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 :/

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$\exp(\pm it)=\cos\;t\pm i\sin\;t$ – J. M. Nov 28 '10 at 12:46
In view of my answer below, note that $\cos(-x)=\cos(x)$ and $\sin(-x) = - \sin(x)$. – Shai Covo Nov 28 '10 at 13:17

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

$$\begin{eqnarray*} e^{-0.5t}+e^{0.5t} &=&\cos \left( -0.5t\right) +i\sin \left( -0.5t\right) +\cos \left( 0.5t\right) +i\sin \left( 0.5t\right) \\ &=&\left( \cos \left( -0.5t\right) +\cos \left( 0.5t\right) \right) +i\left( \sin \left( -0.5t\right) +\sin \left( 0.5t\right) \right) \\ &=&\left( \cos \left( 0.5t\right) +\cos \left( 0.5t\right) \right) +i\left( -\sin \left( 0.5t\right) +\sin \left( 0.5t\right) \right) \\ &=&2\cos \left( 0.5t\right) +i\cdot 0 \\ &=&2\cos \left( 0.5t\right) \end{eqnarray*}$$

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Thanks for this solution:) – mathNewbie89 Nov 28 '10 at 15:16
You are welcome. – Américo Tavares Nov 28 '10 at 15:29

Use $e^{ia} = \cos a + i\sin a$, and find that the answer is slightly different from what you wrote.

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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) + isin(x) \\ e^{-ix} & = cos(x) - isin(x) \\ \end{align}

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