Euler's Formula explanation in $e^{ix} ⋅e^{iy}$

Question

According to Euler's Formula, $e^{ix} = \cos(x) + i\sin(x).$ I'm computing the product $e^{ix} \cdot e^{iy}.$

What is the real part (that is, the term without a factor of $i$)?

Why is it $\cos(x)\cos(y) - \sin(x)\sin(y)??$

Answer 1

Use Euler's formula $$e^{ix}e^{iy} = (\cos(x) + i \sin(x))(\cos(y) + i \sin(y))$$ then expand: $$e^{ix}e^{iy} = \underbrace{(\cos(x)\cos(y) - \sin(x)\sin(y))}_{\text{real part}} + i\underbrace{(\cos(x)\sin(y) + \sin(x)\cos(y))}_{\text{imaginary part}}$$ So as @kingW3 comments, the real part is not $\sin(x)\cos(y) - \cos(x)\sin(y)??$, instead it should be $\cos(x)\cos(y) - \sin(x)\sin(y)$.

Answer 2

It is not. You get the formula wrong. The real part is $\cos x\cos y-\sin x \sin y$. This comes from $$\begin{align}e^{ix}e^{iy}&=(\cos x+i\sin x)(\cos y+i\sin y)\\&=\cos x\cos y+i\cos x \sin y+i\sin x\cos y+i^2\sin x\sin y\end{align}$$ Now using $i^2=-1$, you get $$e^{ix}e^{iy}=(\cos x\cos y-\sin x\sin y)+i(\sin x\cos y+\cos x\sin y)$$