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I think Artes comment is the best answer to this question. I post this just as a way to play with this identity. f will take a product of Cos and return sum. I have not tried to pattern search for parts of expressions. f[exp_] := With[{v = Cases[{exp}, Cos[x_] :> x, Infinity]}, Total[Cos[v.#] & /@ Tuples[{1, -1}, Length[v]]]/2^Length[v]] You ...


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Attacking the problem numerically gives a first insight into the solution n(a/b) The parameters are defined as m = 1; Dx = 100; ν = 0.25; b = 1; λ1 = Sqrt[(((m^2)*(π^2))/a^2) + Sqrt[(n*(m^2)*(π^2))/(Dx*(a^2))]]; λ2 = Sqrt[(-(((m^2)*(π^2))/a^2)) + Sqrt[(n*(m^2)*(π^2))/(Dx*(a^2))]]; ω1 = ((λ1^2) - ν*((m^2) + (π^2))/a^2); ω2 = ((λ1^2) - ν*((m^2) + ...


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Obviously what a person thinks is simplified and what a given CAS thinks is simplified will not always be the same. This probably happens more often with functions like the trigonometric functions that admit so many identities. Often it does not matter that much, but when it does, a certain amount of personal intervention may be required. One hump to get ...



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