Find the real and imaginary parts of $\sinh(e^i)$
my attempt:
$\sinh(e^i) = \frac{e^{e^i}-e^{-e^i}}{2}$
$= \frac{e^{\cos(1)+i\sin(1)}-e^{-\cos(1)-i\sin(1)}}{2}$
$= \frac{e^{\cos(1)}e^{i\sin(1)}-e^{-\cos(1)}e^{-i\sin(1)}}{2}$
$= \frac{e^{\cos(1)}(\cos(\sin(1))+i\sin(\sin(1))-e^{-\cos(1)}(\cos(-\sin(1))+i\sin(-\sin(1))}{2}$
Since $\cos(x)$ is even and $\sin(x)$ is odd…
$= \frac{e^{\cos(1)}(\cos(\sin(1))+i\sin(\sin(1))-e^{-\cos(1)}(\cos(\sin(1))-i\sin(\sin(1))}{2}$
$= \frac{\cos(\sin(1))(e^{\cos(1)}-e^{-\cos(1)})}{2} + i\frac{\sin(\sin(1))(e^{\cos(1)}+e^{-\cos(1)})}{2}$
$= \cos(\sin(1))\sinh(\cos(1)) + i\sin(\sin(1))\cosh(\cos(1))$
Is this correct? This is the most reduced I could get the expression…
Best Answer
Yes, this is correct. There's an alternative approach using the addition formula for the hyperbolic sine and the identities $\cosh(ix)=\cos(x),\sinh(ix)=i\sin(x)$: $$\begin{align} \sinh(e^i) &= \sinh(\cos(1)+i\sin(1))\\ &= \sinh(\cos(1))\cosh(i\sin(1))+\cosh(\cos(1))\sinh(i\sin(1))\\ &= \sinh(\cos(1))\cos(\sin(1))+i\cosh(\cos(1))\sin(\sin(1)) \end{align} $$