# Thread: Conversion (Trigonometry Variable To Standard Variable)

1. ## Conversion (Trigonometry Variable To Standard Variable)

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:

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

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

2. 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:

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

What have they done here? Can this be done to sine too? Thanks in advance for the help.
If $\displaystyle n\in\mathbb{Z}$, we can generate a pattern for values of $\displaystyle \cos\left(n\pi\right)$...

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

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

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

A similar thing can be done with the sine function:

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

Does this make sense?

3. Yes, that makes sense. Thanks.

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

4. Originally Posted by Air
But, does that mean that this genralisation only works with $\displaystyle \pi$ because it's a periodic function?
From what I understand, Yes.

5. Originally Posted by Air
Yes, that makes sense. Thanks.

But, does that mean that this genralisation only works with $\displaystyle \pi$ 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 $\displaystyle \cos(3\pi n)$ gives a similar set of values and in fact $\displaystyle \cos((2z+1)\pi n)~~z\in\mathbb{Z}$ gives the same set of values.

6. Hello,

Here is a better way than induction :

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

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

So we have $\displaystyle \cos(n\pi)=\frac 12 \cdot \left((-1)^n+(-1)^n\right)$
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 :
$\displaystyle \cos(x)=\sin \left(\frac \pi 2-x\right)=-\sin \left(x-\frac \pi 2\right)$