1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why does sinx=(1/2i)[e^(ix)-e^(-ix)]?

  1. Dec 4, 2005 #1
    why does sinx=(1/2i)[e^(ix)-e^(-ix)]?
     
  2. jcsd
  3. Dec 4, 2005 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

  4. Dec 4, 2005 #3
    e^ix=cosx+isinx
    e^(-ix)=?
    do you really add a negative sign and it becomes cos(-x)+isin(-x)?
     
  5. Dec 4, 2005 #4

    LeonhardEuler

    User Avatar
    Gold Member

    Yes, and cos(-x)=cos(x), while sin(-x)=-sin(x), because they are even and odd functions, respectively.
     
  6. Dec 4, 2005 #5

    benorin

    User Avatar
    Homework Helper

    yes, and recall that sine is an odd function and cosine is an even function so we have

    [tex]e^{-ix} = \cos(-x) + i\sin(-x)=\cos(x)-i\sin(x)[/tex]

    and from Euler's formula we have [itex]e^{ix}=\cos(x)+i\sin(x)[/itex]

    and so subtracting the first formula from the second gives

    [tex]e^{ix}-e^{-ix} = 2i\sin(x)[/tex]

    hence

    [tex]\sin(x)=\frac{1}{2i} \left( e^{ix}-e^{-ix}\right) [/tex]
     
  7. Dec 5, 2005 #6
    wow~
    amazing...
    thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Why does sinx=(1/2i)[e^(ix)-e^(-ix)]?
  1. 1-2i in Polar form? (Replies: 2)

Loading...