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MarcoB
MarcoB
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WeierstrasssubstitutionA Weierstrass substitution \[Psi]ψ -> 2 ArcTan[u\[Psi]]ArcTan[uψ] is the way to solve these equations:

eqn = {-a r Cos[\[Phi]]Cos[ϕ] == R Cos[\[Psi]]Cos[ψ], 
 r Sin[\[Phi]]Sin[ϕ] == R Sin[\[Psi]]Sin[ψ]} /. \[Psi]ψ -> 2 ArcTan[u\[Psi]]ArcTan[uψ] //TrigExpand

solu\[Psi]soluψ = Solve[eqn, {R, u\[Psi]uψ}]  ;];
sol\[Psi]solψ = solu\[Psi]soluψ /. u\[Psi]uψ -> Tan[\[Psi]Tan[ψ/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2Cos[ϕ]^2 + Sin[\[Phi]]^2]Sin[ϕ]^2],
Tan[\[Psi]Tan[ψ/2] ->a Cot[\[Phi]]Cot[ϕ] -Csc[\[Phi]]Csc[ϕ] Sqrt[a^2 Cos[\[Phi]]^2Cos[ϕ]^2 +Sin[\[Phi]]^2]+Sin[ϕ]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2Cos[ϕ]^2 + Sin[\[Phi]]^2]Sin[ϕ]^2],
Tan[\[Psi]Tan[ψ/2] ->Csc[\[Phi]]>Csc[ϕ] (a Cos[\[Phi]]Cos[ϕ] + Sqrt[a^2 Cos[\[Phi]]^2Cos[ϕ]^2 +Sin[\[Phi]]^2]+Sin[ϕ]^2])}}*)

Knowing

Tan[x] == (2 Tan[x/2])/(1 - Tan[x/2]^2) // FullSimplify (*True*)

we get

tan\[Psi]tanψ = (2 Tan[ \[Psi]ψ/2])/(1 - Tan[ \[Psi]ψ/2]^2) /. sol\[Psi]solψ// FullSimplify
(* {-(Tan[\[Phi]]Tan[ϕ]/a), -(Tan[\[Phi]]Tan[ϕ]/a)}*)

Hope it helps!

Weierstrasssubstitution \[Psi] -> 2 ArcTan[u\[Psi]] is the way to solve these equations

eqn = {-a r Cos[\[Phi]] == R Cos[\[Psi]], 
 r Sin[\[Phi]] == R Sin[\[Psi]]} /. \[Psi] -> 2 ArcTan[u\[Psi]] //TrigExpand

solu\[Psi] = Solve[eqn, {R, u\[Psi]}]  ;
sol\[Psi] = solu\[Psi] /. u\[Psi] -> Tan[\[Psi]/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->a Cot[\[Phi]] -Csc[\[Phi]] Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->Csc[\[Phi]] (a Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2])}}*)

Knowing

Tan[x] == (2 Tan[x/2])/(1 - Tan[x/2]^2) // FullSimplify (*True*)

we get

tan\[Psi] = (2 Tan[ \[Psi]/2])/(1 - Tan[ \[Psi]/2]^2) /. sol\[Psi]// FullSimplify
(* {-(Tan[\[Phi]]/a), -(Tan[\[Phi]]/a)}*)

Hope it helps!

A Weierstrass substitution ψ -> 2 ArcTan[uψ] is the way to solve these equations:

eqn = {-a r Cos[ϕ] == R Cos[ψ], r Sin[ϕ] == R Sin[ψ]} /. ψ -> 2 ArcTan[uψ] //TrigExpand

soluψ = Solve[eqn, {R, uψ}];
solψ = soluψ /. uψ -> Tan[ψ/2]
(*{{R -> -r Sqrt[a^2 Cos[ϕ]^2 + Sin[ϕ]^2],
Tan[ψ/2] ->a Cot[ϕ] -Csc[ϕ] Sqrt[a^2 Cos[ϕ]^2 +Sin[ϕ]^2]}
, 
{R ->r Sqrt[a^2 Cos[ϕ]^2 + Sin[ϕ]^2],
Tan[ψ/2] ->Csc[ϕ] (a Cos[ϕ] + Sqrt[a^2 Cos[ϕ]^2 +Sin[ϕ]^2])}}*)

Knowing

Tan[x] == (2 Tan[x/2])/(1 - Tan[x/2]^2) // FullSimplify (*True*)

we get

tanψ = (2 Tan[ ψ/2])/(1 - Tan[ ψ/2]^2) /. solψ// FullSimplify
(* {-(Tan[ϕ]/a), -(Tan[ϕ]/a)}*)

Hope it helps!

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Ulrich Neumann
Ulrich Neumann
  • 56.8k
  • 2
  • 26
  • 60

Weierstrasssubstitution \[Psi] -> 2 ArcTan[u\[Psi]] is the way to solve these equations

eqn = {-a r Cos[\[Phi]] == R Cos[\[Psi]], 
r Sin[\[Phi]] == R Sin[\[Psi]]} /. \[Psi] -> 2 ArcTan[u\[Psi]] //TrigExpand

solu\[Psi] = Solve[eqn, {R, u\[Psi]}]  ;
sol\[Psi] = solu\[Psi] /. u\[Psi] -> Tan[\[Psi]/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->a Cot[\[Phi]] -Csc[\[Phi]] Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->Csc[\[Phi]] (a Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2])}}*)

Knowing

Tan[x] == (2 Tan[x/2])/(1 - Tan[x/2]^2) // FullSimplify (*True*)

we get

tan\[Psi] = (2 Tan[ \[Psi]/2])/(1 - Tan[ \[Psi]/2]^2) /. sol\[Psi]// FullSimplify
(* {-(Tan[\[Phi]]/a), -(Tan[\[Phi]]/a)}*)

Hope it helps!

Weierstrasssubstitution \[Psi] -> 2 ArcTan[u\[Psi]] is the way to solve these equations

eqn = {-a r Cos[\[Phi]] == R Cos[\[Psi]], 
r Sin[\[Phi]] == R Sin[\[Psi]]} /. \[Psi] -> 2 ArcTan[u\[Psi]] //TrigExpand

solu\[Psi] = Solve[eqn, {R, u\[Psi]}]  ;
sol\[Psi] = solu\[Psi] /. u\[Psi] -> Tan[\[Psi]/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->a Cot[\[Phi]] -Csc[\[Phi]] Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->Csc[\[Phi]] (a Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2])}}*)

Hope it helps!

Weierstrasssubstitution \[Psi] -> 2 ArcTan[u\[Psi]] is the way to solve these equations

eqn = {-a r Cos[\[Phi]] == R Cos[\[Psi]], 
r Sin[\[Phi]] == R Sin[\[Psi]]} /. \[Psi] -> 2 ArcTan[u\[Psi]] //TrigExpand

solu\[Psi] = Solve[eqn, {R, u\[Psi]}]  ;
sol\[Psi] = solu\[Psi] /. u\[Psi] -> Tan[\[Psi]/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->a Cot[\[Phi]] -Csc[\[Phi]] Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->Csc[\[Phi]] (a Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2])}}*)

Knowing

Tan[x] == (2 Tan[x/2])/(1 - Tan[x/2]^2) // FullSimplify (*True*)

we get

tan\[Psi] = (2 Tan[ \[Psi]/2])/(1 - Tan[ \[Psi]/2]^2) /. sol\[Psi]// FullSimplify
(* {-(Tan[\[Phi]]/a), -(Tan[\[Phi]]/a)}*)

Hope it helps!

Source Link
Ulrich Neumann
Ulrich Neumann
  • 56.8k
  • 2
  • 26
  • 60

Weierstrasssubstitution \[Psi] -> 2 ArcTan[u\[Psi]] is the way to solve these equations

eqn = {-a r Cos[\[Phi]] == R Cos[\[Psi]], 
r Sin[\[Phi]] == R Sin[\[Psi]]} /. \[Psi] -> 2 ArcTan[u\[Psi]] //TrigExpand

solu\[Psi] = Solve[eqn, {R, u\[Psi]}]  ;
sol\[Psi] = solu\[Psi] /. u\[Psi] -> Tan[\[Psi]/2]
(*{{R -> -r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->a Cot[\[Phi]] -Csc[\[Phi]] Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2]}
, 
{R ->r Sqrt[a^2 Cos[\[Phi]]^2 + Sin[\[Phi]]^2],
Tan[\[Psi]/2] ->Csc[\[Phi]] (a Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 +Sin[\[Phi]]^2])}}*)

Hope it helps!