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} …