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Homework Statement
Find where cos(x) = acos(x)
Homework Equations
Trig identities?
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...
Think about the period and phase relationships between y1 = cos (x) and y2 = a cos (x).Homework Statement
Find where cos(x) = acos(x)
Homework Equations
Trig identities?
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...
Does acos(x) mean a * cos(x) or arccos(x)?Homework Statement
Find where cos(x) = acos(x)
irishetalon00 said:Homework Equations
Trig identities?
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...
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.I presume the acos(x) is arccos(x).
No. The correct equation would be ##x = \cos( \cos x))##.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.
@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.No. The correct equation would be ##x = \cos( \cos x))##.
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.@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.
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...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.