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Re: FunctionExpand problem in version 6

• To: mathgroup at smc.vnet.net
• Subject: [mg82032] Re: [mg82000] FunctionExpand problem in version 6
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 10 Oct 2007 04:22:26 -0400 (EDT)
• References: <200710090935.FAA19705@smc.vnet.net>

Developer`TrigToRadicals can do a lot more than FunctionExpand ever  
could:

Developer`TrigToRadicals[Cos[2*(Pi/13)]]
(1/6)*((1/2)*(-1 + Sqrt[13]) +
       (6 + (1/2)*(-1 + Sqrt[13])*
              ((1/2)*(-1 + I*Sqrt[3]) +
                 (1/4)*(-1 + I*Sqrt[3])^2) +
            (1/2)*(-1 - Sqrt[13])*
              (1 + (1/2)*(-1 + I*Sqrt[3]) +
                 (1/4)*(-1 + I*Sqrt[3])^2))/
         (16 + 2*(-1 + I*Sqrt[3]) + (-1 + I*Sqrt[3])^2 +
              (1/2)*(-1 + Sqrt[13])*(2 + 2*(-1 + I*Sqrt[3]) +
                   (7/4)*(-1 + I*Sqrt[3])^2) +
              (1/2)*(-1 - Sqrt[13])*(10 + 5*(-1 + I*Sqrt[3]) +
                   (7/4)*(-1 + I*Sqrt[3])^2))^(1/3) +
       (16 + 2*(-1 + I*Sqrt[3]) + (-1 + I*Sqrt[3])^2 +
            (1/2)*(-1 + Sqrt[13])*(2 + 2*(-1 + I*Sqrt[3]) +
                 (7/4)*(-1 + I*Sqrt[3])^2) +
            (1/2)*(-1 - Sqrt[13])*(10 + 5*(-1 + I*Sqrt[3]) +
                 (7/4)*(-1 + I*Sqrt[3])^2))^(1/3))


However, if you want to be really impressed try:

Developer`TrigToRadicals[Cos[2*(Pi/11)]]

Andrzej Kozlowski



On 9 Oct 2007, at 18:35, michael.p.croucher at googlemail.com wrote:

> Hi All
>
> In version 5.2 If I do
>
> FunctionExpand[Sin[Pi/13]]
>
> I get a very complicated result involving lots of square roots.  This
> is also mentioned here:
>
> http://mathworld.wolfram.com/TrigonometryAnglesPi13.html
>
> But if I do the same thing in version 6.0 then I just get the result
>
> Sin[Pi/13]
>
> Which is not nearly as impressive.  What has changed with the
> FunctionExpand function in v6 that causes this and how do I get the
> original behaviour?
>
> Best Regards,
> Mike
>
>

• References:
  • FunctionExpand problem in version 6
    • From: michael.p.croucher@googlemail.com