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$$\geq$$1 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
[tex]\int_{-\pi}^\pi} sin(mx)sin(nx)dx[/itex] for m= n and for [itex]m\ne n[/tex].

 

[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!