# Thread: is this true?

1. ## is this true?

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

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

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

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

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

6. But when you touch me like this
And you hold me like that
I just have to admit
That it's all coming back to me
When I touch you like this
And I hold you like that
It's so hard to believe but
It's all coming back to me
(It's all coming back, it's all coming back to me now)

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