Simplify formula $e^{-i0.5t}+e^{i0.5t}$ I want to ask, how can I simplify this formula ?
$ e^{-i0.5t}+e^{i0.5t} $
I know that it can be simplify to $\cos(0.5t)$, but I don't know how :/
 A: Use $e^{ia}  = \cos a + i\sin a$, and find that the answer is slightly different from what you wrote.
A: By the Euler's formula
$$e^{ix}=\cos x+i\sin x$$
for $x=-0.5t$ , we obtain
$$e^{-0.5t}=e^{i\left( -0.5t\right) }=\cos \left( -0.5t\right) +i\sin \left(
-0.5t\right) $$
and for $x=0.5t$,
$$e^{0.5t}=\cos \left( 0.5t\right) +i\sin \left( 0.5t\right) .$$
Since $\cos$ is an even function and $\sin$ an odd function, we obtain
$$\begin{eqnarray*}
e^{-0.5t}+e^{0.5t} &=&\cos \left( -0.5t\right) +i\sin \left( -0.5t\right)
+\cos \left( 0.5t\right) +i\sin \left( 0.5t\right)  \\
&=&\left( \cos \left( -0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(
\sin \left( -0.5t\right) +\sin \left( 0.5t\right) \right)  \\
&=&\left( \cos \left( 0.5t\right) +\cos \left( 0.5t\right) \right) +i\left(
-\sin \left( 0.5t\right) +\sin \left( 0.5t\right) \right)  \\
&=&2\cos \left( 0.5t\right) +i\cdot 0 \\
&=&2\cos \left( 0.5t\right) 
\end{eqnarray*}$$
A: The more general answer to that is:
\begin{align}
\forall x \in \mathbb{R},\  \cos(x) & = \frac{e^{ix} + e^{-ix}}{2} \\\\
\sin(x) & = \frac{e^{ix} - e^{-ix}}{2i}
\end{align}
These equations can in turn be obtained by summing and subtracting the following two equations:
\begin{align}
\forall x \in \mathbb{R},\  e^{ix} & = \cos(x) + i\sin(x) \\\\
e^{-ix} & = \cos(x) - i\sin(x) \\\\
\end{align}
