Orthogonality of the trigonometric system

Formulas

\(\begin{flalign*} &\int_{-\pi}^\pi \cos nx \cos mx = \begin{cases} 0 & (n \neq m) \\ \pi & (n = m \neq 0) \\ 2\pi & (n = m = 0) \end{cases}\\ \\ &\int_{-\pi}^\pi \sin nx \sin mx = \begin{cases} 0 & (n \neq m) \\ \pi & (n = m \neq 0) \\ 0 & (n = m = 0) \end{cases}\\ \\ &\int_{-\pi}^\pi \sin nx \cos mx = 0 \end{flalign*}\)

Derivation

(i)
First, we can apply trigonometric addition formulas.

\[\begin{flalign*} & \cos nx \cos mx = \frac{1}{2}\cos(n-m)x + \frac{1}{2}\cos(n+m)x\\ \\ & \Rightarrow \int_{-\pi}^\pi \cos nx \cos mx =\frac{1}{2}\int_{-\pi}^\pi\cos(n-m)x + \frac{1}{2}\int_{-\pi}^\pi\cos(n+m)x \end{flalign*}\]

By the symmetry of the cosine function about the x-axis, we can derive the result as follows.

\[\begin{flalign*} \int_{-\pi}^\pi \cos nx \cos mx = \begin{cases} 0 & (n \neq m) \\ \pi & (n = m \neq 0) \\ 2\pi & (n = m = 0) \end{cases} && \end{flalign*}\]

(ii)
Derivation of (ii) is similar to the one in (i).

\[\begin{flalign*} & \sin nx \sin mx = \frac{1}{2}\cos(n-m)x - \frac{1}{2}\cos(n+m)x\\ \\ & \Rightarrow \int_{-\pi}^\pi \sin nx \sin mx =\frac{1}{2}\int_{-\pi}^\pi\cos(n-m)x - \frac{1}{2}\int_{-\pi}^\pi\cos(n+m)x\\ \\ & \Rightarrow \int_{-\pi}^\pi \sin nx \sin mx = \begin{cases} 0 & (n \neq m) \\ \pi & (n = m \neq 0) \\ 0 & (n = m = 0) \end{cases} \end{flalign*}\]

(iii)
Applying trignometric addtion formulas is same.

\[\begin{flalign*} &\sin nx \cos mx = \frac{1}{2}\sin(n+m)x + \frac{1}{2}\sin(n-m)x \\ \\ &\Rightarrow \int_{-\pi}^\pi \sin nx \cos mx = \frac{1}{2}\int_{-\pi}^\pi\sin(n+m)x + \frac{1}{2}\int_{-\pi}^\pi\sin(n-m)x\\ \end{flalign*}\]

However, by the symmetry of the sine function about the origin, the result of the formula is always 0.

\[\begin{flalign*} \int_{-\pi}^\pi \sin nx \cos mx = 0 && \end{flalign*}\]