Using the definitions of the corresponding functions, prove…

Using the definitions of the corresponding functions, prove: $\ \sin(x+iy)=\sin x\cosh y+i\cos x\sinh y$.

Step 1 · Apply the definition

$$ \sin(x+iy)=\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}=\frac{e^{-y}e^{ix}-e^{y}e^{-ix}}{2i} $$

Step 2 · Apply Euler's formula and group

$$ =\frac{e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)}{2i}=\frac{\cos x(e^{-y}-e^{y})+i\sin x(e^{-y}+e^{y})}{2i} $$

Step 3 · Split and simplify each term

$$ =\cos x\cdot\frac{e^{-y}-e^{y}}{2i}+\sin x\cdot\frac{e^{-y}+e^{y}}2 $$

Since $\dfrac1i=-i$: $\ \dfrac{e^{-y}-e^y}{2i}=-i\cdot\dfrac{-(e^y-e^{-y})}2=i\cdot\dfrac{e^y-e^{-y}}2=i\sinh y$. So:

$$ \sin(x+iy)=\sin x\cdot\frac{e^y+e^{-y}}2+i\cos x\cdot\frac{e^y-e^{-y}}2=\sin x\cosh y+i\cos x\sinh y $$

Checks at special cases

$y=0$: $\ \sin x\cosh0+i\cos x\sinh0=\sin x$ ✓. $\ x=0$: $\ \sin0\cosh y+i\cos0\sinh y=i\sinh y$, confirming $\sin(iy)=i\sinh y$ ✓.