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MathGroup Archive 2000

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Re: ArcCos[]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24793] Re: [mg24730] ArcCos[]
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Thu, 10 Aug 2000 00:33:01 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

on 8/9/00 8:31 AM, Gianluca Gorni at gorni at dimi.uniud.it wrote:

> 
> Hello!
> 
> I have just come across one more example that shows how there is
> room for improving Mathematica's trig functions.
> 
> With Mathematica 4:
> 
> v = Cos[ Pi/8 ] // FunctionExpand   gives   Sqrt[2 + Sqrt[2]]/2
> 
> Still, neither
> 
> ArcCos[v] // FunctionExpand   nor    ArcCos[v] // FullSimplify
> 
> give  Pi/8, as I would expect, but just
> 
> ArcSec[2/Sqrt[2 + Sqrt[2]]]
> 
> %%%%%%%%%%%%%%%
> 
> An unrelated problem: the following instructions consistently crash
> my Mac Mathematica 4 kernel:
> 
> a = Root[-t + 2*#1 + 2*t^2*#1 + #1^3 & , 1];
> b = D[a, t];
> Solve[b == 0, t]
> 
> %%%%%%%%%%%%%%%
> 
> Best regards,
> 
> Gianluca Gorni

I am not an expert on this sort of thing,  but "mathematical common sense"
suggest to me that this may not be easy. Let's consider carefully the
problem of finding a "radical" expression for Cos[Pi/8] (or indeed any
Tr[Pi*m] where Tr is a trigonometric function and m a rational). Although
they do not look like it at first sight these expressions are in fact
algebraic numbers.  The point is that Cos[Pi/8] is just the real part of the
cyclotomic number Cos[2*Pi/16]+I*Sin[2*Pi/16] which is just Root[#^16 - 1 &,
16]  as you can see from:

In[54]:=
(Root[#^16 - 1 &, 16] - Cos[2*Pi/16] - I*Sin[2*Pi/16]) // FullSimplify
Out[54]=
0

So we can write Cos[Pi/8] as 1/2(Root[#^16 - 1 &, 16] + 1/Root[#^16 - 1 &,
16]), which is, of course,  an algebraic number. Using RootReduce we can get

In[58]:=
RootReduce[1/2(Root[#^16 - 1 &, 16] + 1/Root[#^16 - 1 &, 16])]
Out[58]=
             2       4
Root[1 - 8 #1  + 8 #1  & , 4]

and then 

In[60]:=
ToRadicals[%] // FullSimplify
Out[60]=
Sqrt[2 + Sqrt[2]]
-----------------
        2

Of course in this case we could have much easier got this expression
starting from the known values of the trigonometric functions of Pi/4 and
then using half angle formulas. But my point is that there is a procedure
that can be tried in general in such cases.

However, conversely:  given a radical expresion  e.g. v=((-1 + Sqrt[5]))/8 +
(Sqrt[(3*(5 + Sqrt[5]))/2])/4 how do you go about deciding if it is the real
or complex part of a cyclotomic number?   In fact,  v= Cos[Pi/15] but I
doubt that there is any algorithm which has a reasonable chcance of
determining if an arbitrary radical expression (with value between -1 and 1,
say)  is the value of a trigonometric function of some rational multiple of
Pi.
(If I am wrong I would like to hear about this).

 
-- 
Andrzej Kozlowski
Toyama International University, JAPAN

For Mathematica related links and resources try:
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