Answer: F(x)= cosx

$fn(x)=cosnx ,\\fm(x)=cosmx$

$\therefore\ \int_{-\pi}^{\pi} f_m(x).f_n(x)= \int_{-\pi}^{\pi} cosnx.\ sinmx.dx\\ -\dfrac{1}{2}\int_0^{\pi} [\ cos(m+n)x+\ cos(m-n)x].dx\\ =-\dfrac{1}{2} [\dfrac{\ sin(m+n)x}{m+n}+\dfrac{\ sin(m-n)x}{m-n}] _{-\pi }^{\pi}\\ Now,two\ case1\ if\ \ if\ m\ne n ,then\\ \int _{-\pi}^{\pi}f_n(x).f_n(x)=-\dfrac{1}{2} [0+0]\\=0 \\ case 2 \ if\ m= n,then\\ \int _{-\pi}^{\pi}f_m(x).f_n(x )=\int _{-\pi}^{\pi} cos^2nx$

$\int_{-\pi}^{\pi} f_n(x).f_n(x)dx=\int_0^{\pi/2}\ cos^2nx dx\\ \therefore \int_{-\pi}^{\pi} [f_n(x)]^2 dx= \int_{-\pi}^{\pi}(\dfrac{1-cos(2n)x}{2})dx\\ =[\dfrac{x}{2}-\dfrac {\ sin2n )}{4n)}]_{-\pi}^{\pi}\\ =\dfrac{\pi}{2}+\dfrac{\pi}{2} =\pi \ne0$

$since\ \int_{-\pi}^{\pi} f_m(x). f_m(x) dx =o \ if\ m\ne 0 \ and\ \ne0 \ if\ m=0\\ The\ given \ set \ of \ function\ is\ orthogonal $

$If\ the\ set \ to\ be\ orthonormal\ then\ we\ should\ have$

$\int_{-\pi}^{\pi} [f_n]^2.dx =\pi \\ we\ divide\ by\ \pi\\ \int_{-\pi}^{\pi} \dfrac{1}{\pi} [f_n]^2.dx =\dfrac{\pi}{\pi}.\\ =\int_0^{\pi/2} \dfrac{2}{\sqrt\pi} f_n(x) \dfrac{2}{\sqrt\pi} f_n(x)dx=1 \\ Hence \ orthonormal\ set,where\ \phi_n (x)=\dfrac{2}{\sqrt\pi} .\ cosnx\\ The\ required \ orthonormal\ set\ of\ function is\\\dfrac{2}{\sqrt\pi}\ cosx, \dfrac{2}{\sqrt\pi}\ cos2x, \dfrac{2}{\sqrt\pi}\ cos3x, ...... $