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Find where cos(x) = acos(x)

  1. Jun 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Find where cos(x) = acos(x)

    2. Relevant equations
    Trig identities?

    3. The attempt at a solution
    I'm not sure how to proceed with this analytically. Cos(cos(x))=x? that doesn't seem to help...
     
  2. jcsd
  3. Jun 18, 2016 #2

    SteamKing

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    Think about the period and phase relationships between y1 = cos (x) and y2 = a cos (x).

    Where are the only values of x located where each function y1 and y2 could possibly intersect?

    Hint: It's OK to make a sketch of these two functions.
     
  4. Jun 18, 2016 #3

    Mark44

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    Does acos(x) mean a * cos(x) or arccos(x)?
     
  5. Jun 18, 2016 #4

    Charles Link

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    I presume the acos(x) is arccos(x). The x would be measured in radians when taking the cosine. Your graphical solution and or tabulate the two as a function of x might work best. I think you will find the solution near ## x=\pi/4 ## for which ## \cos(\pi/4)=.7071... ##. On the other hand if "a" is simply a constant, see @SteamKing post #2. editing...besides a tabulation, Taylor series of the two functions (near the point(s) above) can provide for a more analytical solution.
     
    Last edited: Jun 18, 2016
  6. Jun 18, 2016 #5

    Mark44

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    As do I, based on what the OP wrote about cos(cos(x)) in post 1. Still, a question should be posted so that it isn't ambiguous. acos(x) is notation that is used in programming, and not as much in a mathematics context.
     
  7. Jun 18, 2016 #6

    Charles Link

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    Upon solving this one, (see post #4), it appears the problem could also be stated as find x (angle in radians) such that x=cosx, because x=cos(x)=arccos(x) is a direct consequence of the first equality. Thereby, you just need to graph y=x and y=cos(x) and find the point of intersection.
     
  8. Jun 18, 2016 #7

    Ray Vickson

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    No. The correct equation would be ##x = \cos( \cos x))##.
     
  9. Jun 18, 2016 #8

    Charles Link

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    @Ray Vickson Please look more closely at my post #6. I think I have this one correct. If x=cos(x) then (taking arccos of both sides) arccos(x)=x and thereby cos(x)=arccos(x). I think the converse is true for this case: if cos(x)=arccos(x) I think x=cos(x). .. editing...if you take x=cos(cos(x)) and suppose a solution of cos(x)=x then it reads (after substituting in the inner term) x=cos(x) and you have consistency.
     
    Last edited: Jun 18, 2016
  10. Jun 18, 2016 #9

    Ray Vickson

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    Take the ##\cos## on both sides of ##\cos(x) = \arccos(x)## and you get ##x = \cos(\cos x))##. Apparently, we must have ##\cos x = \cos( \cos x)## at the solution. So, your way is correct, but so is the alternative.
     
  11. Jun 19, 2016 #10

    Charles Link

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    The graphical solution of cos(x)=arccos(x) is interesting. Graphing y=cos(x) and y=arccos(x) and finding where they intersect. The second graph is x=cos(y) so they have a symmetry to them and will meet along y=x, which is the result from the alternative method (post #6). We still need to hear from the OP...
     
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