Let $X \in \text{Random} \{ -1, 1 \}$ be a D5075: Random integer such that

(i)	\begin{equation} \mathbb{P}(X = -1) = \mathbb{P}(X = 1) = \frac{1}{2} \end{equation}

Let $t \in \mathbb{R}$ be a D993: Real number.

Since $X$ takes values in $\{ - 1, 1 \}$, then $e^{i t X}$ takes values in $\{ e^{i t}, e^{- i t} \}$. Applying results

(i)	R1814: Expectation of discrete random euclidean real number
(ii)	R1608: Cosine function is convex combination of complex exponential functions

then gives \begin{equation} \begin{split} \mathbb{E}(e^{i t X}) & = e^{i t \cdot 1} \mathbb{P}(X = 1) + e^{i t \cdot (-1)} \mathbb{P}(X = -1) \\ & = \frac{1}{2} e^{i t} + \frac{1}{2} e^{- i t} \\ & = \frac{1}{2} (e^{i t} + e^{- i t}) \\ & = \cos(t) \end{split} \end{equation} $\square$