Bill Gosper <billgosper@gmail.com> wrote:
[...]
In[116]:= DeleteCases[ Flatten[Outer[ Equal, {Cos[((8*Pi)/41)]*Cos[((10*Pi)/41)] + Sin[((7*Pi)/82)]*Sin[((19*Pi)/82)], Sin[((3*Pi)/82)]*Cos[((7*Pi)/41)] + Sin[Pi/82]*Sin[((9*Pi)/82)], Sin[Pi/82]*Cos[((10*Pi)/41)] + Sin[((9*Pi)/82)]*Cos[((8*Pi)/41)], -Sin[((9*Pi)/82)]* Cos[((9*Pi)/41)] - Sin[Pi/82]*Cos[Pi/41], -Sin[((5*Pi)/82)]*Cos[((9*Pi)/41)] - Cos[Pi/41]*Cos[((2*Pi)/41)]}, Squint[9 - 84 x - 80 x^2 + 1024 x^3 - 256 x^4 - 1024 x^5]]], False]
Out[116]= {Cos[(8 \[Pi])/41] Cos[(10 \[Pi])/41] + Sin[(7 \[Pi])/82] Sin[(19 \[Pi])/82] == -(1/ 20) + (41 (981 - 25 Sqrt[5] - 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^( 1/5)/(20 2^( 2/5)) + (41 (981 - 25 Sqrt[5] + 15 I Sqrt[2 (305 + 109 Sqrt[5])]))^(1/5)/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] - 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^(-((2 I \[Pi])/5)))/( 20 2^(2/5)) + ((41 (981 + 25 Sqrt[5] + 15 I Sqrt[2 (305 - 109 Sqrt[5])]))^(1/5) E^((2 I \[Pi])/5))/( 20 2^(2/5)),[...]}
Note that Out[125]= Sin[a] Sin[b] + Sin[c] Sin[d] == Cos[a/2 - b/2 + c/2 - d/2] Cos[a/2 - b/2 - c/2 + d/2] - Cos[a/2 + b/2 - c/2 - d/2] Cos[a/2 + b/2 + c/2 + d/2] == Sin[a/2 + b/2 + c/2 - d/2] Sin[a/2 + b/2 - c/2 + d/2] - Sin[a/2 - b/2 - c/2 - d/2] Sin[a/2 - b/2 + c/2 + d/2] In[126]:= Simplify[%] Out[126]= True so there are three ways to write each of those five identities.
Apologies. And thanks for trying it!
(Anyone else?)
--rwg
Dear Bill, I was wondering what Mathema[t]ica's take and research on quintics relates to what you're doing? Best, John
Because being able to solve quintics intoxicates me with power! Seriously, they come up all the time, e.g. with D. Cantrell's disk packings, Corey's denesting adventures, recent offlist discussion with Ed Fredkin,... And they're intrinsically interesting. Rich points out that we can write a 1/5th angle formula sin a/5 = quinticsolution(trigs(a)) I've long been amazed and frustrated by the absence of quintic solvers from available CASes. Apparently, Mathematica still lacks GaloisGroup. I guess Mathematica's take is that anybody in his right mind uses Root. --rwg