Using the definitions of the corresponding functions, prove…
Using the definitions of the corresponding functions, prove: $\ \sin(z_1+z_2)=\sin z_1\cos z_2+\cos z_1\sin z_2$.
Explanation of Method
Expand the right-hand side using the exponential definitions of $\sin$ and $\cos$, and show the result collapses to $\sin(z_1+z_2)$.
Step 1 · Left-hand side directly
$$ \sin(z_1+z_2)=\frac{e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}}{2i}=\frac{e^{iz_1}e^{iz_2}-e^{-iz_1}e^{-iz_2}}{2i} $$
Step 2 · Expand the right-hand side
$$ \sin z_1\cos z_2+\cos z_1\sin z_2=\frac{(e^{iz_1}-e^{-iz_1})(e^{iz_2}+e^{-iz_2})+(e^{iz_1}+e^{-iz_1})(e^{iz_2}-e^{-iz_2})}{4i} $$
Step 3 · Expand both products
$$ (e^{iz_1}-e^{-iz_1})(e^{iz_2}+e^{-iz_2})=e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{i(z_2-z_1)}-e^{-i(z_1+z_2)} $$
$$ (e^{iz_1}+e^{-iz_1})(e^{iz_2}-e^{-iz_2})=e^{i(z_1+z_2)}-e^{i(z_1-z_2)}+e^{i(z_2-z_1)}-e^{-i(z_1+z_2)} $$
Step 4 · Add: the middle terms cancel
$$ \text{sum}=2e^{i(z_1+z_2)}-2e^{-i(z_1+z_2)}=2\left(e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}\right) $$
Step 5 · Divide by $4i$
$$ \sin z_1\cos z_2+\cos z_1\sin z_2=\frac{2\left(e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}\right)}{4i}=\frac{e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}}{2i}=\sin(z_1+z_2) $$
Result
Both sides reduce to the same exponential expression $\dfrac{e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}}{2i}$, so the identity is an algebraic consequence of the exponential definitions rather than a separate trigonometric fact.