \left(\frac{\pi x}{3}\right) \sin \left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi x}{12}\right)+\sin \left(\frac{5 \pi x}{12}\right)+\sin \left(\frac{7 \pi x}{12}\right)-\sin \left(\frac{13 \pi x}{12}\right)$$ It would be much less funny do add the next term; the common factor being $\frac {\pi x}{60}$, we should have $$8\sin \left(\frac{\pi x}{2}\right) \sin \left(\frac{\pi x}{3}\right) \sin \left(\frac{\pi x}{4}\right)\left(\frac{\pi x}{5}\right)=\cos \left(\frac{7 \pi x}{60}\right)+\cos \left(\frac{13 \pi x}{60}\right)-$$ $$\cos \left(\frac{17 \pi x}{60}\right)+\cos \left(\frac{23 \pi x}{60}\right)-\cos \left(\frac{37 \pi x}{60}\right)-\cos \left(\frac{47 \pi x}{60}\right)-\cos \left(\frac{53 \pi x}{60}\right)+\cos \left(\frac{77 \pi x}{60}\right)$$
All the trick is to expand $\sin(nt)$ in terms of powers of $\sin(t)$ and $\cos(t)$ and then the power reduction formulae (look here).
Claude Leibovici
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