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

$cis(\theta)$ 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.