Re: simplify a trig expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg65438] Re: [mg65415] simplify a trig expression*From*: Adam Strzebonski <adams at wolfram.com>*Date*: Sat, 1 Apr 2006 05:38:57 -0500 (EST)*References*: <200603311109.GAA15029@smc.vnet.net> <E2726B55-498F-412D-871F-5B1CEE7D1175@mimuw.edu.pl>*Reply-to*: adams at wolfram.com*Sender*: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote: > This is one of those cases where FullSimplify will not work because it > lacks a suitable transformation function. In this particular case the > transformation function is of the form: > > f[n_*Log[a_]] := Log[a^n] > > Of course this is only valid with various assumptions on n and a, but I > won't bother with this here. Indeed, the two expressions are not equal for all values of x. In[1]:= {2 Log[Cos[x/2] + Sin[x/2]], Log[Sin[x] + 1]} /. x->5Pi/2 Log[2] Out[1]= {2 (I Pi + ------), Log[2]} 2 If we assume that 2 Log[Cos[x/2] + Sin[x/2]] is real we can use the following transformation to simplify it. In[2]:= g[x_]:=Log[TrigReduce[E^x]] In[3]:= g[2 Log[Cos[x/2] + Sin[x/2]]] Out[3]= Log[1 + Sin[x]] > Anyway, observe that: > > > FullSimplify[Integrate[Cos[x]/(Sin[x] + 1), x], > TransformationFunctions -> {Automatic, f}] > > > Log[Sin[x] + 1] > > Note also that Simplify will not work even when you add f. This is because f itself does not make the expression simpler. Simplify does not use TrigReduce, FullSimplify does. Best Regards, Adam Strzebonski Wolfram Research > > I am not sure if there are good reasons for adding a version of f > (taking account of suitable assumptions) to the default transformation > functions of FullSimplify. It may however be a good idea to have > another possible value for the option TransformationFunctions besides > only Automatic and user defined ones. In fact I have suggested in the > past one or two other useful TransformationFunctions; perhaps it might > be a good idea to define more and collect them into a single option > value or maybe several. > > Andrzej Kozlowski > > > > > > > On 31 Mar 2006, at 13:09, Murray Eisenberg wrote: > >> A direct substitution (with paper and pencil) gives that the integral of >> Cos[x]/(Sin[x] + 1) is Log[Sin[x] + 1]. This is valid provided Sin[x] >> is not -1. >> >> Mathematica gives: >> >> Integrate[Cos[x]/(Sin[x] + 1), x] >> 2 Log[Cos[x/2] + Sin[x/2]] >> >> Is there some simple way to coerce the latter Mathematica-supplied >> result into the paper-and-pencil answer? >> >> The closest I could get is: >> >> Log[TrigExpand[Expand[(Cos[x/2] + Sin[x/2])^2]]] /. >> {Sin[x/2] -> Sqrt[(1 - Cos[x])/2], >> Cos[x/2] -> Sqrt[(1 + Cos[x])/2]} >> Log[1 + Sqrt[1 - Cos[x]]*Sqrt[1 + Cos[x]]] >> >> Am I not seeing some easier TrigExpand or TrigReduce method? >> >> -- >> Murray Eisenberg murray at math.umass.edu >> Mathematics & Statistics Dept. >> Lederle Graduate Research Tower phone 413 549-1020 (H) >> University of Massachusetts 413 545-2859 (W) >> 710 North Pleasant Street fax 413 545-1801 >> Amherst, MA 01003-9305 >> >