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1

Try this: This makes everything b-dependent: Map[TrigExpand, expr] /. {Sin[a] -> n*Sin[b], Cos[a] -> Sqrt[1 - n^2*Sin[b]^2]} // Simplify (* -((R Csc[b] (n - n^3 Cos[2 b] + Cos[b] Sqrt[4 - 2 n^2 + 2 n^2 Cos[2 b]] + Sqrt[4 - 2 n^2 + 2 n^2 Cos[2 b]] Cos[3 b] - n^2 Sqrt[4 - 2 n^2 + 2 n^2 Cos[2 b]] Cos[3 b] - 3 n Cos[4 b] + ...


0

{WolframAlpha["Expand Cos[4 x]", {{"Result", 1}, "Output"}], WolframAlpha["Expand Cos[4 x]", {{"Result", 2}, "Output"}]} yields {HoldComplete[Cos[x]^4 - 6 Cos[x]^2 Sin[x]^2 + Sin[x]^4], HoldComplete[2 Sin[[Pi]/4 - 2 x] Sin[[Pi]/4 + 2 x]]} Is this what you're seeking?


13

Disclaimer: This is not a full answer, but perhaps it's a start. From an algebraic stand point this seems like a very hard problem. I attacked it with a more brute force approach. I guess a basis and use LatticeReduce to try to find a Diophantine relation. Note this code only tries to identify roots as the product of integral powers of trig. If it returns ...



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