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!

Using complex numbers to find trig identities

  1. Jul 2, 2013 #1
    I can find for example Tan(2x) by using Euler's formula for example

    Let the complex number Z be equal to 1 + itan(x)

    Then if I calculate Z2 which is equal to 1 + itan(2x) I can find the identity for tan(2x) by the following...

    Z2 =(Z)2 = (1 + itan(x))2 = 1 + (2i)tan(x) -tan(x)2 = 1 -tan(x)2 + i(2tan(x))

    now tan(2x) = Im(Z2)/Re(Z2) = 2tan(x)/(1 - tan2(x)).

    QUESTION:
    Is there a method (still using complex numbers) to find half angle identities?
     
  2. jcsd
  3. Jul 2, 2013 #2
    Sure. Say you want to find [itex] \sin(x/2) [/itex]. [tex] \sin(x/2)= \frac{e^\frac{ix}{2}-e^{-\frac{ix}{2}}}{2i} [/tex] Square both sides and simplify using [itex] \frac{e^{ix}+e^{-ix}}{2}=cos(x) [/itex] to get [tex] \sin^2(x/2)=\frac{1-cos(x)}{2} [/tex]
     
  4. Jul 2, 2013 #3
    would it work the same for sin(x/n) where n>2
     
  5. Jul 2, 2013 #4
    I don't think so. If you were to expand [tex](\frac{e^\frac{ix}{n}-e^{-\frac{ix}{n}}}{2i})^n [/tex] you would get a lot of terms that look like [itex] e^{aix/n} [/itex] that don't look so easy to simplify.
     
    Last edited: Jul 2, 2013
  6. Jul 2, 2013 #5
    Thanks for your help
     
  7. Jul 3, 2013 #6

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    If for any function f, you have a formula f(2x) = g[f(x)]

    Then f(x) = g[f(x/2)], and f(x/2) = g-1[f(x)]

    Which you can obtain provided you can invert g, which in this case you can - it is solving a quadratic.

    You will surely not obtain any different results by whatever other method.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Using complex numbers to find trig identities
  1. Complex Trig Identity (Replies: 7)

Loading...