三角函数组具有正交性 k 1·cos-,ax=0(k≠0) k 1·sndx=0. k COS COS dx=0(k≠n) (5.1.4) k SIn SIn dx=0(k≠n), COS SIn dx=0 因此」a ∫,()cos2a 5.1.5) k f(s)sin kx与三角函数组具有正交性      − − − − −  =  =   =   =  =  l l l l l l l l l l dx l n x l k x dx k n l n x l k x dx k n l n x l k x dx l k x dx k l k x cos sin 0. sin sin 0 ( ), cos cos 0 ( ), 1 sin 0, 1 cos 0 ( 0),         (5.1.4) 因此        = =   − − ( )sin . 1 ( ) cos , 1          d l k f l b d l k f l a l l k l l k k (5.1.5)