is this true?

Defunkt · #2 $Old$ November 6th, 2009, 11:50 AM

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

Drexel28 · #3 $Old$ November 6th, 2009, 12:39 PM

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 $(-1)^n=e^{\pi\cdot i\cdot n}=\cos(n\pi)+ i\sin(n\pi)$ .

billym · #4 $Old$ November 6th, 2009, 12:46 PM

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

Plato · #5 $Old$ November 6th, 2009, 01:26 PM

Quote:

Originally Posted by billym $View Post$

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

You don't know that?

billym · #6 $Old$ November 6th, 2009, 01:44 PM

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)

Drexel28 · #7 $Old$ November 6th, 2009, 02:05 PM

Quote:

Originally Posted by billym $View Post$

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 $\cos(n\pi)=\mathfrak{R}\left[\left(-1\right)^n\right]=\mathfrak{R}\left[e^{n\pi i}\right]=\mathfrak{R}\left[\cos(n\pi)+i\sin(n\pi)\right]=\cos(n\pi)$