Explaination on solving trigonmetric equations for theta.
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Tony Yu
el 8 de Ag. de 2020
Respondida: Walter Roberson
el 8 de Ag. de 2020
I have these two equation and I am trying to solve for theta 1 and theta 2, can someone explain why there are 2 solution for each theta?
syms th1 th2 px py
eq1 = px == cos(th1+th2)+cos(th1)
eq2 = py == sin(th1+th2)+sin(th1)
sol = solve(eq1,eq2,th1,th2)
syms th1 th2 px py
eq1 = px == cos(th1+th2)+cos(th1)
eq2 = py == sin(th1+th2)+sin(th1)
sol = solve(eq1,eq2,th1,th2)
th1 = simplify(sol.th1)
th2 = simplify(sol.th2)
th1 =
2*atan((2*py + (- px^4 - 2*px^2*py^2 + 4*px^2 - py^4 + 4*py^2)^(1/2))/(px^2 + 2*px + py^2))
2*atan((2*py - (- px^4 - 2*px^2*py^2 + 4*px^2 - py^4 + 4*py^2)^(1/2))/(px^2 + 2*px + py^2))
th2 =
-2*atan((-(px^2 + py^2)*(px^2 + py^2 - 4))^(1/2)/(px^2 + py^2))
2*atan((-(px^2 + py^2)*(px^2 + py^2 - 4))^(1/2)/(px^2 + py^2))
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Image Analyst
el 8 de Ag. de 2020
Rescued two of these from spam quarantine, so there may be two almost identical questions.
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Walter Roberson
el 8 de Ag. de 2020
There are not really two solutions for each theta variable: there are two families of solutions, with the other solutions being separated by 2*Pi
As to "why": well, why not ? You can rewrite the first equation in terms of cos(th1) equalling something, and there are generally two solutions to that, reflecting the two quadrants that cos has the same sign.
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