# Simple complex number problem

1. Aug 5, 2007

### nirvana1990

1. The problem statement, all variables and given/known data
Show that: (cosx+isinx)^2= cos2x + isin2x

2. Relevant equations
i^2=-1

3. The attempt at a solution

Well, here's my attempt!
(cosx+isinx)^2=(cosx+isinx)(cosx+isinx)
=(cos^2x)+(2[isinxcosx])+(i^2sin^2x)
=(cos^2x)+(2[isinxcosx])-sin^2x

p.s. when i wrote cos^2x, for example, I meant cos squared, multiplied by x.

2. Aug 5, 2007

### Archduke

Have a look at the basic double angle formulas for cos2x and sin2x, and all will be revealed!

3. Aug 5, 2007

### nirvana1990

Ooh thanks that was quite helpful but now i seem to have over-simplified somehow!
I got: cos^2x-sin^2x+2isin2xcos2x
=cos2x+2isin2xcos2x (by using cos^2x-sin^2x=cos2x)
Then would you divide by cos2x to give: 1+2isin2x? Or should I use sin2x=2sinxcosx somewhere?!?!

4. Aug 6, 2007

### HallsofIvy

Staff Emeritus
No, you did not have that before- you are getting ahead of yourself!
You had cos2x- sin2x+ i(2 sin x cos x), NOT "2cos 2x sin 2x.

Since you do have 2 sin x cos x, it should be obvious exactly where to use that!

5. Aug 6, 2007

### nirvana1990

Yes thanks I realised my error this morning after doing many numerical examples!
cos^2x-sin^2x+ 2(isinxcosx)= cos2x+i(2sinxcosx)=cos2x+isin2x

thanks for the help!