
#1
Dec2208, 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 ?? 



#3
Dec2208, 06:53 AM

P: 774

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



#5
Dec2208, 07:19 AM

Math
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PF Gold
P: 38,879

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




#6
Dec2208, 07:23 AM

P: 774

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



#7
Dec2208, 08:29 AM

Math
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PF Gold
P: 38,879

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




#8
Dec2308, 09:33 AM

P: 774

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



#9
Dec2308, 10:22 AM

Mentor
P: 20,962





#10
Dec2308, 10:58 AM

Math
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Thanks
PF Gold
P: 38,879





#11
Dec2308, 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 



#12
Dec2408, 05:25 AM

Math
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PF Gold
P: 38,879

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




#13
Dec2408, 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) 



#14
Dec2408, 04:18 PM

Math
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PF Gold
P: 38,879

cos(2pi)= cos(0+ 2pi)= cos(0)= 1 cos(4pi)= cos(2pi+ 2pi)= cos(2pi)= 1 cos(6pi)= cos(4pi+ 2pi)= cos(4pi)= 1 



#15
Dec2408, 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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