how do you substitute all iterations of symbolic sin^2theta cos^2theta with 1 in this symbolic MATLAB matrix to have a more simplified output?

clear
clc

% declaration of given variable
syms A;
syms E;
syms I;
syms L;

% c = cos(theta) and s = sin(theta)
% syms c;
% syms s;
syms theta;

c = cos(theta);
s = sin(theta);

% creation of global stiffness matrix
k = sym(zeros(6,6));
k(1,1) = (A*E)/L;
k(1,4) = -(A*E)/L;
k(2,2) = (12*E*I)/(L^3);
k(2,3) = (6*E*I)/(L^2);
k(2,5) = -(12*E*I)/(L^3);
k(2,6) = (6*E*I)/(L^2);
k(3,2) = (6*E*I)/(L^2);
k(3,3) = (4*E*I)/(L);
k(3,5) = -(6*E*I)/(L^2);
k(3,6) = (2*E*I)/(L);
k(4,1) = -(A*E)/L;
k(4,4) = (A*E)/L;
k(5,2) = -(12*E*I)/(L^3);
k(5,3) = -(6*E*I)/(L^2);
k(5,5) = (12*E*I)/(L^3);
k(5,6) = -(6*E*I)/(L^2);
k(6,2) = (6*E*I)/(L^2);
k(6,3) = (2*E*I)/(L);
k(6,5) = -(6*E*I)/(L^2);
k(6,6) = (4*E*I)/(L);

% Creation of the transformation matrix
T = sym(zeros(6,6));
T(1,1) = c;
T(1,2) = s;
T(2,1) = -s;
T(2,2) = c;
T(3,3) = 1;
T(4,4) = c;
T(4,5) = s;
T(5,4) = -s;
T(5,5) = c;
T(6,6) = 1;

% generating the global stiffness matrix
t = inv(T);

K = mtimes(t,k);
K = mtimes(K,T);

fprintf('The Stiffness matrix is:\n%s\n\n', char(K));
fprintf('Note that c^2 + s^2 = 1');

Output

The Stiffness matrix is:

matrix([[(A*E*cos(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)) + (12*E*I*sin(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)), (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)), -(6*E*I*sin(theta))/(L^2*(cos(theta)^2 + sin(theta)^2)), - (A*E*cos(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*sin(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)), (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)), -(6*E*I*sin(theta))/(L^2*(cos(theta)^2 + sin(theta)^2))], [(A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)), (12*E*I*cos(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)) + (A*E*sin(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)), (6*E*I*cos(theta))/(L^2*(cos(theta)^2 + sin(theta)^2)), (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)), - (12*E*I*cos(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*sin(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)), (6*E*I*cos(theta))/(L^2*(cos(theta)^2 + sin(theta)^2))], [-(6*E*I*sin(theta))/L^2, (6*E*I*cos(theta))/L^2, (4*E*I)/L, (6*E*I*sin(theta))/L^2, -(6*E*I*cos(theta))/L^2, (2*E*I)/L], [- (A*E*cos(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*sin(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)), (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)), (6*E*I*sin(theta))/(L^2*(cos(theta)^2 + sin(theta)^2)), (A*E*cos(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)) + (12*E*I*sin(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)), (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)), (6*E*I*sin(theta))/(L^2*(cos(theta)^2 + sin(theta)^2))], [(12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)), - (12*E*I*cos(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)) - (A*E*sin(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)), -(6*E*I*cos(theta))/(L^2*(cos(theta)^2 + sin(theta)^2)), (A*E*cos(theta)*sin(theta))/(L*(cos(theta)^2 + sin(theta)^2)) - (12*E*I*cos(theta)*sin(theta))/(L^3*(cos(theta)^2 + sin(theta)^2)), (12*E*I*cos(theta)^2)/(L^3*(cos(theta)^2 + sin(theta)^2)) + (A*E*sin(theta)^2)/(L*(cos(theta)^2 + sin(theta)^2)), -(6*E*I*cos(theta))/(L^2*(cos(theta)^2 + sin(theta)^2))], [-(6*E*I*sin(theta))/L^2, (6*E*I*cos(theta))/L^2, (2*E*I)/L, (6*E*I*sin(theta))/L^2, -(6*E*I*cos(theta))/L^2, (4*E*I)/L]])

Note that c^2 + s^2 = 1>>

Answer 1

[James Tursa]

simplify(K)