Hello,

Still on the thematic on simplification of trigonometric expression.

I would like to simplify this equation. Normally, for a mecanical point of view, this equation could be simplified a lot and namely the psi[1](t) and theta[1](t) variables should disappear.

The difference with the former posts is the fact that now each term (for example  2*sin(gamma0(t))*z0(t)*cos(beta0(t))*xb[1]) can regroup 2 terms in factor with the trigonometric part.

eq:=l2[1]^2 = 2*sin(gamma0(t))*z0(t)*cos(beta0(t))*xb[1]-2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*xb[1]+2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*zb[1]-2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*zb[1]+2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*zb[1]-2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*zb[1]-2*cos(gamma0(t))*z0(t)*cos(beta0(t))*zb[1]+2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*zb[1]+2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*xb[1]-2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*xb[1]+2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*xb[1]-2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*xb[1]+2*y0(t)*cos(alpha0(t))*cos(beta0(t))*yb[1]-2*yp[1](t)*cos(alpha0(t))*cos(beta0(t))*yb[1]-2*x0(t)*sin(alpha0(t))*cos(beta0(t))*yb[1]+2*xp[1](t)*sin(alpha0(t))*cos(beta0(t))*yb[1]-2*sin(psi[1](t))*cos(theta[1](t))*l3[1]*xb[1]+2*sin(psi[1](t))*sin(theta[1](t))*l3[1]*zb[1]-2*cos(theta[1](t))*cos(psi[1](t))*l3[1]*zb[1]-2*cos(psi[1](t))*sin(theta[1](t))*l3[1]*xb[1]-2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*z0(t)*cos(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*z0(t)*cos(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+yb[1]^2+xb[1]^2+zb[1]^2+l3[1]^2+z0(t)^2+zp[1](t)^2+y0(t)^2+yp[1](t)^2+x0(t)^2+xp[1](t)^2+2*z0(t)*sin(beta0(t))*yb[1]-2*zp[1](t)*sin(beta0(t))*yb[1]-2*z0(t)*zp[1](t)-2*y0(t)*yp[1](t)-2*x0(t)*xp[1](t)-2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*xb[1]+2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*xb[1]+2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*xb[1]-2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*xb[1]+2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*zb[1]-2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*zb[1]-2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*zb[1]+2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*zb[1]-2*sin(gamma0(t))*z0(t)*cos(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*z0(t)*cos(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]

Do you have some ideas so as to simplify this equation ?

N.B : Former posts on the topic of trigonometric simplification

http://www.mapleprimes.com/questions/209884-Simplification-Of-Trigonometric-Expression-II

http://www.mapleprimes.com/questions/209721-Simplification-Of-Trigonometric-Expressions

I put a worksheet attached in order to facilitate the troubleshooting.

Thanks a lot for your help

trigonometric_simplification.mw

Solution.mw 
 
 

Download Solution.mw

Hi all, if there anyone eho couyld help me with this difficult problem. I couldn't solve the attached nonhomogeneous equation...

But I found one series solution which doesn't satisfy the command: odetest (Solution, ode)

Hi. I'm hacing trouble writing a maple procedure for the question below, can anyone help?

Write a maple procedure which takes as its input the vectoeat u1 and u2 and the eigenvectors lambda1 and lambda2 where u1,u2 are element of R^2 and the lambdas are real numbers.

If u1,U2 is linearly independent then the output is the matrix A an element of R^2x2 with the property that Au1= lambda1u1 and AU2=lambda2u2;

if u1,u2 is linearly dependent then the output is the statement "not an eigenbasis".

I I then have two inputs which I have to do but I'm not sure on how to write the procedure. Any help will be much appreciated.  

Thanks :)

Hi dear Maple masters:

Excuse-me. I have a stupid question to ask: how to extract symbolic coefficient in maple? For example, I would like to get the coefficient before sin(Ωt) and cos(Ωt) in the following equation:

eq := (-Omega^2*a*A[2]-Omega^2*m*B[1]+Omega*A[1]*c[1]+B[1]*k[1])*cos(Omega*t)+(Omega^2*a*B[2]-Omega^2*m*A[1]-Omega*B[1]*c[1]+A[1]*k[1])*sin(Omega*t) = 0;

Thank you in adavance for taking a look, wish you a nice day!

Best regards,

Zihan

Hi dear friends. Recently, I am troubled by a question of how to draw an exictation-response diagram in Maplesoft.

I would like to draw an excitation-response diagram with following equation, Ω is greatrer or equal to 0:

(0.1299e-4*(17.1740*Omega^2-1.570200000*10^6))*Omega^2/(-196.1270800*Omega^4+3.954121290*10^7*Omega^2-1.877174100*10^12);

But, when I would like to use Bode diagram in Maple to draw it, it looks very differently.

Maybe I use a wrong function to draw it.

In the book, the diagram takes the form:

Thanks a lot for taking a look.

Looking everywhere for this.  Hard to find.  Trying to shade under the curve of $2=x^{.5}+y^{.5}$ from x=0 to 1.  Here is what I found and tried:

A:=implicitplot(x^.5+y^.5=2,x=-1..2,y=-1..2,thickness=3,color=blue):
B:=implicitplot(x^.5+y^.5=2,x=0..1,filled=true,color=yellow,view[0..1,-1..1]):
plots[display](A,B, scaling=constrained);

no idea what the view does.  Any help would be much appreciated. 

Nick

I am trying to setup a Dual Quaternion Multiplication Table. I found the table on Wikki. I  need some help here.

Have set

x1  =1   x2 = i   x3  =j   x4   =k   x5 =e   x6 = ei   x7 = ej   x8 =ek

Download Dual_Quaternion_Defining_Algebra.mw

Hello,

I have use a sparsematrixplot to identify the patterns in a matrix.

In order to facilitate the readability of my sparsematrixplot, i would like to change the labels of the rows/ columns axis.

Instead of the numeric graduation, i would like to add :

- for my rows, the labels : [eq1,eq2, ..., ]
- for my columns, the labels : [q1,q2,q3,...,]

Do you have ideas how I can do that ?

Thank you for your help

Dear all,

I developed a program to solve f(x, y) = 0 and g(x, y) = 0, I obtained as results (x=2.726, y=2.126) . running the same program another time it gives (x=2.762, y=1.992). how to explain this?

> fsolve({f(x, y) = 0, g(x, y) = 0}, {x = 0 .. infinity, y= 0 .. infinity});

Thanks in advance.

Hello,

In the post :

http://www.mapleprimes.com/questions/209721-Simplification-Of-Trigonometric-Expressions

you have help me to build a procedure so as to simplify trigonometric expressions of the following form, that is to say where each trigonometric expression is in factor with a term :

x0(t)+sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))*xb[1]-sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))*zb[1]-sin(alpha0(t))*cos(beta0(t))*yb[1]+cos(alpha0(t))*sin(gamma0(t))*zb[1]+cos(alpha0(t))*cos(gamma0(t))*xb[1]-l2[1]*(sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-l3[1]*(sin(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))+cos(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))-sin(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*sin(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-xp[1](t) = 0

From a mechanical point of view, this form of equations comes from the constraint equations obtained with a vectorial closure.

Now I would like to silmplify the constraint equations which come form angular closure.

The equations are in the form :

sin(gamma0(t))*cos(beta0(t)) = -(sin(psi[1](t))*cos(theta[1](t))*cos(gamma[1](t))+sin(psi[1](t))*sin(theta[1](t))*sin(gamma[1](t))-cos(theta[1](t))*cos(psi[1](t))*sin(gamma[1](t))+cos(psi[1](t))*sin(theta[1](t))*cos(gamma[1](t)))*cos(beta[1](t))

i try to treat the right side with the following code :

applyrule([
cos(u::anything)*cos(v::anything)-sin(u::anything)*sin(v::anything)=cos(u+v),
cos(u::anything)*sin (v::anything)+sin(u::anything)*cos(v::anything)=sin(u+v),
sin(u::anything)*sin(v::anything)-cos(u::anything)*cos(v::anything)=-cos(u+v),
-sin(v::anything)*cos(u::anything)-sin(u::anything)*cos(v::anything)=-sin(u+v)], simplify(-(sin(psi[1](t))*cos(theta[1](t))*cos(gamma[1](t))+sin(psi[1](t))*sin(theta[1](t))*sin(gamma[1](t))-cos(theta[1](t))*cos(psi[1](t))*sin(gamma[1](t))+cos(psi[1](t))*sin(theta[1](t))*cos(gamma[1](t)))*cos(beta[1](t)), size))

The result is :

(-sin(theta[1](t)+psi[1](t))*cos(gamma[1](t))+sin(gamma[1](t))*cos(theta[1](t)+psi[1](t)))*cos(beta[1](t))

It seems that the result is not simplified enough. I would like to obtain this expression :

cos(beta[1](t))*sin(gamma[1](t)-theta[1](t)-psi[1](t))

Have you a idea why the simplification is not conducted once more ? Do you have ideas so as to simplify the equation so as to obtain the result mentioned ?

Thanks a lot for your help

Dear All

How can we collect coefficient wrt certain differential ration in an expression?

See for detail:

Download Coefficients_of_Fractions.mw

Regards

Dear All

In student numerical package, for in case of Secant method, how can we force Maple to print procedure and result of every iteration like we do in calculations by hand??

It gives direct output like:

*What if, **want to

Download Secant_Method.mw

Regards

So have to calculate frequency response of RLC-circuit
Long story short - I can't get how to solve something like this
solve(abs(I+x) = 1)
Obv - the answer is 0 or -I, but i get 1-I, -1-I instead

In a post of April 15, 2013 by Kitonum, the procedure named Picture accepts a list of polygon segments, creates a plot of these as a 2D polygon's boundaries and fills the polygon with a color.

The code below attempts to modify Picture to produce a 3D filled polygon in a plane parallel to the xy plane.

When invoked by the code below the procedure, the filling color conforms to the straight line boundaries but overflows the curved, parabolic boundary. How can this be corrected?

Picture:=proc(L, C, N::posint:=100, Boundary::list:=[linestyle=1])

 local i, var, var1, var2,e, e1, e2,e3, P, h ;

 global Q,Border;

 for i to nops(L) do    

#` set P`[i] = list of points for each segment.    

#` for a segment defined as a list of points, P[i] = the segment's definition`

#` for a curve definition, approximate it with a list of [x,y] points of its function evaluated at N even intervals in its

# range`  

  if type(L[i],listlist(algebraic))  then P[i]:=op(L[i]);   else  

  #` for curve def'n, set var = def'n and h= `(variable range)/(2)

  var:=lhs(L[i,2]);  var1:=lhs(rhs(L[i,2]));  var2:= rhs(rhs(L[i,2])); h:=(var2-var1)/(N);

  #` for function def'n, set e=function`

 if type(L[i,1], algebraic) then  e:=L[i,1];

  #` for polar function r=f(t) create N values of the [cos*r,sin*r] i.e. the equivalent [x,y] values for r valued at N even

  # divisions of its range`  

 if nops(L[i])=3 then P[i]:=seq(subs(var=var1+h*i,[e*cos(var), e*sin(var)]), i=0..N);  else

    #` for non-polar function y=f(x) create N values of [x,y] for x values at N even divisions of its range`  

 P[i]:=seq([var1+h*i, subs(var=var1+h*i,e)], i=0..N)  fi;  else

 #` for parametric function [f`(t),g(t)] create N values of [f(t),g(t)] for t values at N even divisions of its range.

     e1:=L[i,1,1];  e2:=L[i,1,2];

#` P`[i]:=seq(subs(var=var1+i*h,[e1, e2]), i=0..N):

 P[i]:=seq([subs(var=var1+i*h,e1), subs(var=var1+i*h,e2),0], i=0..N) fi; fi; od;  #`  MODIFIED FOR 3 D `[f(t), g(t), 0] 

  Q:=[seq(P[i], i=1..nops(L))];

 Border:=plottools[curve]([op(Q), Q[1]],  op(Boundary));

     #` the shaded figure is a polygon whose vertices are Q, whose interior color is C`  

 #` return a list of the polygon and its border`

   [plottools[polygon](Q, C),  Border];

 end proc: 

L := [[[0, 0, 0], [0, 1, 0]], [[x, x^2+1, 0], x = 0 .. 2], [[2, 5, 0], [2, 2, 0]], [[x, x, 0], x = 2 .. 0]]:

plots[display](Picture(L, color = yellow), axes = normal, scaling = constrained)

Hello,

I would like to extract from a equation only the list of variables depending of time.

eq:= x0(t)+sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))*xb[1]-sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))*zb[1]-sin(alpha0(t))*cos(beta0(t))*yb[1]+cos(alpha0(t))*sin(gamma0(t))*zb[1]+cos(alpha0(t))*cos(gamma0(t))*xb[1]-l2[1]*(sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-l3[1]*(sin(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))+cos(theta[1](t))*sin(psi[1](t))*sin(alpha0(t))*sin(gamma0(t))*sin(beta0(t))-cos(theta[1](t))*cos(psi[1](t))*sin(alpha0(t))*sin(beta0(t))*cos(gamma0(t))-sin(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*sin(gamma0(t))+sin(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*sin(psi[1](t))*cos(alpha0(t))*cos(gamma0(t))+cos(theta[1](t))*cos(psi[1](t))*cos(alpha0(t))*sin(gamma0(t)))-xp[1](t) = 0;

I think the indets function should help be.

indets(lhs(eq)-rhs(eq), function(t))

The last argument of the indets is wrong but should enable to explain my need.

Thanks a lot for your help