1. ## fourier

Hello all

I have the function

$s(x)= sin(x)cos^2(x)$

and i need to show that it is a $2\pi$-periodic trigonometric function,

My first idea was simply something along the lines of $sin(x)cos^2(x)=sin(2\pi x)cos^2(2\pi x)$

Furthermore i need to determine the fourrier coefficients for all $n \in Z$

$1/2pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx$

Any helps or hints is very much appreciated

2. Originally Posted by Zaph
Hello all

I have the function

$s(x)= sin(x)cos^2(x)$

and i need to show that it is a $2\pi$-periodic trigonometric function,

My first idea was simply something along the lines of $sin(x)cos^2(x)=sin(2\pi x)cos^2(2\pi x)$
No, those are NOT the same. What you want is $sin(x)cos^2(x)= sin(x+ 2\pi)cos^2(x+ 2\pi)$.

Furthermore i need to determine the fourrier coefficients for all $n \in Z$

$1/2pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx$

Any helps or hints is very much appreciated
That will be the same as
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx$
and
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx$

You can use trigonometric identities for those.

3. Bleh brainfart whilst tex'ing, i meant ofcourse
$sin(x)cos^2(x)= sin(2\pi+x)cos^2(2\pi+x)$
But thanx for the heads up, now i know i wasnt all that wrong

For the other half, ill look into what you suggested, and thanx alot for the quick reply

4. Is...

$\sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x = \frac{1}{4} \sin x + \frac{1}{4} \sin 3x$

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
Is...

$\sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x = \frac{1}{4} \sin x + \frac{1}{4} \sin 3x$

Kind regards

$\chi$ $\sigma$
I'm not quite following you here, care to elaborate?

6. Is...

$\sin^{2} x + \cos^{2} x =1 \Rightarrow \cos^{2} x = 1 - \sin^{2} x$ (1)

... and...

$\sin^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3x$ (2)

... so that...

$\sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x =\frac{1}{4} \sin x + \frac{1}{4} \sin 3x$ (3)

Kind regards

$\chi$ $\sigma$

7. Originally Posted by HallsofIvy

That will be the same as
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx$
and
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx$

You can use trigonometric identities for those.
So i tried for the first one, and i'd like someone to check if im going in the right direction.

$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx$ $= \frac{1}{2}\int_{-\pi}^\pi e^{-ix} e^{-ix^2}e^{-inx}= \frac{1}{2}\int_{-\pi}^\pi e^{-ix} e^{-inx}= ...$

[/tex]

8. Originally Posted by HallsofIvy
No, those are NOT the same. What you want is $sin(x)cos^2(x)= sin(x+ 2\pi)cos^2(x+ 2\pi)$.

That will be the same as
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx$
and
$\frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx$

You can use trigonometric identities for those.
Would you care to elaborate a bit on this? Thanx in advance

9. Do you need to show that the function is $2\pi$-periodic and has no smaller period? Some trig functions built up out of your basic sin and cos have periods smaller than the original. $|\cos(x)|$ is a simple one that comes to mind; the same with $\sin^{2}(x)$.

10. Originally Posted by Ackbeet
Do you need to show that the function is $2\pi$-periodic and has no smaller period? Some trig functions built up out of your basic sin and cos have periods smaller than the original. $|\cos(x)|$ is a simple one that comes to mind; the same with $\sin^{2}(x)$.
Yes, i need to show that is is $2\pi$-periodic nothing more there, and besides that i need to determine the fourier coefficients:

$\frac{1}{2}\pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx$