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Solving trigonometric equation (decoupling)

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HN
HN 2020 年 8 月 19 日
編集済み: HN 2020 年 8 月 19 日
I wanted to solve ϕ as a function of other variables
Is is possible to decouple this equation ?
Even matlab symbolic gave me some kind of log function. I don't understand what that mean actually.
any help is apperciated

採用された回答

Alan Stevens
Alan Stevens 2020 年 8 月 19 日
How about just defining the function:
phi = @(psi, theta) atan( sin(psi).*sin(theta)./(cos(psi) + cos(theta)) );
  4 件のコメント
Alan Stevens
Alan Stevens 2020 年 8 月 19 日
Yes, I did it by hand; it was a fairly obvious solution! I've no idea why Matlab came up with a complicated solution.
If you add cos(theta)sin(phi) to both sides, then factor out the sin(phi) on the left hand side; then divide both sides by cos(phi) and both sides by cos(psi)+cos(theta), you have tan(phi) = ... .
HN
HN 2020 年 8 月 19 日
Thank you and apperciated !

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その他の回答 (1 件)

KSSV
KSSV 2020 年 8 月 19 日
編集済み: KSSV 2020 年 8 月 19 日
Hand calculation is also possible.....
syms phi psi theta
eqn = cos(psi)*sin(phi)+cos(theta)*sin(phi)-cos(phi)*sin(psi)*sin(theta)==0
s = solve(eqn,psi)
  5 件のコメント
KSSV
KSSV 2020 年 8 月 19 日
The same answer syms also gives with futher simplifications.
HN
HN 2020 年 8 月 19 日
編集済み: HN 2020 年 8 月 19 日
Yeah, I just got it as soon as I see Alan Stevens 's answer.
First , I used matlab as I usually do along with many other equations and matlab's complicated result set my mind that this problem has complicatation.
And another funny experiance, a Paper with 400 over citation has mislead me in this problem. Eq.15 (in the paper) is what I've been trying to look at.
Thank you very much both of you !

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