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x-ch-sinx-developp-at-fourier-serie-




Question Number 145939 by Mathspace last updated on 09/Jul/21
Ψ(x)=ch(sinx)  developp Ψ at fourier serie
$$\Psi\left({x}\right)={ch}\left({sinx}\right) \\ $$$${developp}\:\Psi\:{at}\:{fourier}\:{serie} \\ $$

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