How Can I Simplify the Expression cos(2n*pi)?

[sara_87]

Homework Statement

what is cos(2n*pi)

Homework Equations

The Attempt at a Solution

I understand that cos(npi)=(-1)^n
so is cos(2n*pi)=2(-1)^n ??

 

[cristo]

Make the substitution m=2n in cos(mπ)=(-1)^m

 

[sara_87]

so you mean cos(2n^2)=(-1)^2n
??

 

[cristo]

No, I mean cos(2n pi)=(-1)^{2n}

 

[HallsofIvy]

Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?

 

[sara_87]

yep cos0=1
yep, cos is periodic with period 2pi

 

[HallsofIvy]

So cos(2n pi)= cos(0+ n(2pi))= ?

 

[sara_87]

oh right,
so =(-1)^(n+1)
is that right?

 

[Mark44]

oh right,
so =(-1)^(n+1)
is that right?

No. Look at posts 4 and 7.

 

[HallsofIvy]

Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?

yep cos0=1
yep, cos is periodic with period 2pi

So cos(2n pi)= cos(0+ n(2pi))= ?

oh right,
so =(-1)^(n+1)
is that right?

Okay, what does "periodic" mean?

 

[boombaby]

...I think it is necessary to know the graph of cos(x), which may help a lot. so, find one.

edit (:shy: trying not to be ambiguous)
...I think it is necessary for one to know the graph of cos(x), which may also help a lot. (regardless of this particular problem)...
"periodic" is really the key

 

[HallsofIvy]

It might help. It is not necessary. All that is necessary is to know what "periodic" means. No computation is required.

 

[sara_87]

I understand that periodic means that cosine function repeats after multiples of 2 pi. but how would that have anything to do with writing cos(2n*pi) ?
cos (npi)=(-1)^n because as long as n is an integer, the value will alternate from -1 and 1 (clearly form the graph)

 

[HallsofIvy]

I understand that periodic means that cosine function repeats after multiples of 2 pi. but how would that have anything to do with writing cos(2n*pi) ?

Would it be easier if it were written n*(2pi) rather than 2n*pi? This is about multiples of 2pi!

cos(2pi)= cos(0+ 2pi)= cos(0)= 1

cos(4pi)= cos(2pi+ 2pi)= cos(2pi)= 1

cos(6pi)= cos(4pi+ 2pi)= cos(4pi)= 1

cos (npi)=(-1)^n because as long as n is an integer, the value will alternate from -1 and 1 (clearly form the graph)

 

[sara_87]

oh right! so cos(n2pi) has to always be 1...i feel very stupid, i should have known that. for all n, cos(2npi) must be 1 as long as n is an integer.

thank you very much