How to solve the equation to get the result in atan2 form?

조회 수: 4 (최근 30일)
Shiv Karpoor
Shiv Karpoor 2022년 3월 21일
댓글: Shiv Karpoor 2022년 3월 23일
Hello MATLAB Community,
I am trying to solve an equation in MATLAB, but I don't know which function or how to solve it.
syms phi theta psi
ROT = [cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi), -cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi), cos(psi)*sin(theta); ...
cos(psi)*sin(phi), cos(psi)*cos(phi), -sin(psi); ...
-sin(theta)*cos(phi)+sin(psi)*cos(theta)*sin(phi), sin(theta)*sin(phi)+sin(psi)*cos(theta)*cos(phi), cos(psi)*cos(theta)];
eqn = ROT(1,2) == ROT(2,1)
I tried to use functions like solve, simplify, etc. but I am not getting it in the desired form as shown below:
psi = atan2(sin(phi)*sin(theta),cos(phi)+cos(theta))
Can anyone please help me with any suggestions.
Thank you in advance!!
I really appreciate your help.
Kind regards,
Shiv
  댓글 수: 5
Shiv Karpoor
Shiv Karpoor 2022년 3월 22일
Matlab Solution
Mohammed Hamaidi
Mohammed Hamaidi 2022년 3월 23일
Be careful. The equation has solution under some strict assumptions.
Example:
sin(x)+cos(x)==4
, has no solution,
Matlab gives:
>> solve(cos(x)+sin(x)==4,x)
ans =
-log(- 28i^(1/2)/2 + (2 + 2i))*1i
-log(28i^(1/2)/2 + (2 + 2i))*1i
So, you have to give the assumptions/conditions

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답변 (1개)

Torsten
Torsten 2022년 3월 21일
편집: Torsten 2022년 3월 21일
syms phi theta psi
ROT = [cos(theta)*cos(phi)+sin(psi)*sin(theta)*sin(phi), -cos(theta)*sin(phi)+sin(psi)*sin(theta)*cos(phi), cos(psi)*sin(theta); ...
cos(psi)*sin(phi), cos(psi)*cos(phi), -sin(psi); ...
-sin(theta)*cos(phi)+sin(psi)*cos(theta)*sin(phi), sin(theta)*sin(phi)+sin(psi)*cos(theta)*cos(phi), cos(psi)*cos(theta)];
eqn = ROT(1,2) == ROT(2,1);
sol = solve(eqn,psi)
  댓글 수: 11
Torsten
Torsten 2022년 3월 23일
편집: Torsten 2022년 3월 23일
You didn't get David's point.
Your formula to calculate psi as
psi = atan2(sin(phi)*sin(theta),cos(phi)+cos(theta))
seems to be wrong.
But now you know the solution to write psi in terms of phi and theta in symbolic form (the expression from octave). Why don't you simply copy it and use it in your code or whereever you need it ?
Shiv Karpoor
Shiv Karpoor 2022년 3월 23일
Oh, Okay.
Might be wrong I can't tell, because I got it from journal paper.
I will do as suggested,
Thanks Torsten, I appreciate your help!

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