Proof I don't even know how to start

[EstimatedEyes]

Homework Statement

Let f(x) = a_{1}\sinx + a_2\sin(2x) + ... + a_Nsin(Nx) where N\geq1 is an integer and a_1, ... , a_N \in\Re. Prove that for every n = 1, ... , N we have
a_n = \frac{1}{\pi}\int{f(x)\sin(nx)dx}
with the integral going from -\pi to \pi (sorry I don't know how to write definite integrals in LaTeX)

For some reason, it's not showing the integral sign. Before \sin(nx)dx there should be an integral sign followed by f(x), but it's not showing up.

Homework Equations

The Attempt at a Solution

I have no idea how to even start it. I'm not looking for the solution, just a push in the right direction. Your answers to all of my other questions of late have been spot on and for that I thank everyone who has responded. Thanks in advance for your help with this problem!

 

[jbunniii]

It shows up OK on my screen. Try refreshing your browser.

Try substituting the expression for f(x) into

\frac{1}{\pi} \int f(x) sin(nx) dx

and using trig identities. You will save yourself some work if you keep in mind that

\int_{-\pi}^{\pi} sin(ax) dx = 0

for any real number a.

 

[HallsofIvy]

First is f(x) really equal to what you have? f(x)= a_1 sin(x)+ a_2 sin(2x)+ \cdot\cdot\cdot a_N sin(Nx) would make more sense!

Assuming that is what it should be, look at
\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin(nx)dx= \frac{1}{\pi}\int_{-\pi}^\pi (a_1+a_2sin(2x)+ \cdot\cdot\cdot+ a_Nsin(Nx))sin(nx) dx
= \frac{1}{\pi}a_1\int_{-\pi}^\pi} sin(x)sin(nx)dx+ \frac{1}{\pi}a_2\int_{-\pi}^\pi}sin(2x)sin(nx)dx+ \cdot\cdot\cdot+ \frac{1}{\pi}a_N \int_{-\pi}^\pi} sin(Nx))sin(nx)dx



And also consider the value of
\int_{-\pi}^\pi} sin(mx)sin(nx)dx[/itex] for m= n and for m\ne n.

 

[EstimatedEyes]

Oh, I proved the second part of your post in part a of the problem and did not realize that it was involved in any way; thanks!