Inverse formula of a difference of cosines

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Profet

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Hi, my problem is the following:

I need to find the inverse formula of this equation to get the "t" inside of the cosine.
So, to be as clear as possible, in the end i should have the equation: "t = ...".

[MATH]x = cos(t) - cos(15t)[/MATH]
Thanks to anyone will try to help me.
 
Profet said:
Hi, my problem is the following:

I need to find the inverse formula of this equation to get the "t" inside of the cosine.
So, to be as clear as possible, in the end i should have the equation: "t = ...".

[MATH]x = cos(t) - cos(15t)[/MATH]
Thanks to anyone will try to help me.
Click to expand...
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment.

I have not worked it out - but if I were to do this problem, I would start with:

(15 + 1)/2 = 8 and (15-1)/2 = 7

then:

cos(t) = cos(8t - 7t)

cos(15t) = cos(8t + 7t)

and continue......
 
Thanks for your reply

The equation I've written above is already the maximum simplification of what I've done.
I don't know how to continue at this point..

I thougth to use the arccosine function to get the inverse formula of the cosine, but I can't use it because I have a difference of cosines.

I understand what you mean, but if I did as you suggest, then I wouldn't know what to do next.
 
Have you tried graphing y = cos(x) - cos(15x)? It is about as far from one-to-one as you can get. See https://www.wolframalpha.com/input/?i=plot+y+=+cos(x)-cos(15x)

Which of the many solutions to cos(t) - cos(15t) = x would you want your "inverse" to provide for a given value of x?

It's entirely possible that the inverse function can't be expressed in closed form, even if you could make that choice. You're right that changing it to 2sin(8t)sin(7t) = x is no easier to work with.

Can you put the question into context, so we could see if something other than the nonexistent full inverse would be of use?
 
The initial equation was
[MATH] x(t) = 2π \cdot [\cos(t) - \cos(15t[/MATH][MATH])] \cdot π^2 \cdot \dfrac{\sqrt{2}}{2} [/MATH]
Then i simplified it in

[MATH] \dfrac{x}{π^3 \cdot \sqrt{2}} = \cos(t) - \cos(15t) [/MATH]
Now, I've temporarily replaced the entire first term with the "x" trying to semplify even more the equation and focus my attention to the second term (the difference of cosines).

The variable "x" can assume values from 0 to infinite as it represents the time.
Instead the variable "t" can assume values from 0 to 360 as it is an angle.

I don't know if this is possible to do...

I hope that what I said is understandable.
 
Just for the record, if t is an angle it does NOT mean that 0<t<360. t can go between -infinity and + infinity. Did the problem state that 0<t<360??? Please post the entire problem.
 
Profet said:
The initial equation was
[MATH] x(t) = 2π \cdot [\cos(t) - \cos(15t[/MATH][MATH])] \cdot π^2 \cdot \dfrac{\sqrt{2}}{2} [/MATH]
Then i simplified it in

[MATH] \dfrac{x}{π^3 \cdot \sqrt{2}} = \cos(t) - \cos(15t) [/MATH]
Now, I've temporarily replaced the entire first term with the "x" trying to semplify even more the equation and focus my attention to the second term (the difference of cosines).

The variable "x" can assume values from 0 to infinite as it represents the time.
Instead the variable "t" can assume values from 0 to 360 as it is an angle.

I don't know if this is possible to do...

I hope that what I said is understandable.
Click to expand...
Ignoring the constants, temporarily is fine.
 
Profet said:
The variable "x" can assume values from 0 to infinite as it represents the time.
Instead the variable "t" can assume values from 0 to 360 as it is an angle.

I don't know if this is possible to do...

I hope that what I said is understandable.
Click to expand...

Well, no, it isn't possible to do. Did you look at the graph I showed?

It is not true that x can take any positive value; and for a specific value of x, there may be many values of t, even with your restriction.

I suspect that if you showed the actual problem you are working on, we would find that you have made a mistake before writing this equation.
 
The initial function [MATH] x(t) = 2π \cdot[\cos(t) - \cos(15t)] \cdot π^2 \cdot \frac{\sqrt2}{2} [/MATH] isn't mine. I've taken it from a university professor and I can confirm that is correct.

Maybe I've expressed myself badly: It just calculates the x coordinate on the chart, nothing more, nothing less.

If we look at the initial and complete function, for any value of "t" there's always a value for "x" as it depends on "t".

So what I wanted from the beginning was trying to get the function that calculates "t" for any value of "x" ([MATH]t(x) = ...[/MATH]).

That was my starting point and this is the actual and the entire problem, so there's nothing else that I've omitted.

Hope that this will clear any possible misunderstandings.
Thanks
 
Profet said:
The initial function [MATH] x(t) = 2π \cdot[\cos(t) - \cos(15t)] \cdot π^2 \cdot \frac{\sqrt2}{2} [/MATH] isn't mine. I've taken it from a university professor and I can confirm that is correct.

Maybe I've expressed myself badly: It just calculates the x coordinate on the chart, nothing more, nothing less.

If we look at the initial and complete function, for any value of "t" there's always a value for "x" as it depends on "t".

So what I wanted from the beginning was trying to get the function that calculates "t" for any value of "x" ([MATH]t(x) = ...[/MATH]).

That was my starting point and this is the actual and the entire problem, so there's nothing else that I've omitted.

Hope that this will clear any possible misunderstandings.
Thanks
Click to expand...

You seem to be missing the point.

First, no, this is not the entire problem; it clearly comes from some larger problem, or you could not confirm that it is the "correct" function. (What conditions must the function satisfy? What is the "chart" for which it gives the x-coordinate as a function of the angle t?)

But here's the point: for any "x", there are typically many values of "t". Here is the graph I asked you to look at; the horizontal axis is t (in radians here, rather than degrees), and the vertical axis is x:
FMH120520.png

For x = 50, I see 12 possible values of t.

It may be the correct equation for whatever it is that you are doing, but asking for "the" value of t corresponding to a given x is the wrong question. What we need to figure out is why you are asking it, and what you should be asking instead.
 
Such I've said before the function calculates only the x coordinate on the chart, so for a given value of t, I have an x on the horizontal axis.

There is the same function that calcutes only the y coordinate on the chart (it has the sine instead of cosine in the formula), so for a given value of t, I have an y on the vertical axis.

The resulting x and y coordinates, depending on t, create the complete graph.

This is the part that I've not said before because it's not the problem at which I was focusing myself and honestly a I didn't think that this could help me to find what I was looking for. Hope i was wrong and this can help you to better understand.
 
So what does "the chart" look like? I take it to be a parametric graph where x and y are functions of t, which you call an angle (but this wouldn't be an angle on the chart itself). From your description (replacing cosine with sine for y), I get this:

FMH120520 par.png

Is that what you are talking about? This context definitely does help me see something of where you are coming from, though the graph I showed before (x as a function of t) should have been enough to answer your question.

It should be clear from this graph that for any value of x alone, there will be many values of t, just as I said before; but for any pair (x,y) there might be 0, 1, or 2 values of t (between 0 and 2pi) -- the curve doesn't pass through every point, but passes through some points twice.

Now, we still have the question of why you want to find "the" value of t for a given value of x. Do you understand yet that that can't be done?
 
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