In the study of Fourier Series, you will find that every continuous function $f$ on an interval $[-L,L]$ can be expressed on that interval as an infinite series of sines and cosines. For example, if the interval is $[-\pi,\pi]$, \[f(x)=A_0+\sum^{\infty}_{k=1} [A_k\cos (kx)+B_k\sin (kx)]\] where the constants are given by integrals involving $f$.

The theory of Fourier series relies on the fact that the functions \[1,\quad \cos x,\quad \sin x,\quad \cos 2x,\quad \sin 2x,\quad \ldots,\quad \cos nx,\quad \sin nx,\quad \ldots\] form an orthogonal set:

Here, we will verify this fact.

We will use the following trigonometric identities: \begin{eqnarray*} \sin A\sin B =\frac{1}{2}[\cos (A-B)-\cos (A+B)] \phantom{.}\\ \cos A\cos B =\frac{1}{2}[\cos (A-B)+\cos (A+B)] \phantom{.}\\ \sin A\cos B =\frac{1}{2}[\sin (A-B)+\sin (A+B)]. \end{eqnarray*} We have six general integrals to evaluate to prove the orthogonality of the set $\{1, ~\cos x, ~\sin x, \ldots\}$. In each of the following, we assume $m$ and $n$ are distinct positive integers.

1. $\quad\displaystyle\int^\pi_{-\pi} 1\cdot \cos (nx)\, dx=\left.\frac{1}{n}\sin (nx)\right|^{\pi}_{-\pi}=0.$
   
   
2. $\quad\displaystyle\int^\pi_{-\pi} 1\cdot \sin (nx)\, dx=\left.-\frac{1}{n}\cos (nx)\right|^{\pi}_{-\pi}=0.$
   
   
3. $\quad\displaystyle\int^\pi_{-\pi} \sin (nx)\cos (nx)\, dx=\left.\frac{\sin^2 (nx)}{2n}\right|^{\pi}_{-\pi}=0.$
   
   
4.   \begin{eqnarray*} \int^\pi_{-\pi} \sin (mx)\sin (nx)\,dx&=&\int^{\pi}_{-\pi}\frac{1}{2}[\cos (m-n)x-\cos (m+n)x]\, dx\\ &=&\left.\left(\frac{\sin[(m-n)x]}{2(m-n)}- \frac{\sin[(m+n)x]}{2(m+n)}\right)\right|^{\pi}_{-\pi}\\ &=&0. \end{eqnarray*}
   
   
5.   \begin{eqnarray*} \int^\pi_{-\pi} \cos (mx)\cos (nx)\,dx&=&\int^{\pi}_{-\pi}\frac{1}{2}[\cos (m-n)x+\cos (m+n)x]\, dx\\ &=&\left.\left(\frac{\sin[(m-n)x]}{2(m-n)}+ \frac{\sin[(m+n)x]}{2(m+n)}\right)\right|^{\pi}_{-\pi}\\ &=&0. \end{eqnarray*}
   
   
6.   \begin{eqnarray*} \int^\pi_{-\pi} \sin (mx)\cos (nx)\,dx&=&\int^{\pi}_{-\pi}\frac{1}{2}[\sin (m-n)x+\sin (m+n)x]\, dx\\ &=&\left.\left(-\frac{\cos[(m-n)x]}{2(m-n)}- \frac{\cos[(m+n)x]}{2(m+n)}\right)\right|^{\pi}_{-\pi}\\ &=&0. \end{eqnarray*}

We have now shown that $\{1, ~\cos x, ~\sin x, ~\cos 2x, ~\sin 2x, \ldots\}$ is indeed an orthogonal set of functions!

In the following Exploration, graph functions $\sin(mx)\sin(nx)$, $\sin(mx)\cos(nx)$, and $\cos(mx)\cos(nx)$ for various values of m and n and observe the interesting curves that result.

The integral of the product of any $2$ of these functions over $[-\pi,\pi]$ is $0$.