I am trying to simplify the following expression in sage:
(sqrt(3)/3*cos(x)+1/3*sin(x))the resulting expression should be:
2/3*cos(pi/6-x).simplify_full(), trig_reduce(), or simplify_trig() cannot produce this simplification.
Is sage currently capable of doing this?
With the algorithm by Fu et al, implemented as fu.py in Sympy:
import sympy as sy
f = sqrt(3)/3*cos(x)+1/3*sin(x)
f_reduced = SR(sy.fu(sy.sympify(f))); f_reducedgives 2/3*sin(1/3*pi + x) (which is π2-close to the answer you expected!).
From my own experience, this type of functionality is very useful (1). I'd define this reduction as: "try to get a formula with the smaller number of terms as possible". This is the type of simplification that often looks nicer, or simpler (to a physicist, say), and it is what I often want from a CAS -- though in many real cases a trained human does better and faster. A step forward would be that the program introduces new useful auxiliary variables, I ignore if this exists in some form already.
In the case at hand, there is a single specific reduction which works, sy.FU['TR10i'](expr), while using sy.fu(expr) runs through several reductions and picks the one which gives a "simpler" expression, as explained in the doc.
(1) The use case is solving physics/engineering problems that involve formulas with many trig functions/complex exponentials. As a concrete example, consider a determinant in Fourier space arising from the normal modes of a discrete system.
Thanks, I will have to look more into the fu.py algorithm. That gets me where I need to be.
Well, it is hard to decide which is a simplified version of the other. For many, the first expression is simpler than the second since the involved trigonometric functions are eveluated on x, not pi/6-x. But i agree that something might be missing to express what kind of modification the user expect from Sage.
However, Sage somehow knows that the two expressions are equal:
sage: bool((sqrt(3)/3*cos(x)+1/3*sin(x)) == 2/3*cos(pi/6-x))
TrueSee also:
sage: (sqrt(3)/3*cos(x)+1/3*sin(x)).full_simplify()
1/9*sqrt(3)*(sqrt(3)*sin(x) + 3*cos(x))
sage: ( 2/3*cos(pi/6-x)).full_simplify()
1/9*sqrt(3)*(sqrt(3)*sin(x) + 3*cos(x))Thanks for your response. I understand that sage knows they are equal. However, getting the answer in the single term form helps me see a relationship more easily.
EDIT : A better (systematic) solution is available at this repost question.
Well... this is not as easy as it should, since Sage doesn't (yet) do partial substitutions (e. replace a two-factors product in a multiple-factors expression).
But it can be done, if one can follow a bit of intuition :
sage: f=sqrt(3)/3*cos(x)+1/3*sin(x)
sage: f*3/2
1/2*sqrt(3)*cos(x) + 1/2*sin(x)Aha ! ISTR that cosπ3=12 and that sinπ3=√32. Our f*3/2* would then be :
sage: bool((f*3/2)==sin(pi/3)*cos(x)+cos(pi/3)*sin(x))
TrueAha again. Now, this looks like sin(π3+x). Let's check :
sage: bool(f*3/2==sin(pi/3+x))
TrueAgain a distant memory stirs in my obsolescent neurons :
sage: bool(sin(x)==cos(pi/2-x))
TrueTherefore, since π2−(π3+x)=π6−x, we have :
sage: bool(f*3/2==cos(pi/6-x))
True
sage: bool(f==cos(pi/6-x)*2/3)
TrueQ. E. bloody D...
Here, apart from the already mentioned problem of Sage having difficulties with partial substituition, Sage doesn't have methods for going from/to sums to products of trig/hyperbolic expressions. It can be done with subs using a relevant dictionary (using SR wildcards) such as one of those :
sage: S2PW
[cos($1) + cos($0) == 2*cos(1/2*$1 + 1/2*$0)*cos(-1/2*$1 + 1/2*$0),
-cos($1) + cos($0) == -2*sin(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0),
sin($1) + sin($0) == 2*cos(-1/2*$1 + 1/2*$0)*sin(1/2*$1 + 1/2*$0),
-sin($1) + sin($0) == 2*cos(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0)]
sage: P2SW
[cos($1)*sin($0) == 1/2*sin($1 + $0) + 1/2*sin(-$1 + $0),
cos($0)*sin($1) == 1/2*sin($1 + $0) - 1/2*sin(-$1 + $0),
cos($1)*cos($0) == 1/2*cos($1 + $0) + 1/2*cos(-$1 + $0),
sin($1)*sin($0) == -1/2*cos($1 + $0) + 1/2*cos(-$1 + $0)]
sage: SQSW
[cos($0)^2 == 1/2*cos(2*$0) + 1/2,
sin($0)^2 == -1/2*cos(2*$0) + 1/2,
tan($0)^2 == -(cos(2*$0) - 1)/(cos(2*$0) + 1),
cos($0) + 1 == 2*cos(1/2*$0)^2,
-cos(2*$0) + 1 == 2*sin($0)^2]Those are not very difficult to derive in Sage, but adding them to Sage (possibly as an optional package) would be useful...
Asked: 8 years ago
Seen: 2,638 times
Last updated: Jul 19 '18
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