How can I get analytical solution of trigonometric equations?

Question

Mukul 2018-6-20

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/406537-how-can-i-get-analytical-solution-of-trigonometric-equations

评论： Walter Roberson 2018-6-28

    the constants are:
k11 = (16*V1*V1)/(n^3*(pi)^2*(2*pi*f)*L)
k22 = (16*V2*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k33 = (16*V3*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k12 = (8*V1*V2)/(n^3*(pi)^2*(2*pi*f)*L)
k13 = (8*V1*V3)/(n^3*(pi)^2*(2*pi*f)*L)
k23 = (8*V2*V3)/(n^3*(pi)^2*(2*pi*f)*L)
 The equations are:
P1  = (k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))
P2  = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*sin(x(4)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(5)-x(4))*pi/180))
P3  = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*sin(x(5)*pi/180))+(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*sin((x(4)-x(5))*pi/180))
Q1 = (k11.*cos(x(1)*pi/360).*cos(x(1)*pi/360))-(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))-(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))
Q2 = -(k12.*cos(x(1)*pi/360).*cos(x(2)*pi/360).*cos(x(4)*pi/180))+(k22.*cos(x(2)*pi/360).*cos(x(2)*pi/360))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))
Q3 = -(k13.*cos(x(1)*pi/360).*cos(x(3)*pi/360).*cos(x(5)*pi/180))-(k23.*cos(x(2)*pi/360).*cos(x(3)*pi/360).*cos((x(5)-x(4))*pi/180))+(k33.*cos(x(3)*pi/360).*cos(x(3)*pi/360))

How can I solve for the angles x(1), x(2), x(3), x(4) and x(5)? Can anyone please help me to solve these equations?

8 个评论
显示 6更早的评论隐藏 6更早的评论

Mukul 2018-6-22

在 MATLAB Online 中打开

Hi John. Thanks for your comments. Yes I have tried several times using many functions such as fmincon, fsolve and isqnonlin. It has been confirmed that the solutions I am getting are same for each function. For example, for

L = 22.5e-6; V1 = V2 = V3 = 40, n=1 and f=20e3;
and 
P1 = 400; P2 = -200; P3 = -200;  Q1 = 193; Q2 = 96.86; Q3 = -96.86;

fmincon gives (angles are in degrees)

x(1)=-0.0000 x(2)=28.2228 x(3)=71.3840 x(4)=27.8753 x(5)=29.6215

Isqnonlin gives

x(1)=0.0000 x(2)=28.2228 x(3)=71.3840 x(4)=27.8753 x(5)=29.6215

fsolve gives

x(1)=0.0182 x(2)=28.2228 x(3)=71.3840 x(4)=27.8753 x(5)=29.6215

so these are the numerical solutions and I think that these are correct as confirmed by three different functions but I want the exact analytical solution. I tried with the solve function but end up with

Here is my code:

if true
syms x1 x2 x3 x4 x5 V1 V2 V3 f L k11 k22 k33 k12 k13 k23 P1 P2 P3 Q1 Q2 Q3
k11 = (16*V1*V1)/((pi)^2*(2*pi*f)*L)
k22 = (16*V2*V2)/((pi)^2*(2*pi*f)*L)
k33 = (16*V3*V3)/((pi)^2*(2*pi*f)*L)
k12 = (8*V1*V2)/((pi)^2*(2*pi*f)*L)
k13 = (8*V1*V3)/((pi)^2*(2*pi*f)*L)
k23 = (8*V2*V3)/((pi)^2*(2*pi*f)*L)
eqn_P1  = (k12.*cos(x1*pi/360).*cos(x2*pi/360).*sin(x4*pi/180))+(k13.*cos(x1*pi/360).*cos(x3*pi/360).*sin(x5*pi/180)) == P1
eqn_P2  = -(k12.*cos(x1*pi/360).*cos(x2*pi/360).*sin(x4*pi/180))+(k23.*cos(x2*pi/360).*cos(x3*pi/360).*sin((x5-x4)*pi/180)) == P2
eqn_P3  = -(k13.*cos(x1*pi/360).*cos(x3*pi/360).*sin(x5*pi/180))+(k23.*cos(x2*pi/360).*cos(x3*pi/360).*sin((x4-x5)*pi/180)) == P3
eqn_Q1 = (k11.*cos(x1*pi/360).*cos(x1*pi/360))-(k12.*cos(x1*pi/360).*cos(x2*pi/360).*cos(x4*pi/180))-(k13.*cos(x1*pi/360).*cos(x3*pi/360).*cos(x5*pi/180)) == Q1
eqn_Q2 = -(k12.*cos(x1*pi/360).*cos(x2*pi/360).*cos(x4*pi/180))+(k22.*cos(x2*pi/360).*cos(x2*pi/360))-(k23.*cos(x2*pi/360).*cos(x3*pi/360).*cos((x5-x4)*pi/180)) == Q2
eqn_Q3 = -(k13.*cos(x1*pi/360).*cos(x3*pi/360).*cos(x5*pi/180))-(k23.*cos(x2*pi/360).*cos(x3*pi/360).*cos((x5-x4)*pi/180))+(k33.*cos(x3*pi/360).*cos(x3*pi/360)) == Q3
S = solve(eqn_P1, eqn_P2, eqn_P3, eqn_Q1, eqn_Q2, eqn_Q3, -pi/2<x1, x1<pi/2, -pi/2<x2, x2<pi/2, -pi/2<x3, x3<pi/2, -pi/2<x4, x4<pi/2, -pi/2<x5, x5<pi/2, 'ReturnConditions', true);
end

and the results comes up as:

    Warning: Unable to find explicit solution. For options, see help. 
> In solve (line 317)
  In test_5 (line 22)

Walter Roberson 2018-6-26

There is not necessarily any error in your code. The system is just difficult to solve.

My work so far shows that for each x1 there are two x2, and that for each x2 there are four x3. Computation is slow, so I have not gotten further than that quite yet.

Mukul 2018-6-26

Ok Walter, please let me inform what you get when the computation would finish

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2018-6-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/406537-how-can-i-get-analytical-solution-of-trigonometric-equations#answer_325755

在 MATLAB Online 中打开

Analytic solution:

x(1) = 180 + 360*Z1
x(2) = 180 + 360*Z2;
x(3) = 180 + 360*Z3;
x(4) and x(5) arbitrary (that is, the above 3 together solve all 5 equations)

Here, Z1, Z2, and Z3 represent arbitrary integers

6 个评论
显示 4更早的评论隐藏 4更早的评论

Mukul 2018-6-22

在 MATLAB Online 中打开

Dear Walter, Thank you so much for your effort to get this.

Can you please tell me how you get this and what is the expression of Z1, Z2 and Z3? I believe these should be in terms of other variables in the equations.

If you follow this post, I have solved these equations for angles using three different functions for specific values of variables. But I want exact analytical solution/expression of the angles that you have given.

I tried with the solve function and I can't get to your solution and here is my code:

.
syms x1 x2 x3 x4 x5 V1 V2 V3 f L k11 k22 k33 k12 k13 k23 P1 P2 P3 Q1 Q2 Q3
k11 = (16*V1*V1)/((pi)^2*(2*pi*f)*L)
k22 = (16*V2*V2)/((pi)^2*(2*pi*f)*L)
k33 = (16*V3*V3)/((pi)^2*(2*pi*f)*L)
k12 = (8*V1*V2)/((pi)^2*(2*pi*f)*L)
k13 = (8*V1*V3)/((pi)^2*(2*pi*f)*L)
k23 = (8*V2*V3)/((pi)^2*(2*pi*f)*L)
eqn_P1  = (k12.*cos(x1*pi/360).*cos(x2*pi/360).*sin(x4*pi/180))+(k13.*cos(x1*pi/360).*cos(x3*pi/360).*sin(x5*pi/180)) == P1
eqn_P2  = -(k12.*cos(x1*pi/360).*cos(x2*pi/360).*sin(x4*pi/180))+(k23.*cos(x2*pi/360).*cos(x3*pi/360).*sin((x5-x4)*pi/180)) == P2
eqn_P3  = -(k13.*cos(x1*pi/360).*cos(x3*pi/360).*sin(x5*pi/180))+(k23.*cos(x2*pi/360).*cos(x3*pi/360).*sin((x4-x5)*pi/180)) == P3
eqn_Q1 = (k11.*cos(x1*pi/360).*cos(x1*pi/360))-(k12.*cos(x1*pi/360).*cos(x2*pi/360).*cos(x4*pi/180))-(k13.*cos(x1*pi/360).*cos(x3*pi/360).*cos(x5*pi/180)) == Q1
eqn_Q2 = -(k12.*cos(x1*pi/360).*cos(x2*pi/360).*cos(x4*pi/180))+(k22.*cos(x2*pi/360).*cos(x2*pi/360))-(k23.*cos(x2*pi/360).*cos(x3*pi/360).*cos((x5-x4)*pi/180)) == Q2
eqn_Q3 = -(k13.*cos(x1*pi/360).*cos(x3*pi/360).*cos(x5*pi/180))-(k23.*cos(x2*pi/360).*cos(x3*pi/360).*cos((x5-x4)*pi/180))+(k33.*cos(x3*pi/360).*cos(x3*pi/360)) == Q3
S = solve(eqn_P1, eqn_P2, eqn_P3, eqn_Q1, eqn_Q2, eqn_Q3, -pi/2<x1, x1<pi/2, -pi/2<x2, x2<pi/2, -pi/2<x3, x3<pi/2, -pi/2<x4, x4<pi/2, -pi/2<x5, x5<pi/2, 'ReturnConditions', true);
end

and the results I am getting is something like:

    Warning: Unable to find explicit solution. For options, see help. 
> In solve (line 317)
  In test_5 (line 22)

could you please explain it bit more?

Walter Roberson 2018-6-26

You can solve the first equation set, eqn1, for the first variable x1. Substitute that into the remaining equations, getting eqn2, one shorter than before. Solve the first equation of eqn2 for the second variable x2, getting two solutions. Substitute the first solution of x2 into the remaining variables for eqn2, getting eqn3 that is one shorter still.

Now, solving the first equation of eqn3 (which was the third equation of the original set) with respect to x3, x4, or x5 pretty much fails, so we do not get anywhere with that.

Solve the second equation of eqn3 (which was the fourth equation of the original set) with respect to x3. Substitute that into eqn3([1,3, 4]), getting eqn4 with three equations.

At this point, eqn4(1) is the same equation that we failed to solve for x3, x4, or x5, and we have made it even more complicated by the substitution of x3, so we will not be able to get anywhere easily by solving it.

We can now try to solve eqn4(2) [which used to be the fifth equation] for x4. But it will take a long time. I do not know yet if it can work; my system has been computing it for hours so far.

If eventually we do get a solution, take the first value, substitute it into eqn4(3) [which used to be the sixth equation] and try to solve for x5. This will leave eqn4(1) to substitute x5 into, and as all variables would now have been substituted in, you would check to see if the equation is consistent.

If the equation is not consistent, then back up to the last place you had a choice about which of multiple solutions you substituted, and try the next possibility and move forward from there.

If all goes well, eventually you get a consistent solution.

It is not likely that you will get a consistent solution for six trig equations in five variables.

If you are trying to solve for analytical equations of motion of a multi-jointed arm, then be advised that most such systems people try to solve are analytically inconsistent or else have multiple fairly different solutions.

Mukul 2018-6-28

Dear Walter,

I am waiting to solutions you have got so far.

Would you prefer solving these equations using solve function or any other way you suggest for me?

Walter Roberson 2018-6-28

Solving for x(4) and x(5) both failed at the place I was indicating was taking a long time. I did not go back to try substituting in the other choices.

请先登录，再进行评论。

How can I get analytical solution of trigonometric equations?

8 个评论
显示 6更早的评论隐藏 6更早的评论

采纳的回答

6 个评论
显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

How can I get analytical solution of trigonometric equations?

8 个评论 显示 6更早的评论隐藏 6更早的评论

采纳的回答

6 个评论 显示 4更早的评论隐藏 4更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

8 个评论
显示 6更早的评论隐藏 6更早的评论

6 个评论
显示 4更早的评论隐藏 4更早的评论