The function $(-1)^{-x}$ 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.
 A: 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
$$
\Re((-1)^{-x})=\cos(x\pi) \text{ and } \Im((-1)^{-x})=-\sin(\pi x)
$$
A: This has been asked several times already, I am just too bored to look for duplicates.
In general, for $x,y \in \Bbb C$, one defines $x^y$ as $\Bbb e ^{y \ln x}$. Of course, one has to tell what the natural logarithm means for complex numbers. The usual procedure is to eliminate a half-line from the complex plane, in order to be able to unambigously define a "principal argument". Fine, but what half-line to choose? Most people choose to eliminate the subset $\{ z \in \Bbb C \mid \Re z \le 0 \}$, which means that with this convention one may not speak about $\ln y$ for real $y \le 0$. In particular, $(-1)^{-x}$ would no longer make sense under this convention.
Of course, other half-lines may be eliminated, in which case one might be able to define $(-1)^{-x}$ (which would be a complex number even for $x \in \Bbb R$!), but this approach is less common. Assuming that you choose to go this way, you could remove the half-line $\{ z \in \Bbb C \mid \Re z \ge 0 \}$ (this being just one possibility among many others), define the principal argument to be $\arg z$ to be the angle formed by the segment $0z$ with the $x$-axis (measured from the $x$-axis counter-clockwisely), and then define $\ln z = \ln |z| + \Bbb i \arg z$. It becomes clear that, with this convention, $\ln (-1) = \ln 1 + \Bbb i \pi = \Bbb i \pi$, so that
$$(-1)^{-x} = \Bbb e^{-x \ln (-1)} = \Bbb e^{\pi \Bbb i (-x)} = \cos (-\pi x) + \Bbb i \sin (-\pi x) = \cos (\pi x) - \Bbb i \sin (\pi x)$$
so $\Re (-1)^{-x} = \cos (\pi x), \quad \Im (-1)^x = -\sin (\pi x)$, which explains the shape of the curves in your plot.
(I have used Euler's formula $\Bbb e ^{\Bbb i x} = \cos x + \Bbb i \sin x$ and the fact that $\cos (-x) = \cos x$ and $\sin (-x) = \sin x$.)
A: I assume that you are wondering why this pattern emerges,
It follows from Euler's identity that $ (-1)^x = \cos(\pi x) + i\sin(\pi x) = \text{cis}(\pi x) $, which explains the sine-like behavior.
Thus, $ (-1)^{-x} = \text{cis}(-\pi x) = e^{-i x} = \cos(\pi x) - i \sin(\pi x) $.
It is worth noting, that it really has to do with arbitrary, although convinient, conventions.
A: $(-1)^x=e^{x\cdot\ln(-1)}$. There is no unique value of $\ln(-1)$, however if we chose the principal value of the logarithm, $\ln(-1)=i\pi$, yielding $(-1)^x=e^{ix\pi}=\cos(\pi x)+i\sin(\pi x)$. Weird results such as $1^x=e^{2i\pi x}$ are obtained by taking $\ln(1)=2i\pi$, which lies on another branch.
It is a matter of convention which branch we chose.
