Given f(x) = sin(2n + 1)x

A set of functions $f_{1}(x), f_{2}(x),.....f_{n}(x)$ is orthogonal on (a,b) if

$\int\limits_{a}^{b}f_{m}(x)f_{n}(x) = 0$ if $m \neq n$ &

$\int\limits_{a}^{b}[f_{n}(x)]^2 dx \neq 0 $ if m = n

A set of functions $f_{1}(x), f_{2}(x),.....f_{n}(x)$ is orthonormal on (a,b) if

$\int\limits_{a}^{b}f_{m}(x)f_{n}(x) = 0$ if $m \neq n$ &

$\int\limits_{a}^{b}[f_{m}(x)]^2 dx = 1 $ if m = n

In this case

$\int\limits_{0}^{\frac{\pi}{2}} f_{m}(x) f_{n}(x) = \int\limits_{0}^{\frac{\pi}{2}} sin(2m + 1)x sin(2n + 1)x$

$\therefore \int\limits_{0}^{\frac{\pi}{2}} f_{m}(x) f_{n}(x)dx = -\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}} cos(2m + 2 + 2)x - cos(2m - 2n)xdx$

$[\because sina sinb = \frac{1}{2}[cos(a - b) - cos(a + b)] = -\frac{1}{2}[cos(a + b) - cos(a - b)]]$

$\int\limits_{0}^{\frac{\pi}{2}}f_{m}(x)f_{n}(x) = -\frac{1}{2}\bigg[\frac{sin(2m + 2n + 2)x}{2m + 2n + 2} - \frac{sin(2m - 2n)x}{2m - 2n}\bigg]_{0}^{\frac{\pi}{2}}$

Now if m $\neq$ n

$\int\limits_{0}^{\frac{\pi}{2}}f_{m}(x)f_{n}(x) = 0$

If m = n

$\int\limits_{0}^{\frac{\pi}{2}}[f_{n}(x)]^2 = \int\limits_{0}^{\frac{\pi}{2}} sin^2(2n + 1)xdx$

$\int\limits_{0}^{\frac{\pi}{2}}[f_{n}(x)]^2 = \int\limits_{0}^{\frac{\pi}{2}} \frac{1 - cos(2n + 1)x}{2}dx$

$\int\limits_{0}^{\frac{\pi}{2}}[f_{n}(x)]^2 = \frac{1}{2}\bigg[x - \frac{sin2(2n + 1)x}{2n + 1}\bigg]_{0}^{\frac{\pi}{2}}$

$\int\limits_{0}^{\frac{\pi}{2}}[f_{n}(x)]^2 = \frac{1}{2}\bigg[\frac{1}{2} - 0\bigg]$

$\int\limits_{0}^{\frac{x}{2}}[f_{n}(x)]^2 = \frac{\pi}{4} \neq 0$..........(A)

Since,

$\int\limits_{0}^{\frac{\pi}{2}} f_{m}(x)f_{n}(x) = 0$ if m $\neq$ n

$\int\limits_{0}^{\frac{\pi}{2}} f_{m}(x)f_{n}(x) \neq 0$ if m = n

$\therefore$ The given set of functions is orthogonal over $[0, \pi/2]$

Now if the set is to be orthonormal, then we should have

$\int\limits_{0}^{\frac{x}{2}}[f_{n}(x)]^2 = 1$

For this we divide equation (A) by $\frac{\pi}{4}$

$\int\limits_{0}^{\frac{x}{2}}\frac{[f_{n}(x)]^2}{\frac{\pi}{4}} = \frac{\pi}{4} * \frac{4}{\pi}$

$\int\limits_{0}^{\frac{x}{2}}\frac{4}{\pi}[f_{n}(x)]^2 = 1$

$\therefore$ i.e.

$\int\limits_{0}^{\frac{x}{2}}\frac{2}{\sqrt{\pi}}f_{n}(x)\frac{2}{\sqrt{\pi}}f_{n}(x)dx = 1$

The above is now orthonormal set

$\phi_{n}(x) = \frac{2}{\sqrt{\pi}}sin(2n + 1)x$

Here the required orthonormal set of functions is

$\frac{2}{\sqrt{\pi}}sinx$, $\frac{2}{\sqrt{\pi}}sin3x$, $\frac{2}{\sqrt{\pi}}sin5x$....