Show that $f_n = \sin(n\pi x)$ for $n\ge 1$ forms an orthonormal set in $C[0,1]$ with respect to the $L^2[0,1]$ norm.

Question

Show that $f_n = \sin(n\pi x)$ for $n\ge 1$ forms an orthonormal set in $C[0,1]$ with respect to the $L^2[0,1]$ norm.

My attempt,

$$\langle \sin(m\pi x), \sin(n \pi x) \rangle^2 = \int^1_0 (\sin(m\pi x) \sin(n \pi x))dx$$ for $m,n \ge1$ boils down to $$\left(\frac{1}{4} - \frac{1}{4}\frac{\sin(2m\pi)}{2m\pi} - \frac{1}{4}\frac{\sin(2n\pi)}{2n\pi} + \frac{1}{8}\frac{\sin(2(m-n)\pi)}{2(m-n)\pi} + \frac{1}{8}\frac{\sin(2(m+n)\pi)}{2(m+n)\pi}\right)$$ but I don't understand how it equals $0$.

Any help would be appreciated, thanks

Answer 1

Your inner product (is) was not correctly defined. If should be $$ (f_m,f_n)=\int_{0}^1\sin(m\pi x)\sin(n\pi x)dx. $$

Then use the trigonometric formula that transforms the product of two sines into...

You can find it here under "Product to sum".

Answer 2

Treat the cases of $m=n$ and $m\ne n$ separately, and recall that $\sin(k\pi)=0$ for $k\in\mathbb{Z}$.