Proving cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

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Homework Help Overview

The discussion revolves around proving the equation cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2), which involves concepts from trigonometry and complex numbers.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the definition of cis and its relation to sine and cosine. Some express skepticism about the validity of the original statement, suggesting that it might lead to contradictions. Others propose using trigonometric identities and the properties of sine and cosine to explore the problem further.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being discussed. Some participants have offered insights into the relationships between sine and cosine, while others have pointed out potential flaws in the original equation. The conversation remains open-ended without a clear consensus.

Contextual Notes

Participants note that certain assumptions about the values of x1 and x2 may affect the validity of the statements being discussed. There is also mention of the need for clarity in the definitions and properties being used in the proof.

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Homework Statement


Prove that
cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

The Attempt at a Solution



Cis is a sin-and-cos summation. Shortly,
cisx = cosx + isinx

How can you prove the statement?
 
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You're going to have a hard time proving that, unless my math is bad today (a very real possibility... I've been making some pretty bad gaffes lately...)

I get that it should come out to 2isin(x1-x2).
 
Use the addition formulas for trigonometric functions:

cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

sin(A + B) = sin(A)cos(B) + sin(B)cos(A)

EDIT: AUMathTutor is right.
 
dx, do you get the answer the OP is looking for when you work it out that way?

I used the rules cos(x) = cos(-x) and sin(x) = -sin(-x) and got it my way.
 
In fact,

cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

Is clearly wrong. try x1 = x2 = 0. You get

cis(0) - cis(0) = 2cos(0) = 2

so 0 = 2. I think I'm alright on this one today.
 
You're right AUMathTutor, the answer is 2isin(x1-x2).
 
AUMathTutor said:
You're going to have a hard time proving that, unless my math is bad today (a very real possibility... I've been making some pretty bad gaffes lately...)

I get that it should come out to 2isin(x1-x2).

Could you please, write your methods down.

The following must be true to prove the statement

cos(x1 - x2) = cos(x2 - x1)
and
sin(x1 - x2) = -sin(x1 - x2)

If the above statements hold, then the original statement can be proven true.
 
Here'w what you need to do:

1. Write down the problem in terms of cis.
2. Rewrite the problem in terms of sin and cos using the definition of cis.
3. Apply the equalities sin(x) = -sin(-x) and cos(x) = cos(-x).
4. Collect like terms and/or cancel out terms.
5. See whether you get what you wanted.

The problem is incorrect. The answer is 2isin(x1-x2). You'll see the cosines cancel out. This is just some simple algebra.

You can also use the rules dx posted, and then recombine your answer to get the same thing. It's a few more lines of math, but probably a little more clear.
 
It can also be seen very easily by drawing them in the complex plane, if you've been taught that.
 
  • #10
dx said:
It can also be seen very easily by drawing them in the complex plane, if you've been taught that.

I got the same answer as you.
I can now draw easily the result to a complex plane. I can also draw the result on LHS in the first post.
This shows me that the initial statement must be false.

However, I am not use how you can use complex plane without expanding cis parts.
I personally need to see the imag and real parts to draw the results on the plane.
 
  • #11
[itex]cis(\theta)[/itex] is the point where the unit circle crosses the line through (0,0) that makes angle [itex]\theta[/itex] with the positive x-axis. Surely that is not hard to find on a plane.
 
  • #12
soopo said:
Could you please, write your methods down.

The following must be true to prove the statement

cos(x1 - x2) = cos(x2 - x1)
and
sin(x1 - x2) = -sin(x1 - x2)

If the above statements hold, then the original statement can be proven true.

The two statements do not hold for all values of x1 and x2.
The first equation is identically true, because cos(x) = cos(-x) for all x.
The second equation is true only when x1 - x2 = 0.
 
  • #13
"The second equation is true only when x1 - x2 = 0."

Any reasonable person can see he meant to write x2 - x1.
 
  • #14
I'm a reasonable person, by my own estimation, but I am unable to see into his/her mind to see what he/she means to do.
 
  • #15
Mark44 said:
I'm a reasonable person, by my own estimation, but I am unable to see into his/her mind to see what he/she means to do.

@Mark: Thank you for your correction!

I tried to get the above result by "brute force" without considering the situation in the unit circle.
 

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