Where is the mistake in my proof? I tried to prove
$$ \sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$$
but instead I proved $ \sin(x - iy) = \sin x \cosh y + i \cos x \sinh y$.

Please could someone check my proof and tell me where my mistake is?

$$ 
\begin{align}
\sin x \cosh y + i\cos x \sinh y &= \sin x {e^{-y}+e^y\over 2} + \cos x {e^{y}-e^{-y}\over 2}\\
&= \sin x {e^{-y{i\over i}}+e^{y{i\over i}}\over 2} + \cos x {e^{y{i\over i}}-e^{-y{i\over i}}\over 2}\\\\
&=\sin x \cos {y\over i} + i \cos x \sin {y\over i}\\
&=\sin x \cos {-iy} + i \cos x \sin {-iy}\\
&=\sin(x-iy)
\end{align}
$$
where the last equality follows from $\sin(x-y) = \sin x \cos y - \sin y \cos x$.

$
\sin(x + iy) = \sin(x) \cos(iy) + \cos(x) \sin(iy) \\
             = \sin(x) \cosh(y) + i \cos(x) \sinh(y) [\because sinh(a) = -i sin(ia), cosh(a) = cos(ia)] \\
$
Also
$
e^{ix} = \cos(x) + i \sin(x) \ [\text{ Euler identity}] \\
e^{i(ix)} = \cos(ix) + i \sin(ix) \\
e^{i(-ix)} = \cos(ix) - i \sin(ix) \\ 
e^{i(ix)} + e^{-i(ix)} = 2\cos(ix) \\
e^{-x} + e^{x} = 2\cos(ix) \ [\because i^2 = -1] \\
\frac{e^{-x} + e^{x}}{2} = \cos(ix) \\
cosh(x) = \cos(ix)
$
I will leave the proof of 
$\sinh(x) = -i \sin(ia)$ for you to work
