# Math Help - how do you represent cos in power form?

1. ## how do you represent cos in power form?

for example, $cos(m\pi)$ = $(-1)^m$ because cos alternates between 1 and -1 every $m\pi$.

so how can you represent $cos(\frac{m\pi}{2})$ in power form?
this one is a bit confusing to me because when:

m=1, $cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})$= 0
m=2, $cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})$= -1
m=3, $cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})$= 0
m=4, $cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})$= 1

2. Originally Posted by mathfied
for example, $cos(m\pi)$ = $(-1)^m$ because cos alternates between 1 and -1 every $m\pi$.

so how can you represent $cos(\frac{m\pi}{2})$ in power form?
this one is a bit confusing to me because when:

m=1, $cos(\frac{m\pi}{2})=cos(\frac{1\pi}{2})$= 0
m=2, $cos(\frac{m\pi}{2})=cos(\frac{2\pi}{2})$= -1
m=3, $cos(\frac{m\pi}{2})=cos(\frac{3\pi}{2})$= 0
m=4, $cos(\frac{m\pi}{2})=cos(\frac{4\pi}{2})$= 1
First off you have to state that $m\in\mathbb{N}$

But how about...hmm...Let me think here...you need...ok

Well after some thought I have come up with

$\cos\left(\frac{n\pi}{2}\right)=\frac{1}{2}(-1)^{\frac{n(n+1)}{2}}\bigg[(-1)^n+1\bigg]\quad{n\in\mathbb{N}}$

3. I will include my thought process as well..

$\cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1$

$\cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2$

$\cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}$

Now we see that when an even m is a multiple of 4, its a + else it is a -

We can capture this idea by $(-1)^{\frac{m}2}$

When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose $(-1)^{\lfloor\frac{m}2\rfloor}$

Where I have used the floor function on $\frac{m}2$.

Thus $\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}$

Note that functionally, $\sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2$

Thus a simplified form is:

$\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)$

4. Originally Posted by Isomorphism
I will include my thought process as well..

$\cos m\pi = (-1)^m = 2\cos^2\left(\frac{m\pi}2\right) - 1$

$\cos^2\left(\frac{m\pi}2\right) = \frac{(-1)^m + 1}2$

$\cos\left(\frac{m\pi}2\right) = \pm\sqrt{\frac{(-1)^m + 1}2}$

Now we see that when an even m is a multiple of 4, its a + else it is a -

We can capture this idea by $(-1)^{\frac{m}2}$

When m is odd, anyway the expression is 0. The sign wont matter then. Thats why I choose $(-1)^{\lfloor\frac{m}2\rfloor}$

Where I have used the floor function on $\frac{m}2$.

Thus $\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\sqrt{\frac{(-1)^m + 1}2}$

Note that functionally, $\sqrt{\frac{(-1)^m + 1}2} = \frac{(-1)^m + 1}2$

Thus a simplified form is:

$\cos\left(\frac{m\pi}2\right) = (-1)^{\lfloor\frac{m}2\rfloor}\left(\frac{(-1)^m + 1}2\right)$
Nice!

I should probably include my thought process as well.

I did not work off of the trig, rather I worked off of the sequence $0,-1,0,1\cdots$

I firstly saw that

$\frac{1}{2}\left((-1)^n+1\right)$ gave $0,1,0,1$

So I wanted then that for even value of n that ever other was positive.

So I saw that this was described by $(-1)^{\frac{n(n+1)}{2}}$

A little less eloquent than Iso

5. Another nice formula is: $\cos\left(\frac{n\,\pi}2\right)=\frac12\big(i^n+(-i)^n\big)$ where $i=\sqrt{-1}, n \in \mathbb N$.

Also, the generating function is $\frac1{1+x^2}$.

Edit: Also $\cos\left(\frac{n\,\pi}2\right)=\Re\{i^n\}$