20 20 sinkotd(at)==mI cos kot k 20 (1-cosk丌) kT 当k为奇数时,Cosk丌=-1,b kn 当k为偶数时,COsk丌=1,bk 40 由此可得()=-n(snot+2sin3ot+sin5ot+ + sin kot+…)(k为奇数) k (1 cos ) 2 cos 2 sin ( ) 2 0 0 k k U k t k U k t d t U m m m = − = = − cos 1, 0 4 cos 1, = = = − = k m k k b k U k b 当k为奇数时, 当k为偶数时, 由此可得 k t k为奇数) k t t t U u t m sin ) ( 1 sin 5 5 1 sin 3 3 1 (sin 4 ( ) + + = + + +