1. May 14, 2009

### soopo

1. The problem statement, all variables and given/known data
Prove that
cis(x1 - x2) - cis(x2 - x1) = 2cos(x1 - x2)

3. The attempt at a solution

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

How can you prove the statement?

2. May 14, 2009

### AUMathTutor

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).

3. May 14, 2009

### dx

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.

4. May 14, 2009

### AUMathTutor

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.

5. May 14, 2009

### AUMathTutor

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.

6. May 14, 2009

### dx

You're right AUMathTutor, the answer is 2isin(x1-x2).

7. May 14, 2009

### soopo

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.

8. May 14, 2009

### AUMathTutor

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.

9. May 14, 2009

### dx

It can also be seen very easily by drawing them in the complex plane, if you've been taught that.

10. May 15, 2009

### soopo

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. May 15, 2009

### HallsofIvy

Staff Emeritus
$cis(\theta)$ is the point where the unit circle crosses the line through (0,0) that makes angle $\theta$ with the positive x-axis. Surely that is not hard to find on a plane.

12. May 15, 2009

### Staff: Mentor

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. May 15, 2009

### AUMathTutor

"The second equation is true only when x1 - x2 = 0."

Any reasonable person can see he meant to write x2 - x1.

14. May 15, 2009

### Staff: Mentor

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. May 16, 2009

### soopo

@Mark: Thank you for your correction!

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