1. ## is this true?

$\displaystyle cos(n\Pi)=(-1)^n$

2. Yes. Simply look at the unit circle and you should see this immediately.

3. Of course assuming that $\displaystyle n\in\mathbb{N}$. Otherwise, not so much. If you don't make that restriction you need to say that $\displaystyle \cos(n\pi)=\mathfrak{R}\left((-1)^n\right)$ which is really just a highfalutin way to say $\displaystyle \cos(n\pi)$. This is of course because $\displaystyle (-1)^n=e^{\pi\cdot i\cdot n}=\cos(n\pi)+ i\sin(n\pi)$.

4. I take it $\displaystyle sin(n\pi)=0$ then?

5. Originally Posted by billym
I take it $\displaystyle sin(n\pi)=0$ then?
You don't know that?

6. But when you touch me like this
And you hold me like that
I take it $\displaystyle sin(n\pi)=0$ then?
Whether or not that is true, what I said is still valid. To see this merely note that $\displaystyle \cos(x),\sin(x)$ are both mappings from $\displaystyle \mathbb{R}\mapsto\mathbb{R}$. So $\displaystyle \cos(n\pi),\sin(n\pi)$ is real for any $\displaystyle n\in\mathbb{R}$. Also, it should be apparent that if both $\displaystyle a,b\in\mathbb{R}$ that $\displaystyle \mathfrak{R}\left[a+bi\right]=a$. Consequently, it follows that $\displaystyle \cos(n\pi)=\mathfrak{R}\left[\left(-1\right)^n\right]=\mathfrak{R}\left[e^{n\pi i}\right]=\mathfrak{R}\left[\cos(n\pi)+i\sin(n\pi)\right]=\cos(n\pi)$