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I'm new to Mathematica and I'm trying to solve this set of trig equations (I'm not exactly sure if the syntax looks correct):

Solve[Tan[(delta + beta)/2] == nz*Tan[theta/2] && ny*Tan[(delta - beta)/2] == nx && Cos[(delta + beta)/2]*Cos[gamma/2] == Cos[theta/2], {delta, beta, gamma}, Reals]

The goal is to solve for delta, beta and gamma. I want to substitute different values of nx,ny,nz,theta into these equations and get the solutions. How can I substitute those values in there? Thanks a lot for the help:)

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  • $\begingroup$ Do you have conditions on the variables? If so maybe you can stick these into the system of equations and locate a possible general solution. Additionally, if you have all numerical inputs, then you should use NSolve. $\endgroup$ Commented Mar 15, 2021 at 20:49
  • $\begingroup$ @CA Trevillian Thanks for the comment! Yeah I'm hoping to make all the variables range from [-pi,pi] i.e. principal values. Is there a way I can do that? $\endgroup$ Commented Mar 15, 2021 at 20:53
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    $\begingroup$ So you will want a solution that is valid for this region? If you wish to go numeric route you’ll need to choose a level of discretization. Look up Assumptions If you want to stay analytical and NSolve with Table, if you want to use numerics. $\endgroup$ Commented Mar 15, 2021 at 20:58

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Assuming that you have nz, nx and ny reals, try this:

sol1 = Simplify[ Solve[Tan[(delta + beta)/2] == nz*Tan[theta/2] /. delta + beta -> x, x][[1, 1]], {C[1] == 0}] /. x -> delta + beta (* beta + delta -> 2 ArcTan[nz Tan[theta/2]] *) sol2 = Simplify[ Solve[ny*Tan[(delta - beta)/2] == nx /. delta - beta -> y, y][[1, 1]], {C[1] == 0}] /. y -> delta - beta (* -beta + delta -> 2 ArcTan[nx/ny] *) Simplify[Solve[Cos[(delta + beta)/2]*Cos[gamma/2] == Cos[theta/2], gamma][[1, 1]], {C[1] == 0}] /. sol1 (* gamma -> -2 ArcCos[Cos[theta/2] Sqrt[1 + nz^2 Tan[theta/2]^2]] *)

And then

eq1 = Equal @@ sol1; eq2 = Equal @@ sol2; Solve[{eq1, eq2}, {beta, delta}] (* {{beta -> -ArcTan[nx/ny] + ArcTan[nz Tan[theta/2]], delta -> ArcTan[nx/ny] + ArcTan[nz Tan[theta/2]]}} *)

Have fun!

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  • $\begingroup$ Thanks so much! That's really helpful:) $\endgroup$ Commented Mar 17, 2021 at 2:30

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