找到一个方法,跟各位分享下,如果有更好的办法,希望各位大侠指教,下面是我的解决办法:
# N5 B R. {/ \. c8 G! }* Mcos(sym(pi/2)),这样运行后结果就是零了,之前的矩阵结果是:T40 =
8 I6 P7 K$ ~; U) X$ W% X$ p3 ~
5 w! _7 G* \- J0 {9 F[ (2^(1/2)*3^(1/2)*sin(s1))/3, (3^(1/2)*sin(s1)^2)/3 - cos(s1)^2, cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3, a1*cos(s1) - a3*((3^(1/2)*sin(s1)^2)/3 - cos(s1)^2) + d4*(cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3) + (2^(1/2)*3^(1/2)*d3*sin(s1))/3]
2 J, g3 H6 s" s7 t& l[ -(2^(1/2)*3^(1/2)*cos(s1))/3, - cos(s1)*sin(s1) - (3^(1/2)*cos(s1)*sin(s1))/3, sin(s1)^2 - (3^(1/2)*cos(s1)^2)/3, d4*(sin(s1)^2 - (3^(1/2)*cos(s1)^2)/3) + a1*sin(s1) + a3*(cos(s1)*sin(s1) + (3^(1/2)*cos(s1)*sin(s1))/3) - (2^(1/2)*3^(1/2)*d3*cos(s1))/3]9 |$ z Q c8 Y
[ 3^(1/2)/3, -(2^(1/2)*3^(1/2)*sin(s1))/3, -(2^(1/2)*3^(1/2)*cos(s1))/3, (3^(1/2)*d3)/3 - (2^(1/2)*3^(1/2)*d4*cos(s1))/3 + (2^(1/2)*3^(1/2)*a3*sin(s1))/3] Y! H0 [; `5 {2 \% p
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