hey,
(1-{{x}^{2}}){{y}^{''}}-2x{{y}^{'}}+n(n+1)y=0,\,\,\,\,\,-1\le x\le 1
to convert the legendre equation y(x) into trig form y(cos\theta) is it simply, set x=cos\theta then
(1-{{\cos }^{2}}\theta ){{y}^{''}}-2{{y}^{'}}\cos \theta +n(n+1)y=0 for -\pi \le x\le \pi
{{\sin }^{2}}\theta...