anonymous
  • anonymous
Find cosine series for at function with eulers formula.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I have this function \[f(x)=\cos(3x)\sin^2(x)\] and I need to find the cosine series to the function. With Eulers formula I write it this way \[\frac{ e^{i3x}+e^{-i3x} }{ 2 } * (\frac{ e^{ix}-e^{-ix} }{ 2i })^2\] \[= \frac{ e^{-i3x} }{ 4 }+\frac{ e^{i3x} }{ 4 }-\frac{ e^{-ix} }{ 8 }-\frac{ e^{ix} }{ 8 }-\frac{ e^{-i5x} }{ 8 }-\frac{ e^{i5x} }{ 8 }\] But have do I get from here to at cosine series?
anonymous
  • anonymous
@phi Can you help me?
phi
  • phi
maybe, but what do you mean by a cosine series? what is the answer supposed to look like ?

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More answers

anonymous
  • anonymous
i know the answer is \[-\frac{\cos(x) }{ 4 }-\frac{\cos(5x) }{ 4 }+\frac{ \cos(3x) }{ 2 }\]
phi
  • phi
it looks like you are almost there: \[ \frac{ e^{-i3x} }{ 4 }+\frac{ e^{i3x} }{ 4 }-\frac{ e^{-ix} }{ 8 }-\frac{ e^{ix} }{ 8 }-\frac{ e^{-i5x} }{ 8 }-\frac{ e^{i5x} }{ 8 } \] collect terms: \[ -\frac{1}{4}\left( \frac{ e^{ix} }{2 }+\frac{ e^{-ix} }{ 2 }\right) \] and so on...
phi
  • phi
use the identity \[ \frac{ e^{ix}+ e^{-ix} }{ 2 }=\cos(x) \]
anonymous
  • anonymous
I will try work it out...
anonymous
  • anonymous
Thank you...
phi
  • phi
there is not much to work out \[ -\frac{1}{4}\left( \frac{ e^{ix} }{2 }+\frac{ e^{-ix} }{ 2 }\right) +\frac{1}{2}\left( \frac{ e^{i3x} }{2 }+\frac{ e^{-i3x} }{ 2 }\right) -\frac{1}{4}\left( \frac{ e^{i5x} }{2 }+\frac{ e^{-i5x} }{ 2 }\right)\]
phi
  • phi
the exponentials inside the parens are cosines
anonymous
  • anonymous
I see..
phi
  • phi
\[ -\frac{1}{4}\left( \frac{ e^{ix} + e^{-ix}}{2 }\right) +\frac{1}{2}\left( \frac{ e^{i3x} +e^{-i3x}}{2 }\right) -\frac{1}{4}\left( \frac{ e^{i5x} + e^{i5x}}{2 }\right) \] \[ -\frac{1}{4} \cos(x) +\frac{1}{2}\cos(3x) -\frac{1}{4}\cos(5x)\]
anonymous
  • anonymous
Yes. thank you...

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