Using the definitions of the corresponding functions, prove…

Using the definitions of the corresponding functions, prove: $\ \cos(z_1-z_2)=\cos z_1\cos z_2+\sin z_1\sin z_2$.

Step 1 · Left-hand side directly

$$ \cos(z_1-z_2)=\frac{e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}}2=\frac{e^{iz_1}e^{-iz_2}+e^{-iz_1}e^{iz_2}}2 $$

Step 2 · Expand the right-hand side

$$ \cos z_1\cos z_2+\sin z_1\sin z_2=\frac{(e^{iz_1}+e^{-iz_1})(e^{iz_2}+e^{-iz_2})}4-\frac{(e^{iz_1}-e^{-iz_1})(e^{iz_2}-e^{-iz_2})}4 $$

(using $\dfrac1{(2i)^2}=-\dfrac14$ for the $\sin z_1\sin z_2$ term.)

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 · Subtract: the outer terms cancel

$$ \text{difference}=2e^{i(z_1-z_2)}+2e^{i(z_2-z_1)}=2\left(e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}\right) $$

(using $e^{i(z_2-z_1)}=e^{-i(z_1-z_2)}$.)

Step 5 · Divide by $4$

$$ \cos z_1\cos z_2+\sin z_1\sin z_2=\frac{2\left(e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}\right)}4=\frac{e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}}2=\cos(z_1-z_2) $$