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Education > Math Recreational > Re: An exact si...
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Re: An exact simplification challenge - 57 (arctangents) - Proud

by Vladimir Bondarenko <vb@[EMAIL PROTECTED] > Apr 26, 2008 at 06:50 AM

Indeed, Derive 6.1 handles this the way you have shown.

Actually, by a mistake I quoted the result produced by
Derive 6.0. 8-(

I was wondering if someone can get this nice result
using Maple or Mathematica?


On Apr 26, 5:07=A0am, cliclic...@[EMAIL PROTECTED]
 wrote:
> Vladimir Bondarenko schrieb:
>
>
>
>
>
> > Hummm... your answer is asolutely correct!
>
> > However, in my Derive 6.1 with standard factory settings
> > (the settings by default) it simplifies to
>
> > -(SQRT(1/480-SQRT(5)/2400)+SQRT(5)/
> > 120-1/40)*ATAN((SQRT(1920-384*SQRT(5))-8*S~
> > QRT(5)-8)/(SQRT(640-128*SQRT(5))+8*SQRT(3)*(SQRT(5)+1)))+(SQRT(1/480-
> > SQRT(5)/~
> > 2400)-SQRT(5)/120+1/40)*ATAN((SQRT(1920-384*SQRT(5))+8*SQRT(5)+8)/
> > (SQRT(640-1~
> > 28*SQRT(5))-8*SQRT(3)*(SQRT(5)+1)))-(SQRT(SQRT(5)/2400+1/480)+SQRT(5)/
> > 120+1/4~
> > 0)*ATAN((SQRT(384*SQRT(5)+1920)-8*SQRT(5)+8)/
> > (SQRT(128*SQRT(5)+640)+8*SQRT(3)~
> > *(SQRT(5)-1)))-(SQRT(SQRT(5)/2400+1/480)-SQRT(5)/
> > 120-1/40)*ATAN((SQRT(384*SQR~
> >
T(5)+1920)+8*SQRT(5)-8)/(SQRT(128*SQRT(5)+640)-8*SQRT(3)*(SQRT(5)-1)))+
> > (SQRT(~
> > SQRT(5)/9600+1/1920)+SQRT(5)/240+1/80)*ATAN((SQRT(10-2*SQRT(5))
> > +SQRT(3)*(SQRT~
> > (5)+1))/(SQRT(30-6*SQRT(5))-SQRT(5)-1))+(SQRT(SQRT(5)/9600+1/1920)-
> > SQRT(5)/24~
> > 0-1/80)*ATAN((SQRT(10-2*SQRT(5))-SQRT(3)*(SQRT(5)+1))/
> > (SQRT(30-6*SQRT(5))+SQR~
> > T(5)+1))-(SQRT(1/1920-SQRT(5)/9600)-SQRT(5)/
> > 240+1/80)*ATAN((SQRT(2*SQRT(5)+10~
> > )+SQRT(3)*(SQRT(5)-1))/(SQRT(6*SQRT(5)+30)-SQRT(5)+1))+(SQRT(1/1920-
> > SQRT(5)/9~
> > 600)+SQRT(5)/240-1/80)*ATAN((SQRT(2*SQRT(5)+10)-SQRT(3)*(SQRT(5)-1))/
> > (SQRT(6*~
> > SQRT(5)+30)+SQRT(5)-1))+pi*(SQRT(SQRT(5)/96+5/192)+7*SQRT(5)/120+1/10)
>
> > 8-(
>
> > Are your settings the standard ones?
> > If not what are your settings?
>
> Your result is what I get on Derive 4.07. My Derive 6.10 settings are
> my own perpetuated settings (I didn't think about them or change them
> for your problem as the simplification worked at once):
>
> =A0 Precision:=3DExact
> =A0 Branch:=3DPrincipal
> =A0 Exponential:=3DAuto
> =A0 Logarithm:=3DAuto
> =A0 Trigonometry:=3DAuto
> =A0 Trigpower:=3DAuto
> =A0 Angle:=3DRadian
>
> By the way, the expression produced by Derive 4.07 is still correct,
> and on Derive 6.10 also immediately simplifies to
>
> =A0 Pi*(Sqrt[6*Sqrt[5] + 15]/30 + Sqrt[5]/15 + 1/10).
>
> Martin.- Hide quoted text -
>
> - Show quoted text -
 




 11 Posts in Topic:
An exact simplification challenge - 57 (arctangents) - Proud
Vladimir Bondarenko <v  2008-04-26 01:36:16 
Re: An exact simplification challenge - 57 (arctangents) - Proud
clicliclic@[EMAIL PROTECT  2008-04-26 02:30:57 
Re: An exact simplification challenge - 57 (arctangents) - Proud
clicliclic@[EMAIL PROTECT  2008-04-26 02:36:19 
Re: An exact simplification challenge - 57 (arctangents) - Proud
Vladimir Bondarenko <v  2008-04-26 02:50:25 
Re: An exact simplification challenge - 57 (arctangents) - Proud
clicliclic@[EMAIL PROTECT  2008-04-26 05:07:57 
Re: An exact simplification challenge - 57 (arctangents) - Proud
Vladimir Bondarenko <v  2008-04-26 06:50:47 
Re: An exact simplification challenge - 57 (arctangents) - Proud
Peter Pein <petsie@[EM  2008-04-26 22:55:08 
Re: An exact simplification challenge - 57 (arctangents) - Proud
Peter Pein <petsie@[EM  2008-04-26 22:58:50 
Re: An exact simplification challenge - 57 (arctangents) - Proud
clicliclic@[EMAIL PROTECT  2008-04-26 11:55:37 
Re: An exact simplification challenge - 57 (arctangents) - Proud
SzH <szhorvat@[EMAIL P  2008-04-26 13:45:19 
Re: An exact simplification challenge - 57 (arctangents) - Proud
SzH <szhorvat@[EMAIL P  2008-04-26 13:55:23 

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tan12V112 Sat Aug 30 8:31:39 CDT 2008.