Solution in the first way. We apply the following formulas
sinz=2ieizโeโizโ and sinhz=2ezโeโzโ. Then siniy=2iei(iy)โeโi(iy)โ=2ieโyโeyโ=โi1โ2eyโeโyโ=isinhy, as required.
Solution in the second way. We apply the following formulas
sinz=n=0โ+โโ(2n+1)!(โ1)nโz2n+1 and sinhz=n=0โ+โโ(2n+1)!1โz2n+1. Then
siniy=n=0โ+โโ(2n+1)!(โ1)nโ(iy)2n+1=n=0โ+โโ(2n+1)!(โ1)nโi2n+1y2n+1
=n=0โ+โโ(2n+1)!(โ1)nโi(โ1)ny2n+1=in=0โ+โโ(2n+1)!1โy2n+1=isinhy, as required.
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