Conversion (Trigonometry Variable To Standard Variable)

**Simplicity** · January 10th 2009, 10:37 AM

I came across an example when understanding Fourier Series. There was one part where they converted a trigonometry function to standard variable. They did this:

$\frac{2}{n^2\pi^2}(1-\cos n \pi) = \frac{2}{n^2\pi^2}(1-(-1)^n)$ .

What have they done here? Can this be done to sine too? Thanks in advance for the help.

**Chris L T521** · January 10th 2009, 10:50 AM

Originally Posted by Air

I came across an example when understanding Fourier Series. There was one part where they converted a trigonometry function to standard variable. They did this:

$\frac{2}{n^2\pi^2}(1-\cos n \pi) = \frac{2}{n^2\pi^2}(1-(-1)^n)$ .

What have they done here? Can this be done to sine too? Thanks in advance for the help.

If $n\in\mathbb{Z}$ , we can generate a pattern for values of $\cos\left(n\pi\right)$ ...

...
$\cos\left(-3\pi\right)=\cos\left(3\pi\right)=-1$
$\cos\left(-2\pi\right)=\cos\left(2\pi\right)=1$
$\cos\left(-\pi\right)=\cos\left(\pi\right)=-1$
$\cos\left(0\right)=1$

So, when $n$ is even (or zero), we see that $\cos\left(n\pi\right)=1$ . When $n$ is odd, we see that $\cos\left(n\pi\right)=-1$

So, we can rewrite $\cos\left(n\pi\right)$ as $\left(-1\right)^n;~n\in\mathbb{Z}$ .

A similar thing can be done with the sine function:

I'll let you verify that $\forall n\in\mathbb{Z},~\sin\left(\frac{\left(2n-1\right)\pi}{2}\right)=\left(-1\right)^{n+1}$

Does this make sense?

**Simplicity** · January 10th 2009, 10:58 AM

Yes, that makes sense. Thanks.

But, does that mean that this genralisation only works with $\pi$ because it's a periodic function?

**Chris L T521** · January 10th 2009, 11:39 AM

Originally Posted by Air

But, does that mean that this genralisation only works with $\pi$ because it's a periodic function?

From what I understand, Yes.

**Mathstud28** · January 10th 2009, 07:15 PM

Originally Posted by Air

Yes, that makes sense. Thanks.

But, does that mean that this genralisation only works with $\pi$ because it's a periodic function?

Originally Posted by Chris L T521

From what I understand, Yes.

Kind of. It depends how you interpret that question. If you'll notice $\cos(3\pi n)$ gives a similar set of values and in fact $\cos((2z+1)\pi n)~~z\in\mathbb{Z}$ gives the same set of values.

**Moo** · January 10th 2009, 11:22 PM

Hello,

Here is a better way than induction :

$\cos(n\pi)=\frac{e^{in\pi}+e^{-in\pi}}{2}=\frac{\left(e^{i\pi}\right)^n+\left(e^{-i\pi}\right)^n}{2}$

But $e^{i \pi}=-1=e^{-i\pi}$

So we have $\cos(n\pi)=\frac 12 \cdot \left((-1)^n+(-1)^n\right)$
If n is even, then it's equal to 1. If n is odd, then it's equal to -1 (just take n=2k or n=2k+1)

As for the sine thing Chris gave, you can use this identity :
$\cos(x)=\sin \left(\frac \pi 2-x\right)=-\sin \left(x-\frac \pi 2\right)$

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