Solution in the first way. We apply the following formulas
sinz=2ieiz−e−iz and sinhz=2ez−e−z. Then siniy=2iei(iy)−e−i(iy)=2ie−y−ey=−i12ey−e−y=isinhy, as required.
Solution in the second way. We apply the following formulas
sinz=n=0∑+∞(2n+1)!(−1)nz2n+1 and sinhz=n=0∑+∞(2n+1)!1z2n+1. Then
siniy=n=0∑+∞(2n+1)!(−1)n(iy)2n+1=n=0∑+∞(2n+1)!(−1)ni2n+1y2n+1
=n=0∑+∞(2n+1)!(−1)ni(−1)ny2n+1=in=0∑+∞(2n+1)!1y2n+1=isinhy, as required.
Comments