TITLE: The function $(-1)^{-x}$
QUESTION [7 upvotes]: I was bored so I put functions in Wolfram Alpha. And I got something that looks like a sin function. And in addition to that, the real part was continuous and the imaginary part was a cos function.
It might be obvious to most of you but the only math education I got is from high school.

Thanks.

REPLY [3 votes]: First you need to assume that you're working within the complex numbers, meaning that we assume $-1\in \mathbb C$ - otherwise your expression is simply not defined - and further we assume that $x\in \mathbb{R}$.
Now we perform some basic manipulations on your expression and get
$$
(-1)^{-x}=\exp\{\log((-1)^{-x})\}=\exp\{-x\log(-1)\}
$$
The expression $\log(-1)$ is not defined in $\mathbb R$ but is very much in $\mathbb C$, we are working with the complex logarithm and are using the principal value, we get 
$$
\log(-1)=\log(1)+i\mathrm{arg}(1)=i\pi
$$
Now we put everything back together and we get using the Euler formula for $x\in \mathbb R$
\begin{align}
(-1)^{-x}=\exp\{\log((-1)^{-x})\}=&\exp\{-x\log(-1)\}\\
=&\exp\{-xi\pi\}\\
=&\cos(x\pi)-i\sin(\pi x)
\end{align}
Now you can plot your function, decomposed into an imaginary and real part, and you'll get the desired $\sin(\cdot),\cos(\cdot)$ functions, which are scaled accordingly. That means
$$
\Re((-1)^{-x})=\cos(x\pi) \text{ and } \Im((-1)^{-x})=-\sin(\pi x)
$$