cosine function


by sara_87
Tags: cosine, function
sara_87
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#1
Dec22-08, 05:58 AM
P: 774
1. The problem statement, all variables and given/known data

what is cos(2n*pi)

2. Relevant equations



3. The attempt at a solution

I understand that cos(npi)=(-1)^n
so is cos(2n*pi)=2(-1)^n ??
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cristo
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#2
Dec22-08, 06:19 AM
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Make the substitution m=2n in cos(mπ)=(-1)^m
sara_87
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#3
Dec22-08, 06:53 AM
P: 774
so you mean cos(2n^2)=(-1)^2n
??

cristo
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#4
Dec22-08, 06:54 AM
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cosine function


No, I mean cos(2n pi)=(-1)^{2n}
HallsofIvy
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#5
Dec22-08, 07:19 AM
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Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
sara_87
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#6
Dec22-08, 07:23 AM
P: 774
yep cos0=1
yep, cos is periodic with period 2pi
HallsofIvy
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#7
Dec22-08, 08:29 AM
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So cos(2n pi)= cos(0+ n(2pi))= ?
sara_87
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#8
Dec23-08, 09:33 AM
P: 774
oh right,
so =(-1)^(n+1)
is that right?
Mark44
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#9
Dec23-08, 10:22 AM
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Quote Quote by sara_87 View Post
oh right,
so =(-1)^(n+1)
is that right?
No. Look at posts 4 and 7.
HallsofIvy
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#10
Dec23-08, 10:58 AM
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Quote Quote by HallsofIvy View Post
Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
Quote Quote by sara_87 View Post
yep cos0=1
yep, cos is periodic with period 2pi
Quote Quote by HallsofIvy View Post
So cos(2n pi)= cos(0+ n(2pi))= ?
Quote Quote by sara_87 View Post
oh right,
so =(-1)^(n+1)
is that right?
Okay, what does "periodic" mean????
boombaby
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#11
Dec23-08, 11:05 PM
P: 130
...I think it is necessary to know the graph of cos(x), which may help a lot. so, find one.

edit ( 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
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#12
Dec24-08, 05:25 AM
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It might help. It is not necessary. All that is necessary is to know what "periodic" means. No computation is required.
sara_87
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#13
Dec24-08, 02:48 PM
P: 774
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
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#14
Dec24-08, 04:18 PM
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Quote Quote by sara_87 View Post
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
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#15
Dec24-08, 05:11 PM
P: 774
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


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