Fourier Series Help: Solving sin^2x + sin^3x

[Kuma]

Homework Statement

Hi. I want to find the Fourier series of

sin^2 x + sin ^3x

and sin θ = [e^iθ - e-iθ]/2i

Homework Equations

The Attempt at a Solution

So if i use sin x = [e^ix - e-ix]/2i I will get for the first term:

[e^2ix + e^-2ix -2]/-4

I can do the same for the second term but what do i do from here..? I'm not sure how to integrate these.

 

[sunjin09]

Homework Statement

Hi. I want to find the Fourier series of

sin^2 x + sin ^3x

and sin θ = [e^iθ - e-iθ]/2i

Homework Equations

The Attempt at a Solution

So if i use sin x = [e^ix - e-ix]/2i I will get for the first term:

[e^2ix + e^-2ix -2]/-4

I can do the same for the second term but what do i do from here..? I'm not sure how to integrate these.

Your Fourier coefficients for sin2x are simply ±1/(2i) for exp(±2ix), similar for all the other terms

 

[vela]

You don't need to integrate.

Hint: Use $\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ to rewrite

$$-\frac{e^{i(2x)} + e^{-i(2x)} - 2}{4} = \ ?$$

 

[Kuma]

isn't that just cos^2x ?? It should most likely be something else but that's the only way i see it being re written as cos. That just brings me back to the starting point so yeah..heh

 

[vela]

No, you have to get the algebra right. You should know from trig that
$$\sin^2 x = \frac{1-\cos 2x}{2}$$ Can you see how to get that result from
$$\sin^2 x = -\frac{e^{i(2x)} + e^{-i(2x)} - 2}{4}$$using the hint I mentioned above.