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matlab符号计算syms s1 s2 t f ps1=solve('2*3^(1/2)*cos(t)*sin(f)-

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/12 01:30:59
matlab符号计算
syms s1 s2 t f p
s1=solve('2*3^(1/2)*cos(t)*sin(f)-2+2*cos(t)*cos(f)-3^(1/2)*sin(p)*sin(t)*cos(f)+sin(p)*sin(t)*sin(f)=0','f');
s2=simplify(s1)
结果为什么是2行2列,请各位好手帮忙
这是我算的,为什么和你们不一样,很复杂的结果,而且不是两行一列,用逗号隔开
syms s1 s2 t f p
s1=solve('2*3^(1/2)*cos(t)*sin(f)-2+2*cos(t)*cos(f)-3^(1/2)*sin(p)*sin(t)*cos(f)+sin(p)*sin(t)*sin(f)=0','f');
s2=simplify(s1)
s2 =
2*atan((3^(1/2)*tan(p/2)^2 - 3^(1/2)*tan(p/2)^2*tan(t/2)^2 + 2*tan(p/2)*tan(t/2) - 3^(1/2)*tan(t/2)^2 + 3^(1/2) + (3*tan(p/2)^4*tan(t/2)^4 - 10*tan(p/2)^4*tan(t/2)^2 + 3*tan(p/2)^4 + 6*tan(p/2)^2*tan(t/2)^4 - 4*tan(p/2)^2*tan(t/2)^2 + 6*tan(p/2)^2 + 3*tan(t/2)^4 - 10*tan(t/2)^2 + 3)^(1/2))/(2*tan(p/2)^2 - 2*3^(1/2)*tan(t/2)*tan(p/2) + 2))
2*atan((3^(1/2)*tan(p/2)^2 - 3^(1/2)*tan(p/2)^2*tan(t/2)^2 + 2*tan(p/2)*tan(t/2) - 3^(1/2)*tan(t/2)^2 + 3^(1/2) - (3*tan(1/2*p)^4*tan(1/2*t)^4 - 10*tan(1/2*p)^4*tan(1/2*t)^2 + 3*tan(1/2*p)^4 + 6*tan(1/2*p)^2*tan(1/2*t)^4 - 4*tan(1/2*p)^2*tan(1/2*t)^2 + 6*tan(1/2*p)^2 + 3*tan(1/2*t)^4 - 10*tan(1/2*t)^2 + 3)^(1/2))/(2*tan(p/2)^2 - 2*3^(1/2)*tan(t/2)*tan(p/2) + 2))
>> s1
s1 =
2*atan((3^(1/2)*tan(p/2)^2 - 3^(1/2)*tan(p/2)^2*tan(t/2)^2 + 2*tan(p/2)*tan(t/2) - 3^(1/2)*tan(t/2)^2 + 3^(1/2) + (3*tan(p/2)^4*tan(t/2)^4 - 10*tan(p/2)^4*tan(t/2)^2 + 3*tan(p/2)^4 + 6*tan(p/2)^2*tan(t/2)^4 - 4*tan(p/2)^2*tan(t/2)^2 + 6*tan(p/2)^2 + 3*tan(t/2)^4 - 10*tan(t/2)^2 + 3)^(1/2))/(2*tan(p/2)^2 - 2*3^(1/2)*tan(t/2)*tan(p/2) + 2))
2*atan((3^(1/2)*tan(p/2)^2 - 3^(1/2)*tan(p/2)^2*tan(t/2)^2 + 2*tan(p/2)*tan(t/2) - 3^(1/2)*tan(t/2)^2 + 3^(1/2) - (3*tan(1/2*p)^4*tan(1/2*t)^4 - 10*tan(1/2*p)^4*tan(1/2*t)^2 + 3*tan(1/2*p)^4 + 6*tan(1/2*p)^2*tan(1/2*t)^4 - 4*tan(1/2*p)^2*tan(1/2*t)^2 + 6*tan(1/2*p)^2 + 3*tan(1/2*t)^4 - 10*tan(1/2*t)^2 + 3)^(1/2))/(2*tan(p/2)^2 - 2*3^(1/2)*tan(t/2)*tan(p/2) + 2))
>> whos s1
Name Size Bytes Class Attributes
s1 2x1 176 sym
>> whos s2
Name Size Bytes Class Attributes
s2 2x1 176 sym
>> size(s2)
ans =
2 1
是两行一列的啊!