Answer
See below
Work Step by Step
We are given:
$f_1(x)=1\\
f_2(x)=\sin x\\
f_3(x)=\cos x\\
a=-\frac{\pi}{2}\\
b=\frac{\pi}{2}$
According to Gram-Schmidt we have:
$v_1=f_1(x)=1\\
v_2=f_2(x)-\frac{(f_2(x),v_1)}{||v_1||^2}v_1\\
=\sin x-\frac{(\sin x,1)}{||1||^2}1 \\
=\sin x-\frac{\int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin x.1dx}{\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}dx}.1\\
=\sin x-\frac{ -\cos x|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}{x|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} \\
=\sin x-\frac{0-0}{\frac{\pi}{2}+\frac{\pi}{2}}\\
=\sin x\\
v_3=f_3(x)-\frac{(f_3(x),v_1)}{||v_1||^2}v_1-\frac{(f_3(x),v_2)}{||v_2||^2}v_2\\
=\cos x-\frac{(\cos x,1)}{||1||^2}1- \frac{(\cos x,\sin x)}{||\sin x||^2}\sin x\\
=\cos x-\frac{\int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \cos x.1dx}{\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}dx}-\frac{\int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos x.\sin x dx}{\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin ^2xdx}\sin x\\
=\cos x-\frac{\sin x|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}{x|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}-\frac{ -\frac{1}{4}\cos 2x|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}{\frac{1}{2}(x-\frac{\sin 2x}{2})|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}\sin x \\
=\cos x-\frac{2}{\pi}-\frac{\frac{1}{2}(\cos \pi -\cos (-\pi))}{\frac{1}{4}(2x-\sin 2x)|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}}\sin x\\
=\cos x-\frac{2}{\pi}$
Hence, an orthogonal basic spanned by the given factors is:
$\{1,\sin x, \cos x-\frac{2}{\pi}\}$
