Trigonometric identity via complex exponential

Noting that
$$\text{Re}[z_1z_2] = \text{Re}[z_1]\text{Re}[z_2]-\text{Im}[z_1]\text{Im}[z_2],$$
how can
$$\cos(\alpha+\beta) = \text{Re}\left[e^{j(\alpha+\beta)}\right]$$
be expressed? Give the final answer in a simple form, without complex-valued functions.

Source.

and here is my attempt:

We know that
$$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta).$$
So taking the real part of the cosine function gives
$$\text{Re}\left[\cos(\alpha)\right]\text{Re}[\cos(\beta)]-\text{Im}[\sin(\alpha)]\text{Im}[\sin(\beta)].$$

Now we use Euler's formula to convert sin and cosine to complex exponentials
\begin{align*}
\cos x &= \text{Re}\{e^{ix}\} = \frac{e^{ix}+e^{-ix}}{2}\\
\sin x &= \text{Im}\{e^{ix}\} = \frac{e^{ix}-e^{-ix}}{2i}
\end{align*}
$$ = \left[\frac{e^{j\alpha}+e^{-j\alpha}}{2}\right]\left[\frac{e^{j\beta}+e^{-j\beta}}{2}\right]-\left[\frac{e^{j\alpha}-e^{-j\alpha}}{2j}\right]\left[\frac{e^{j\beta}-e^{-j\beta}}{2j}\right]$$
Now once I do this, I will cancel out some terms, but my final answer is not $e^{\alpha+\beta}$ as it should be.

Source.

Please let me know what it is I am doing wrong.
Thanks

Answer

Hint:

You know that:
$$
\cos (\alpha+\beta)=\mbox{Re}\left[e^{i(\alpha+\beta)}\right]=\mbox{Re}\left[e^{i\alpha}e^{i\beta}\right]

$$
Now use the given rule for the real part of a product and you have done.

---

From your rule you have:
$$
\mbox{Re}\left[e^{i\alpha}e^{i\beta}\right]=\mbox{Re}\left[e^{i\alpha}\right]\mbox{Re}\left[e^{i\beta}\right]-\mbox{Im}\left[e^{i\alpha}\right]\mbox{Im}\left[e^{i\beta}\right]
$$
and: $\mbox{Re}\left[e^{i\alpha}\right]=\cos \alpha$, $\mbox{Im}\left[e^{i\alpha}\right]=\sin \alpha$, $\mbox{Re}\left[e^{i\beta}\right]=\cos \beta$, $\mbox{Im}\left[e^{i\beta}\right]=\sin \beta$