x=L1*sin(v1)+L2*sin(v1+v2) y=L1*cos(v1)+L2*cos(v1+v2) 请问如何用matlab求出v1和v2的表达式?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/26 03:52:01

x=L1*sin(v1)+L2*sin(v1+v2) y=L1*cos(v1)+L2*cos(v1+v2) 请问如何用matlab求出v1和v2的表达式?
x=L1*sin(v1)+L2*sin(v1+v2) y=L1*cos(v1)+L2*cos(v1+v2) 请问如何用matlab求出v1和v2的表达式?

x=L1*sin(v1)+L2*sin(v1+v2) y=L1*cos(v1)+L2*cos(v1+v2) 请问如何用matlab求出v1和v2的表达式?
>> clear
>> syms x y L1 L2 V1 V2
>> [V1,V2]=solve('x=L1*sin(V1)+L2*sin(V1+V2)','y=L1*cos(V1)+L2*cos(V1+V2)','V1','V2')
V1 =
pi
z7
2*atan((2*L1*x - L1^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) - L2^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) + x^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) + y^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) + 2*L1*L2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))/(L1^2 + 2*L1*y - L2^2 + x^2 + y^2))
2*atan((2*L1*x + L1^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) + L2^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) - x^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) - y^2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2) - 2*L1*L2*((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))/(L1^2 + 2*L1*y - L2^2 + x^2 + y^2))
2*atan((L1*i + L2*i)/(L1 + L2))
V2 =
z
pi
-2*atan(((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))
2*atan(((L1^2 + 2*L1*L2 + L2^2 - x^2 - y^2)/(- L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))
2*atan((L1*i + L2*i)/(L1 - L2))

x=L1*sin(v1)+L2*sin(v1+v2) y=L1*cos(v1)+L2*cos(v1+v2) 请问如何用matlab求出v1和v2的表达式? matlab解二元一次方程求助l1=36.5;l2=138.5;l3=100;l4=200;syms x ysolve('l1*cos(x)+l2*cos(y)=l4+l3*cos(-2)','l1*sin(x)+l2*sin(y)=l3sin(-2)','x','y')ans = x:[2x1 sym]y:[2x1 sym] 已知直线L1:Y=X×SINα和直线L2:Y=2X+C,则直线L1与L2()A.通过平移可以重合B.不可能垂直C.可能与X轴围成等腰直角三角形D.通过绕L1上的某一点旋转可以重合 用matlab求二元二次方程组的解,程序如下,十万火急!syms x y L1 L2 L3 L4 a1 a2 L5 a3 a4eq1=solve('(x-(L3*cos(a3)-L1*cos(a1)))^2+(y-(L3*sin(a3)+L1*sin(a1)))^2=L3^2')eq2=solve('(x-(L5+L2*cos(a2)))^2+(y-(L1*sin(a1)))^2=L5^2')[x,y]=solve(eq 已知α属于(π,3π/2),则两条直线l1:x+sinα*y+3=0和l2:(根号下1-cos^2α)*x+y+6=0的位置关系是 MATLAB Index exceeds matrix dimensions问题for i=1:90;lac=lab*cos(th1)+sqrt(lbc^2-(lab*sin(th1))^2);l3=sqrt((l1-lac)^2+(l2)^2);x=(l1-lac)/l2;y=(l3^2+led^2-lcd^2)/(2*l3*led);z=l1-lab*cos(th1)-sqrt(lbc^2-(lab*sin(th1)^2));a=sqrt(z^2-l2^2);z1=lab*sin(t 急!数学证明题(初三)(三角函数)任意凸四边形的两条对角线长分别为L1、L2,两条对角线所夹锐角为α,求证:四边形的面积S=1/2*L1*L2*sinα 已知空间不共面的四个点,与此四个点距离都相等的平面有( )个?已知直线L1:xsinα+2y=1,L2:2x+ysinα=2,若L1到L2的角为60°,则sinα的值为? 解含三角函数的方程组的解法,可以用matlab方程组如下:第一步,L1*sin(θ1)+L2*sin(θ2)-L3*sin(θ3)-0.52=0L1*cos(θ1)+L2*cos(θ2)-0.32-L3*cos(θ3)=0已知L1=0.12,L2=0.5,L3=0.27,求θ2和θ3(用含θ1的式子表示)第二步,θ MATLAB程序运行不出结果v=14;g=10;L1=0.6;L2=0.6;R = @(x) -(v*sin(x(2))*(g*v*cos(x(2))+sqrt(g^2*v^2^cos(x(2))^2+2*L1*g*sin(x(1))+2*L2*g*sin(x(2)))));lb = [20; 20]; % x y的下限ub = [45; 90] ; % x y的上限x0 = [30;60]; % Starting guess at the 哪位高人能利用MATLAB求解如下三角函数方程组,最好有源程序?方程组如下:L1*sin(θ1)-L3*sin(θ3)+A-L5*sin(θ5)=0L1*cos(θ1)-L3*cos(θ3)-B-L5*cos(θ5)=0L1*sin(θ1)+L2*sin(θ2)+A-L4*sin(θ4)-L5*sin(θ5)=0L1*cos(θ1)+L2*cos(θ2) sin(sin(sin(sin(x)=cos(cos(cos(cos(x),X等于多少? 已知直线l1:2x+y-10=0,l2垂直于l1,且l2过点(-10,0),l1与l2的交点坐标是 已知直线L1:X+MY+6=0,L2:(M-2)X+3Y+2M=0,求M得值,使,(1).L1与L2相交(2).L1垂直L2(3)L1与L2重合 设直线l1:(m-2)x+3y+2m=0,l2:x+my+6=0,当____时,l1与l2相交;当_____时,l1‖l2;当_____时,l1⊥l2 傅里叶级数作图f(x)=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si matlab 出错Error: Unbalanced or unexpected parenthesis or bracket.M=sqrt(L1*L2*P)*exp{j*w*(t-T2)}*sin{B1*cos[r*(t-T2)]}*sin{B2*cos[r*(t-T1)]};? M=sqrt(L1*L2*P)*exp{j*w*(t-T2)}*sin{B1*cos[r*(t-T2)]}*sin{B2*cos[r*(t-T1)]}; 用matlab解方程组方程一:L1+L2*COS(TH1)=L3*COS(TH2)-L4*COS(TH3);方程二:L2*SIN(TH1)+L3*SIN()TH2-L4*SIN(TH3),其中把L1,L2,L3,L4,TH1当做已知量求TH2和TH3