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Subject: Re: Simplify trigonometric expressions
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Axel Vogt schrieb:
> 
> On 14.08.2015 18:45, Peter Luschny wrote:
> >
> > OK. So what about these?
> >
> > [1] -1/7+x-(2/7)*cos((2/7)*Pi)+(2/7)*cos((3/7)*Pi)+(2/7)*cos((1/7)*Pi)
> >
> > [2] (4/7)*x*cos((1/7)*Pi)-(2/7)*cos((1/7)*Pi)-(4/7)*x*cos((2/7)*Pi)+(2/7)*cos((2/7)*Pi)+(4/7)*x*cos((3/7)*Pi)-(2/7)*cos((3/7)*Pi)+1/7-(2/7)*x+x^2
> >
> > [3] (2/7)*cos((1/7)*Pi)+(6/7)*x^2*cos((1/7)*Pi)-(6/7)*x*cos((1/7)*Pi)-(2/7)*cos((2/7)*Pi)-(6/7)*cos((2/7)*Pi)*x^2+(6/7)*x*cos((2/7)*Pi)+(2/7)*cos((3/7)*Pi)+(6/7)*x^2*cos((3/7)*Pi)-(6/7)*x*cos((3/7)*Pi)-1/7+(3/7)*x-(3/7)*x^2+x^3
> >
> > [4] -(2/7)*cos((1/7)*Pi)-(12/7)*x^2*cos((1/7)*Pi)+(8/7)*x*cos((1/7)*Pi)+(8/7)*x^3*cos((1/7)*Pi)-(8/7)*cos((2/7)*Pi)*x^3+(2/7)*cos((2/7)*Pi)+(12/7)*cos((2/7)*Pi)*x^2-(8/7)*x*cos((2/7)*Pi)-(2/7)*cos((3/7)*Pi)-(12/7)*x^2*cos((3/7)*Pi)+(8/7)*x*cos((3/7)*Pi)+(8/7)*x^3*cos((3/7)*Pi)+1/7-(4/7)*x+(6/7)*x^2-(4/7)*x^3+x^4
> >
> 
> evalf[20](L): fnormal(%): identify(%); # to have a guess
> 
>                                  2   3   4
>                             [x, x , x , x ]
> 
> convert(L, RootOf): # nun aber in echt ...
> simplify(%);
>                                  2   3   4
>                             [x, x , x , x ]
> 
> I think it is also "what is intended by simplify (and should trig
> survive)?" Thus I included sci.math.symbolic for further answers.
> 
> PS: would you mind to post as list
> 
> PPS: well, it may break down at some degree


Derive 6.10 doesn't need any teaching: your quadruple expression 

[-1/7+x-2/7*COS(2/7*pi)+2/7*COS(3/7*pi)+2/7*COS(1/7*pi),4/7*x*CO~
S(1/7*pi)-2/7*COS(1/7*pi)-4/7*x*COS(2/7*pi)+2/7*COS(2/7*pi)+4/7*~
x*COS(3/7*pi)-2/7*COS(3/7*pi)+1/7-2/7*x+x^2,2/7*COS(1/7*pi)+6/7*~
x^2*COS(1/7*pi)-6/7*x*COS(1/7*pi)-2/7*COS(2/7*pi)-6/7*COS(2/7*pi~
)*x^2+6/7*x*COS(2/7*pi)+2/7*COS(3/7*pi)+6/7*x^2*COS(3/7*pi)-6/7*~
x*COS(3/7*pi)-1/7+3/7*x-3/7*x^2+x^3,-2/7*COS(1/7*pi)-12/7*x^2*CO~
S(1/7*pi)+8/7*x*COS(1/7*pi)+8/7*x^3*COS(1/7*pi)-8/7*COS(2/7*pi)*~
x^3+2/7*COS(2/7*pi)+12/7*COS(2/7*pi)*x^2-8/7*x*COS(2/7*pi)-2/7*C~
OS(3/7*pi)-12/7*x^2*COS(3/7*pi)+8/7*x*COS(3/7*pi)+8/7*x^3*COS(3/~
7*pi)+1/7-4/7*x+6/7*x^2-4/7*x^3+x^4]

is automatically simplified to

[x,x^2,x^3,x^4]

within a fraction of a second. These are the reduction steps for the
first expression:

-1/7+x-2/7*COS(2/7*pi)+2/7*COS(3/7*pi)+2/7*COS(1/7*pi)

" COS(n*pi) -> SIN((1/2-n)*pi) "

-1/7+x-2*SIN(3*pi/14)/7+2*COS(3*pi/7)/7+2*COS(pi/7)/7

" COS(n*pi) -> SIN((1/2-n)*pi) "

-1/7+x-2*SIN(3*pi/14)/7+2*SIN(pi/14)/7+2*COS(pi/7)/7

" SIN(3*pi/14)-SIN(pi/14) -> COS(pi/7)-1/2 "

-1/7+x-2*(COS(pi/7)-1/2)/7+2*COS(pi/7)/7

" one final step "

x

I expect the remainder to be handled in the same manner. But I don't see
why Derive should not fail to simplify similar expressions whose
trigonometric arguments involve larger denominators, as the rule to
handle SIN(3*pi/14) - SIN(pi/14) is not generic.

Martin.