正在加载图片...
Asin( aa+8)=0 Asin(aua)cos&+Acos(aa)sind=0 (1) Asin(aa+8)=0 Asin(a)coso+AcoS(a)sin=0 (2) sinδ=0 (1)+(2) cos(aa)sin=0(3) costa=0 (2)-(1) sin(a)cosS=0 (4) COsS=0 sIna= 两种情况: 由(4)式 .sinδ=0→6=0则cos6=1 sin aa =0 n元 =n元 (n=0,±1,±2,…) E n元 丌h 所以 E E 2 n元 A sina= Asin sin( ) 0 sin( ) 0 + = − + =     A a A a  + = − + = sin( )cos cos( )sin 0 ( 2 ) sin( )cos cos( )sin 0 ( 1 )         A a A a A a A a (1)+(2) cos(a ) sin  = 0 ( 3 ) (2) -(1) sin(a ) cos  = 0 ( 4 )  = = cos 0 sin 0 a   == sin 0 cos 0 a 两种情况: I. sin  = 0   = 0 则 cos  = 1 由( 4)式 sin a = 0 = = ( n = 0, 1, 2, ) an a n     2 E 2 2 因  = n E a n an E  = =   = = 2 2 2 2 2 2 2 2 2 2 2          所 以 x an A x A II n   = sin  = sin
<<向上翻页向下翻页>>
©2008-现在 cucdc.com 高等教育资讯网 版权所有