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Symbolic solution by matching coefficients in trigonometric equation

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Gabriel Droguett on 6 Apr 2021
Commented: Paul on 8 Apr 2021
Hi, I am trying to solve the following symbolic equation:
id*cos(th) - iq*sin(th) == 2*x1*cos(th) - 2*y1*sin(th)
solving for x1 and y1 in terms of id and iq for all angles th. If considering all possible angles the solution should come from equating the coefficients of sines and cosines separetely. So I expected:
x1 = id/2
y1 = iq/2
However the solution I get is:
x1 = (id*cos(th) - iq*sin(th))/(2*cos(th))
y1 = 0
Is there an option that can be added to the solve function to handle this case?
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Paul on 8 Apr 2021
I anticipated solve() returning three parametric solutions: one for th an odd mutiple of pi/2, one for th an even multiple of pi/2, one for th not a multiple of pi/2. However, it only returns one parametric solution:
>> sol = solve(eqn,[x1 y1],'ReturnConditions',true);
>> [sol.x1 sol.y1]
ans =
[ (id*cos(th) - iq*sin(th) + 2*z*sin(th))/(2*cos(th)), z]
>> sol.conditions
ans =
~in(th/pi - 1/2, 'integer')

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Answers (1)

David Goodmanson on 6 Apr 2021
Edited: David Goodmanson on 6 Apr 2021
Hello Gabriel,
here is one way
syms id iq x1 y1 th1 th2
eq1 = id*cos(th1) - iq*sin(th1) == 2*x1*cos(th1) - 2*y1*sin(th1)
eq2 = id*cos(th2) - iq*sin(th2) == 2*x1*cos(th2) - 2*y1*sin(th2)
s = solve(eq1,eq2,x1,y1)
You have to persuade symbolics that the equation obtains for more than just one one angle.
3 CommentsShowHide 2 older comments
Paul on 7 Apr 2021
David's solution yields:
>> [s.x1 s.y1]
ans =
[ id/2, iq/2]
What do you mean by substituting th1 for th2?

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