Using trigonometric identity to compute an inner product through an integral to form an orthogonal sequence of functions.

Question

Consider the inner product space: $(C(0,L),\langle \cdot,\cdot \rangle),$ where: $\langle f,g \rangle = \displaystyle\int_0^L f(x)g(x)\, dx$.

Use the trigonometric identity: $\sin(u)\sin(v) = \dfrac{1}{2} \left[ \cos(u-v) - \cos(u+v) \right]$

to show that the functions: $\left\{ \sin \left( \dfrac{n \pi x}{L} \right) \right\}_{n=1}^{\infty}$ form an orthogonal sequence of functions.

Hint: assume that: $m \neq n$ and show that $\left\langle \sin \left(\dfrac{m \pi x}{L} \right), \sin \left( \dfrac{n \pi x }{L} \right) \right\rangle = 0$

I am confused on the where to start with this problem.

Should I input the inner product: $\left\langle \sin \left(\dfrac{m \pi x}{L} \right), \sin \left( \dfrac{n \pi x }{L} \right) \right\rangle$ into the integral and solve? If so, would I use $u$-substitution as: $u = \left(\dfrac{m \pi x}{L} \right)$?

Any help is much appreciated!

Answer 1

Yes, and you get $$\left\langle \sin \left(\dfrac{m \pi x}{L} \right), \sin \left( \dfrac{n \pi x }{L} \right) \right\rangle = \int_0^L \frac12\left[\cos\left(\frac{(m-n)\pi x}{L}\right)-\cos\left(\frac{(m+n)\pi x}{L}\right)\right]dx=0.$$