Trigonometry, complex numbers

Question

Does $\cos(a+b)=\cos(a)*\cos(b)-\sin(a)*\sin(b)$ apply even for complex values. If so why?

Answer 1

For a complex number $z$, sine and cosine defined by $$\cos z=\frac{e^{iz}+e^{-iz}}{2}\quad\text{and} \quad \sin z=\frac{e^{iz}-e^{-iz}}{2i}$$ therefore you can compute $$\cos(a+b)=\frac{e^{i(a+b)}+e^{-i(a+b)}}{2}$$ and proceed.

---

Edit: From Euler formula $e^{ix}=\cos x+i\sin x$, we have $$\cos x=\frac{e^{ix}+e^{-ix}}{2}\quad\text{and} \quad \sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ for real $x$.

Therefore it leads us to define sine and cosine of a complex variable in that form. Also the Taylor series $$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$$ makes this definition firm.

Answer 2

Let us consider a number $a+ib$.

Then

$\cos(a+ib) = \cos a \cos(ib) - \sin a \sin(ib) = \color{blue}{\cos a \cosh b - i \sin a \sinh b}$

as $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$