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 Author Comment/Response Peter Pein 01/29/12 10:56am Hi, there are some ways to achieve that: In[1]:= ser=Series[Cos[2*ArcTan[Sqrt[(1+e)/(1-e)]*Tan[(A+e*Sin[A]+(1/2)*e^2*Sin[2*A])/2]]],{e,0,2}]; 1.) apply some trigonometric conversions in proper order serCos=TrigReduce/@TrigFactor/@TrigExpand/@ser Out[2]= Cos[A]+(-1+Cos[2 A]) e-9/8 (Cos[A]-Cos[3 A]) e^2+O[e]^3 2.) use FullSimplify with an appopriate ComplexityFunction: In[3]:= Factor/@FullSimplify[TrigExpand/@ser,ComplexityFunction->(Count[#,_Sin|_Tan,\[Infinity]]-Count[#,Cos[_.*A],\[Infinity]]&)] Out[3]= Cos[A]+(-1+Cos[2 A]) e-9/8 (Cos[A]-Cos[3 A]) e^2+O[e]^3 If you want to collect by cos(n A)-Terms: In[4]:= Collect[serCos//Normal,_Cos] Out[4]= -e+(1-(9 e^2)/8) Cos[A]+e Cos[2 A]+9/8 e^2 Cos[3 A] or: In[5]:= FourierCosSeries[ser//Normal,A,3] Out[5]= -e+(1-(9 e^2)/8) Cos[A]+e Cos[2 A]+9/8 e^2 Cos[3 A] Peter URL: ,

 Subject (listing for 'Simplifying a result.') Author Date Posted Simplifying a result. Pigkappa 01/27/12 8:39pm Re: Simplifying a result. Peter Pein 01/29/12 10:56am
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