1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Homework Helper
    Gold Member

    This can be shown from Euler's formula.

    http://mathworld.wolfram.com/EulerFormula.html" [Broken]
    Last edited by a moderator: May 2, 2017
  4. Dec 4, 2005 #3
    do you really add a negative sign and it becomes cos(-x)+isin(-x)?
  5. Dec 4, 2005 #4


    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


    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]


    [tex]\sin(x)=\frac{1}{2i} \left( e^{ix}-e^{-ix}\right) [/tex]
  7. Dec 5, 2005 #6
    thank you very much!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook