Maxima of Cosine Functions with Integer Values of Y

  • Thread starter d6syxx
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In summary, the maxima of the functions cos(x-y) and -cos(x) are located at x values that can be represented as (n/2), (1+n/2), (2n+1/2), n, (2n+3/2), 2n, (2n+1), and (n+1/2), where n is an integer. This is based on the relationship between cos(A-B) and cos(A) given in the problem, and the fact that the x values for when cos(x)=-1 are either 0 or π on a unit circle.
  • #1
d6syxx
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0
For what values of y are the maxima of the functions cos(x-y) and -cos(x) located at the same x values?
(n/2), where n is an integer
(1+n/2), where n is an integer
(2n+1/2), where n is an integer
n , where n is an integer
(2n+3/2), where n is an integer
2n, where n is an integer
(2n+1), where n is an integer
(n+1/2), where n is an integer

I'm no physicist, but in order to graduate I need to take physics. I skipped Trigonometry and went straight to calculus, learning whatever Trig I needed to pass the class. I feel I made a horrible mistake because I can't do this problem or any problems like it. If anyone is willing to help me I would be much appreciative and even offer a cash bonus to anyone that is willing to do so. I have several problems similar to the one above. Thanks.
 
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  • #2
Note that;

[tex]\cos(A-B) = \cos(A)\cos(B)+\sin(A)\sin(B)[/tex]

P.S. This is probably better placed in PreCalc math
 
  • #3
I know that, but how does that relate to this problem?
 
  • #4
Well, where are the maximas of -cos(x) located?
 
  • #5
I have no idea! 1?
 
  • #6
Okay, what are the x values for when -cos(x)=1 (I'll give you a clue, there's only one)
 
  • #7
2pie or zero?

Are you saying, where does the -cos of x = 1? Like on a unit circle?
 
  • #8
d6syxx said:
2pie or zero?
Close, that would be the cases where cos(x)=1
d6syxx said:
Are you saying, where does the -cos of x = 1? Like on a unit circle?
Yes, or in other words, where does cos(x)=-1
 
  • #9
In that case, it would be Pie
 

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