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1

lap = InverseLaplaceTransform[(12*183.26)/((s^2 + 183.26^2) (55* s^2 + (5.5*s^2 (s*24.2525 + 784.243))/ (5.5*s^2 + 24.2525*s + 784.243) + 8000)), s, t]; {re, im} = (Simplify[#@lap, Assumptions -> t ∈ Reals] // ComplexExpand // Chop) & /@ {Re, Im}; re (* 0.0000328276 E^(-0.675566 t) Cos[10.5332 t] - 0.0000164059 E^(-1.74968 t) ...


1

You can use TrigExpand and then Simplify: TrigExpand[Cos[Pi/18 + 2 x] + Cos[Pi/18 - 2 x]] // Simplify (* 2 Cos[\[Pi]/18] Cos[2 x] *)


2

Try this: A = (Cos[Pi/18 + 2 x] + Cos[Pi/18 - 2 x] /. Pi -> pi // Simplify) /. pi -> \[Pi] (* 2 Cos[\[Pi]/18] Cos[2 x] *) Have fun!


1

The main problem here is that Mathematica immediately converts terms such as Sin[x+π/2] to the more simpler Cos[x]. You have to prevent this from occurring. You could do that with Defer or in V10 with Inactivate: Inactivate[ Cos[p x + 2] Sin[z a + q] + Cos[x]^2/(Sin[a + b] - Cos[2 n]) /. Cos[x_] :> Sin[x + π/2], Sin ] or, perhaps better ...


0

ContourPlot[ Tan[ a Sqrt[al]]/Sqrt[al] + (Tan[ (1 - a ) Sqrt[al]]/Sqrt[al] + 10^-10) == 1, {a, -10, 10}, {al, 0, 10}, ContourStyle -> {Thick, Red}, GridLines -> Automatic, AspectRatio -> Automatic] Plot3D[ Tan[ a Sqrt[al]]/Sqrt[al],{a, -5, 5}, {al, 0, 10}] The extended domain plot is sufficiently regular for tan function with vertical ...



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