Question

Find cosine series for at function with eulers formula.

Answer 1

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?

Answer 2

@phi Can you help me?

Answer 3

maybe, but what do you mean by a cosine series? what is the answer supposed to look like ?

Answer 4

i know the answer is \[-\frac{\cos(x) }{ 4 }-\frac{\cos(5x) }{ 4 }+\frac{ \cos(3x) }{ 2 }\]

Answer 5

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...

Answer 6

use the identity
\[ \frac{ e^{ix}+ e^{-ix} }{ 2 }=\cos(x) \]

Answer 7

I will try work it out...

Answer 8

Thank you...

Answer 9

there is not much to work out

\[ -\frac{1}{4}\left( \frac{ e^{ix} }{2 }+\frac{ e^{-ix} }{ 2 }\right) +\frac{1}{2}\left( \frac{ e^{i3x} }{2 }+\frac{ e^{-i3x} }{ 2 }\right) -\frac{1}{4}\left( \frac{ e^{i5x} }{2 }+\frac{ e^{-i5x} }{ 2 }\right)\]

Answer 10

the exponentials inside the parens are cosines

Answer 11

I see..

Answer 12

\[ -\frac{1}{4}\left( \frac{ e^{ix} + e^{-ix}}{2 }\right) +\frac{1}{2}\left( \frac{ e^{i3x} +e^{-i3x}}{2 }\right) -\frac{1}{4}\left( \frac{ e^{i5x} + e^{i5x}}{2 }\right) \]
\[ -\frac{1}{4} \cos(x) +\frac{1}{2}\cos(3x) -\frac{1}{4}\cos(5x)\]

Answer 13

Yes. thank you...