Where Does cos(x) Equal acos(x)?

[irishetalon00]

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...

 

[SteamKing]

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).

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.

 

[Mark44]

Homework Statement

Find where cos(x) = acos(x)

Does acos(x) mean a * cos(x) or arccos(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...

 

[Charles Link]

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.

 

[Mark44]

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

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.

 

[Charles Link]

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]

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.

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

 

[Charles Link]

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

@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.

 

[Ray Vickson]

@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.

 

[Charles Link]

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.

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...