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You might have some success with specifying a different complexity function for FullSimplify. The following statement e.g. will avoid introducing Sec and Csc: FullSimplify[g, ComplexityFunction->(Count[{#}, (Sec|Csc)[__]]&)] Alas, the output might still not be exactly what you are after: $$\left( \begin{array}{cccc} \sin ^2(\zeta ) & 0 & ...


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You can add ComplexExpand to the transformations FullSimplify will try (since a is real). But FullSimplify seeks to minimize the complexity of the expression, and the starting expression is a local minimum, To get it over the hump, we can penalize Abs. Assuming[0 <= a <= π/2, FullSimplify[Abs[Sin[4 a]] == 2 Abs[Cos[2 a]] Sin[2 a], ...


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As I was recently informed by Wolfram support, Simplify is not guaranteed to fully solve equalities or inequalities. Take this example: FullSimplify[Log[x] > 1, Element[x, Reals]] (* Log[x] > 1 *) Reduce[Log[x] > 1, x, Reals] (* x > E *) We can also get your equation to simplify by using Reduce. Reduce[Abs[Sin[4 a]] == 2 Abs[Cos[2 a]] Sin[2 ...


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I would define it as follows. f[x_] := (E^(3 x + 2))HoldForm[(1/Sin[2*x])] If you type in f[x] , it returns e^(3x+2) (1/Sin[2x]) In order to evaluate it at any time, ReleaseHold[ f[x]] /. x -> Pi/4


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Coincidently, I recently made the following routine for the exact task: Clear[TrigShrink]; TrigShrink[exp_, trgt_, lag_: \[Phi]] := Module[{sign, xExp, xRes, xTrig, cCos, cSin, tan, xlag, clag}, xExp = exp // TrigExpand // Collect[#, {Sin[trgt], Cos[trgt]}] &; xRes = xExp /. {Sin[trgt] -> 0, Cos[trgt] -> 0}; xTrig = xExp - xRes; cCos = ...


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Not sure this is what you need : A = 1/6; B = -Sqrt[3]/24; ω = 8 Sqrt[3]; r = Sqrt[A^2 + B^2] ClearAll[t]; sol = Solve[ A Cos[ω t] + B Sin[ω t] == r Cos[ω t - ϕ], ϕ]; t = 0; ϕ /. sol[[1]] (* ConditionalExpression[-ArcCos[4/Sqrt[19]] - 2 π C[1], C[1] ∈ Integers] *)



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