Using Euler's Formula to prove $e^{i\theta}e^{i\alpha}=e^{i(\theta+ \alpha)}$ I have a homework question thats been puzzling me. It says:

Using Euler's Formula, prove the product property of the complex exponential: 
  $$e^{i\theta}e^{i\alpha}=e^{i(\theta+ \alpha)}$$

Besides knowing Euler's formula, I have no idea where to start so any help is appreciated. :)
 A: Okay, so let's recall Euler's formula:
$$e^{it} = \cos(t) + i \sin(t)$$
Apply this to $e^{i\alpha}e^{i\theta}$: we FOIL out the result and group real and imaginary bits together:
$$\begin{align}
e^{i\alpha}e^{i\theta} &= (\cos(\alpha) + i \sin(\alpha))(\cos(\theta) + i \sin(\theta))\\
&= \underbrace{\cos(\alpha)\cos(\theta) - \sin(\alpha)\sin(\theta)}_{\text{real part}} + i \underbrace{( \sin(\theta) \cos(\alpha) + \sin(\alpha)\sin(\theta))}_{\text{imaginary part}}
\end{align}$$
From here, there are a few identities you'll want to use: the sum/difference formulas  which are on page 2 of this reference sheet I have bookmarked.
They should convert the real part into $\cos(\alpha + \theta)$ and the imaginary part into $\sin(\alpha+\theta)$. Then using Euler's formula "backwards" on the resulting formula should give you the desired result.
A: Fix $y$ and let $f(x) = e^{ix} e^{iy}- e^{i(x+y)}$.
Note that $f(0)=0$ and
$f'(x) = (\sin x - i \cos x) (\cos y + i \sin y) - \sin (x+y) +i \cos (x+y) = 0$.
Hence $f(x) = 0$ for all $x$.
A: $e^{i\theta} = (\cos\theta+i\sin\theta),e^{i\alpha} = (\cos\alpha+i\sin\alpha)$
$e^{i\theta}e^{i\alpha} = (\cos\theta+i\sin\theta)(\cos\alpha+i\sin\alpha)\\
=\cos\theta cos\alpha - \sin\theta\sin\alpha +i(\sin\theta cos\alpha + \cos\theta\sin\alpha) = \cos(\theta+\alpha) +i\sin(\theta+\alpha) = e^{i{(\theta+\alpha)}}$
A: I guess if you really had to use Euler's formula, consider the following 3 trig identities to help simplify your product:
$$\begin{align*}
\sin\alpha\cos\beta &= \frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\\
\cos\alpha\cos\beta &= \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\\
\sin\alpha\sin\beta &= \frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}\\
\end{align*}$$
A: By Euler's Formula,$$e^{i\theta} = \cos\theta + i \sin\theta$$
So,
$$\begin{align} 
e^{i\theta}e^{i\alpha} &= (\cos\theta + i \sin\theta)(\cos\alpha + i \sin\alpha) \\
&= \cos\theta \cos\alpha + i \sin\theta \cos \alpha + i \cos\theta \sin\alpha +(i)^2 \sin\theta \sin\alpha \\
&= \cos\theta \cos\alpha  - \sin\theta \sin\alpha + i (\sin\theta \cos \alpha +  \cos\theta \sin\alpha)\\
&= \cos(\theta+\alpha) +i \sin(\theta+\alpha) \\
&= e^{i(\theta+\alpha)}
\end{align}$$
