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Homework Help: 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)).

    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


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    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.
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