# cosine function

by sara_87
Tags: cosine, function
 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 ??
 Mentor P: 8,262 Make the substitution m=2n in cos(mπ)=(-1)^m
 P: 774 so you mean cos(2n^2)=(-1)^2n ??
Mentor
P: 8,262

## cosine function

No, I mean cos(2n pi)=(-1)^{2n}
 PF Patron Sci Advisor Thanks Emeritus P: 38,430 Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
 P: 774 yep cos0=1 yep, cos is periodic with period 2pi
 PF Patron Sci Advisor Thanks Emeritus P: 38,430 So cos(2n pi)= cos(0+ n(2pi))= ?
 P: 774 oh right, so =(-1)^(n+1) is that right?
Mentor
P: 19,798
 Quote by sara_87 oh right, so =(-1)^(n+1) is that right?
No. Look at posts 4 and 7.
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Thanks
Emeritus
P: 38,430
 Quote by HallsofIvy Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
 Quote by sara_87 yep cos0=1 yep, cos is periodic with period 2pi
 Quote by HallsofIvy So cos(2n pi)= cos(0+ n(2pi))= ?
 Quote by sara_87 oh right, so =(-1)^(n+1) is that right?
Okay, what does "periodic" mean????
 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
 PF Patron Sci Advisor Thanks Emeritus P: 38,430 It might help. It is not necessary. All that is necessary is to know what "periodic" means. No computation is required.
 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)
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Emeritus
P: 38,430
 Quote by 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) ?
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)
 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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