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Simple complex number problem

  1. Aug 5, 2007 #1
    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. jcsd
  3. Aug 5, 2007 #2
    Have a look at the basic double angle formulas for cos2x and sin2x, and all will be revealed!
     
  4. Aug 5, 2007 #3
    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?!?!
     
  5. Aug 6, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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!
     
  6. Aug 6, 2007 #5
    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!
     
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