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: <http://www.sstreams.com/Mathematica/>