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> How is sin(npi/2) = (-1)^((n-1)/2) for odd n?
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How is sin(npi/2) = (-1)^((n-1)/2) for odd n?
jaejoon89
Sep19-09, 08:28 AM
I know the MacLaurin series for sin(x) but can't figure out how this reduces to the above (if it does).
JG89
Sep19-09, 08:45 AM
sin(npi/2) is going to be equal to -1 or 1, depending on the value of n, right? So the formula seems plausible. You could prove it through induction.
jaejoon89
Sep19-09, 09:30 AM
thanks
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