Impose an specific Trigonometric expansion with TrigExpand

Question

Is there a way to impose in mathematica an specific trigonometric expansion with TrigExpand?

I have various double-angle expressions for cosine and sine in my problem and i need to rewrite them in terms of cosines and sines functions. With this code you can get the same expressions than me.

ClearAll["Global`*"]
SetOptions[$FrontEndSession, PrintingStyleEnvironment -> "Condensed"]
$PrePrint = # /. {Csc[z_] :> 1/Defer@Sin[z], 
     Sec[z_] :> 1/Defer@Cos[z], 
     Cot[z_] :> Defer@Cos[z]/Defer@Sin[z]} &;
Derivative /: 
 MakeBoxes[Derivative[α__][f1_][vars__], TraditionalForm] := 
 Module[{bb, dd, sp}, 
  MakeBoxes[dd, _] ^= 
   If[Length[{α}] == 1, "\[DifferentialD]", "∂"];
  MakeBoxes[sp, _] ^= "\[ThinSpace]";
  bb /: MakeBoxes[bb[x__], _] := RowBox[Map[ToBoxes[#] &, {x}]];
  TemplateBox[{ToBoxes[bb[dd^Plus[α], f1]], 
    ToBoxes[Apply[bb, 
      Riffle[Map[bb[dd, #] &, 
        Select[({vars}^{α}), (# =!= 1 &)]], sp]]], 
    ToBoxes[Derivative[α][f1][vars]]}, "ShortFraction", 
   DisplayFunction :> (FractionBox[#1, #2] &), 
   InterpretationFunction :> (#3 &), Tooltip -> Automatic]]

Subscript[w, 1] = 
  f[Subscript[r, 1], Subscript[r, 2], α, β, γ, 
   Subscript[Θ, 12]];
Set[J, -(D[Subscript[w, 1], {β, 2}] + 
      1/Sin[β]^2 (D[Subscript[w, 1], {α, 2}] + 
         D[Subscript[w, 1], {γ, 2}]) + 
      Cot[β] D[Subscript[w, 1], β] - 
      2 Cot[β]/
       Sin[β] D[Subscript[w, 1], α, γ])] // 
  TraditionalForm ;
Set[Subscript[a, 1], 
   2 D[Subscript[w, 1], {β, 2}] + 
    D[Subscript[w, 1], {γ, 2}] + J] // TraditionalForm ;
Set[Subscript[a, 2], 
   2 Cot[β]/
     Sin[β] D[Subscript[w, 
      1], {α, 1}] - (1 + 2 Cot[β]^2) D[Subscript[w, 
      1], {γ, 1}] - 
    2/Sin[β] D[Subscript[w, 1], β, α] + 
    2 Cot[β] D[Subscript[w, 1], β, γ]] // 
  TraditionalForm ;
Set[Subscript[f, 1], 
    1/(2 Sin[Subscript[Θ, 12]]^2) (-J + 
        Cos[Subscript[Θ, 
          12]] (Sin[2 γ] Subscript[a, 2] - 
           Cos[2 γ] Subscript[a, 1]) + 
        Sin[Subscript[Θ, 
          12]] (Sin[2 γ] Subscript[a, 1] + 
           Cos[2 γ] Subscript[a, 2])) - 
     D[Subscript[w, 1], Subscript[Θ, 12], γ] - 
     1/2 Cot[Subscript[Θ, 12]] D[Subscript[w, 
       1], {γ, 1}] + (1/4 - 1/(
        2 Sin[Subscript[Θ, 12]]^2)) D[Subscript[w, 
       1], {γ, 2}] ] // ExpandAll // TraditionalForm ;
Set[lapla1, -(1/(
      2 Subscript[μ, 1])) (1/Subscript[r, 1]
        D[Subscript[w, 1] Subscript[r, 1], {Subscript[r, 1], 2}] - 
      1/Subscript[r, 
        1]^2 (D[Subscript[w, 
          1], {Subscript[Θ, 12], 2}] + 
         Cot[Subscript[Θ, 12]] D[Subscript[w, 
           1], {Subscript[Θ, 12], 1}]) - Subscript[f, 
      1]/Subscript[r, 1]^2)] // ExpandAll // TraditionalForm

In this specific case, when I use TrigExpand for lapla1, mathematica does not use a unique relation for doing the expansion of each of the terms and is quite difficult to follow it by hand. So, I would like to impose these two specific trigonometric expansions in order to check the results.

$$\sin(2x)= 2\cos x \sin x$$ $$\cos(2x)= \cos^2 x-\sin ^2x$$

Yo can check that those expressions above can be written in several ways:

Answer 1

One way would be like the following. Let us define the function rule as follows:

Clear[rule];
rule[expr_] := ReplaceAll[ expr, {Sin[2 γ_] -> 2*Sin[γ]*Cos[γ], 
                                  Cos[2 γ_] -> Cos[γ]^2 - Sin[γ]^2}];

and map this function on your expression. For the sake of shortness I take here only a small part of your otherwise a too long expression. The effect is, however, the same, I checked. So, let this:

 expr=(Cos[Subscript[Θ, 12]] Sin[2 γ] f[Subscript[r, 1], Subscript[r, 2], α, β, γ, 
  Subscript[Θ, 12]])/(2 Sin[β] Sin[ Subscript[Θ, 12]]^2 
\!\(\*SubsuperscriptBox[\(r\), \(1\), \(2\)]\) Subscript[μ, 1])

be your expression in the StandardForm. Then this

Map[rule, expr]//TraditionalForm

returning this:

(*  (sin(γ) cos(γ) cos(Subscript[Θ, 12]) f(Subscript[r, 1], Subscript[r, ],α,β,γ, 
  Subscript[Θ, 12]) )/(Subscript[μ, 1] Subsuperscript[r, 1, 2] sin(β) 
  sin(Subscript[Θ, 12])^2)  *)

Though it looks awfully here, this:

is what you see on the screen. And suchlike looks each term.