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0

Well, if it is only for $\frac{1-cos[c x]}{c}$, I tried the following which works perfectly (even without assumptions): FullSimplify[((1 - Cos[c x])/(c x))/ Sinc[c x]] Sinc[c x] which in turn yields: Sinc[c x] Tan[(c x)/2] I am not sure about the bigger context where you want to use that, expanding this technique depends on what you want to do exactly. ...

2

Simplify[(Cos[2*a + b + c]*Tan[a + 2*b + 2*c])/ Sin[(b + c - a)/2]] /. (b -> Pi - a - c) // TrigExpand Tan[a] Simplify[(Sin[(6*Pi)/5 + x])^2 + (Sin[(4*Pi)/5 - x])^2 // Simplify, Pi/5 + x == ArcCos[m^-1]] // Together (2*(-1 + m^2))/m^2

1

You could call a function that fetches all WolframAlpha alternate expression forms: AlternateExpressionForms[expression_]:=Module[{alternateFormData}, alternateFormData={}; alternateFormData=Quiet[Check[TimeConstrained[ReleaseHold[WolframAlpha[ToString[expression,InputForm],{"AlternateForm","Input"}]],60],{}]]; ...

6

Use the TrigFactor[] function, as in TrigFactor[1/2*(Sqrt[3]*Cos[x] - Sin[x])] Output Cos[Pi/6 + x].

7

One idea is to extend the domain with a piecewise function by taking limits at singularities. ExtendFunctionDomain[expr_, vars_] := Module[{domain, antidomain, locassums, lims}, domain = FunctionDomain[expr, vars, Reals] /. { NotElement[f_, S_] :> Not[f == C[1] && Element[C[1], S]] }; antidomain = ReduceToDNF[Reduce[!domain, vars, ...

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Using the value of expr from your question, you can extract the coefficient of the $\cos$ function and the constant part as follows: coscoefficient = Coefficient[expr, Cos[angle n]]; constantpart = expr - coscoefficient * Cos[angle n]; You can check that indeed your expression expr can be reconstructed from those two parts: constantpart + Cos[angle ...

1

One possibility is the following: Suppose you put this into a file called f.m : If[FindFile["FeynCalc"] === \$Failed, (* as explained here: https://github.com/FeynCalc/feyncalc/wiki/Installation *) Import["http://users.ph.tum.de/ga57tah/feyncalc/FeynCalcInstallNightly.m"] ]; Needs["FeynCalc`"]; (* assuming you use Fortran 90 or newer*) ...

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Second update (2015-05-11): Amandeep, you recently left a comment with the following expression: xpr = a1*D11*Cos[n*x] + a0*a1*D11*Cos[n*x] + 1/2*a1*a3*D11*Cos[n*x] + 1/2*a2*a4*D11*Cos[n*x] + a2*a3*a4*Sin[n*x] I believe that you may have left out a multiplication sign on the argument of the last Sin function. Once we add that back in, the approach using ...

0

Have you tried Simplify expr = a1 D11 Cos[n x] + a0 a1 D11 Cos[n x] + 1/2 a1 a3 D11 Cos[n x] + 1/2 a2 a4 D11 Cos[n x]; expr // Simplify 1/2 (a1 (2 + 2 a0 + a3) + a2 a4) D11 Cos[n x] expr2 = a1*D11*Cos[n*x] + a0*a1 D11*Cos[n*x] + 1/2 a1*a3*D11*Cos[n*x] + 1/2*a2*a4 D11*Cos[n*x] + a2*a3*a4*Sin[n*x]; expr2 // Simplify 1/2 (a1 (2 + 2 a0 + ...

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