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Q. $ sin\, (ix) = $

J & K CETJ & K CET 2017Trigonometric Functions

Solution:

$sin \left(ix\right)=\frac{e^{i\left(ix\right)}-e^{-i\left(ix\right)}}{2i}$
$\left[sin \,x=\frac{e^{ix}-e^{-ix}}{2i}\right]$
$=\frac{e^{-x}-e^{x}}{2i}
=-\frac{i}{2}\left(e^{-x}-e^{x}\right)$
$=i\left(\frac{e^{x}-e^{-x}}{2}\right)$
$=i\,sin \,hx$