Orthogonality of Sine and Cosine

Inner Products and Hilbert Space Foundations

Inner product theory tells us that two functions are orthogonal on an interval \([a,b]\) if their inner product is 0. This induces the \(L^2\) norm, turning the function space into a Hilbert Space.

The Inner Product: For real-valued functions on \([a,b]\), the inner product is:

\langle f, g \rangle = \int_a^b f(x)g(x)\,dx

This allows us to define the norm as \( \| f \| = \sqrt{\langle f, f \rangle} \).

Integral Identities

By applying basic calculus and the fact that \(\sin(n\pi) = 0\) for all integers \(n\), we find:

\int_{-\pi}^{\pi}\cos(nx)\,dx = 0, \quad \int_{-\pi}^{\pi}\sin(nx)\,dx = 0 \quad (\forall n \neq 0)

Furthermore, using power-reduction identities, we find the norms of these functions:

\int_{-\pi}^{\pi}\cos^2(nx)\,dx = \pi, \quad \int_{-\pi}^{\pi}\sin^2(nx)\,dx = \pi

Proving Orthogonality

To show that these functions form an orthogonal set, we use product-to-sum trigonometric formulas:

\cos(\alpha)\cos(\beta) = \frac{1}{2}\big[\cos(\alpha + \beta) + \cos(\alpha - \beta)\big]

For \(m \neq n\), the integral of the product evaluates to zero:

\frac{1}{2}\int_{-\pi}^{\pi}\big(\cos[(n+m)x] + \cos[(n-m)x]\big)\,dx = 0 + 0 = 0

Similarly, for \(\sin(nx)\) and \(\cos(mx)\), the integral also vanishes regardless of the values of \(n\) and \(m\):

\int_{-\pi}^{\pi}\sin(nx)\cos(mx)\,dx = 0

The Orthogonal Basis

The set of functions:

\{\,1,\ \sin(nx),\ \cos(nx)\ \mid\ n \in \mathbb{N}\,\}

is mutually orthogonal on the interval \([-\pi, \pi]\). This geometric property is exactly why we can "decompose" signals into their constituent frequencies without them interfering with one another.

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