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Fourier Series

  1. Oct 18, 2015 #1
    Consider the following article:

    At definition, they say that an = An*sin() and bn = An*cos()

    So with these notations you can go from a sum having sin and cos to a sum having only sin but with initial phases.

    Why can I write an = An*sin() and bn = An*cos() ?
    It seems out of the blue.
  2. jcsd
  3. Oct 18, 2015 #2


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    Substitute the 2nd equation to the first equation.
  4. Oct 18, 2015 #3
    I know that by substitution we get from one form to another.
    But my question is why I can write cos(phi) = a/sqrt(a^2+b^2) and sin(phi) = -b/sqrt(a^2+b^2) ?
    I see that by taking cos(phi)^2 + sin(phi)^2 I get 1, so is good.

    But why I can write cos(phi) like that? Writing cos(phi) like that, then from cos(phi)^2 + sin(phi)^2 = 1, I get sin(phi). But why I can write cos(phi) in the first place like that?

    It is just arbitrary? If I write cos(phi) = a, then I find sin(phi)...then, ok. Is fine.I can see that. But writing as a/sqrt(a^2+b^2), it does not seem so straight-forward. Maybe there is a property that for any two numbers a,b then I can write cos(phi) in that way. I do not know.

    Going from the trigonometric Fourier sum to the exponential form, we use Euler's formula to write cos() = 1/2(e^+e^) and sin too. So I have Euler's formula here.
    Last edited: Oct 18, 2015
  5. Oct 18, 2015 #4


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    Expression like ##A_n \cos \phi_n## only depends on the index ##n##, so there is no harm in writing them in a more simple way such as ##a_n##.
    If you want to picture it that way, you first have to draw a right triangle and define which sides ##a## and ##b## correspond to, and which angle ##\phi## corresponds to.
  6. Oct 18, 2015 #5
    Aa, ok. Now makes sense. Thanks.
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