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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).irishetalon00 said: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)?irishetalon00 said: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.Charles Link said:I presume the acos(x) is arccos(x).
Charles Link said: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.Ray Vickson said:No. The correct equation would be ##x = \cos( \cos x))##.
Charles Link said:@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...Ray Vickson said: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.
This equation means that the cosine of a given angle (x) is equal to the inverse cosine of that same angle. In other words, the angle (x) is the same for both the cosine and inverse cosine functions.
To solve for x, you can use a trigonometric identity known as the inverse cosine function (arccos). This function will give you the value of x that satisfies the equation cos(x) = acos(x).
Yes, there can be multiple solutions to this equation. Since the cosine function is periodic, it will repeat itself after every 2π radians (or 360 degrees). This means that there can be infinitely many angles that satisfy cos(x) = acos(x).
You can graph the solutions by plotting the points where the cosine and inverse cosine functions intersect on a coordinate plane. These points will form a straight line at the angle x.
Yes, this equation can be used in various fields such as engineering, physics, and astronomy. For example, it can be used to calculate the phase difference between two oscillating systems or the angle of incidence in optics.