- #1

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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...

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- Thread starter irishetalon00
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- #1

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Find where cos(x) = acos(x)

Trig identities?

I'm not sure how to proceed with this analytically. Cos(cos(x))=x? that doesn't seem to help...

- #2

SteamKing

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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...

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.

- #3

Mark44

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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...

- #4

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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.

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- #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.I presume the acos(x) is arccos(x).

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- #7

Ray Vickson

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No. The correct equation would be ##x = \cos( \cos x))##.

- #8

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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.No. The correct equation would be ##x = \cos( \cos x))##.

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- #9

Ray Vickson

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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.

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.

- #10

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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...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.

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