Re: An exact simplification challenge - 11

On Apr 22, 7:18 am, Daniel Lichtblau <d...@xxxxxxxxxxx> wrote:


DL> In[46]:= FullSimplify[ff]
DL> Out[46]= (1/8)*Pi*Csc[Pi/8]

Have you applied here the development version (that is something
quite close to Mathematica 6) ?

Thanks in advance.

Vladimir Bondarenko
Cyber Tester

On Apr 22, 8:38 am, Vladimir Bondarenko <v...@xxxxxxxxxxxxxxx> wrote:





Hello all CAS knights,

Maple 11 simplifies the following expression

Pi*cos(1/7*Pi)*cos(2/7*Pi)*cos(3/7*Pi)/
sin(Pi*cos(1/7*Pi)*cos(2/7*Pi)*cos(3/7*Pi))

into such a mess...

16*Pi*sin(1/56*Pi)*(131072*cos(1/56*Pi)^20-655360
*cos(1/56*Pi)^18+1409024*cos(1/56*Pi)^16-1703936
*cos(1/56*Pi)^14+1272320*cos(1/56*Pi)^12-605696
*cos(1/56*Pi)^10+183040*cos(1/56*Pi)^8-33792
*cos(1/56*Pi)^6+3500*cos(1/56*Pi)^4-172
*cos(1/56*Pi)^2+3)*cos(1/56*Pi)^2/(64
*cos(1/56*Pi)^6-80*cos(1/56*Pi)^4+24
*cos(1/56*Pi)^2-1)

and all the other modern CASs cannot simplify
this expression straightforwardly...

Is there a simplification wizard who can using
a set of a CAS commands yield the 9-bytes-long
exact answer?

Best wishes,

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/CAS Testing

ff = Pi*Cos[1/7*Pi]*Cos[2/7*Pi]*
Cos[3/7*Pi]/Sin[Pi*Cos[1/7*Pi]*Cos[2/7*Pi]*Cos[3/7*Pi]];

In[46]:= FullSimplify[ff]
Out[46]= (1/8)*Pi*Csc[Pi/8]

This assumes, among other things, that I did the format conversion
correctly.

Daniel Lichtblau
Wolfram Research- Hide quoted text -

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