On 2013-11-22, John Cremona <john.crem...@gmail.com> wrote:
> 1. It is a fact that if a^2+b^2=1 then arcsin(a)+arcsin(b)=pi/2, but
> how can I get this to happen automatically? I have an expression
> which comes out as
>
> arcsin(1/3*sqrt(3)*sqrt(2)) + arcsin(1/3*sqrt(3))
>
> which I would like to get simplified in this way.
I think this can be achieved via simplification rules in Maxima. I don't
know if there are other ways to get the same result elsewhere in Sage
(if so I am interested to hear about it -- I wouldn't mind adopting
some ideas into Maxima). Here's what I have.
matchdeclare (aa, lambda ([e], not atom(e) and op(e) = nounify(asin)));
matchdeclare (xx, lambda ([e], atom(e) or op(e) # nounify(asin)));
tellsimpafter (aa + xx, xx + FOO1 (aa));
FOO1(a):=if op(a) = "+" then FOO2 (a) else a;
FOO2(a):=(a:args(a),
for i thru length(a) do
(for j from i + 1 thru length(a) do
if a[i] # 0
then (FOO3 (a[i], a[j]),
if %% # false then [a[i], a[j]] : [%%, 0])),
apply ("+", a));
FOO3(u,v):=if is (equal (args(u)[1]^2 + args(v)[1]^2, 1)) = true then %pi/2;
If there are any asin terms, call FOO1 on them. If there's more than one,
FOO1 gets an expression like asin(u) + asin(v) + ... (otherwise it's just
asin(u)), so call FOO2 on the "+" expression. FOO2 marches thru the list
of asin terms and tries to combine them two by two (via FOO3).
With that I get the following:
(%i9) asin(sqrt(3)*sqrt(2)/3) + asin(sqrt(3)/3);
(%o9) %pi/2
(%i10) asin(sqrt(3)*sqrt(2)/3) + asin(sqrt(3)/3) - 1;
(%o10) %pi/2-1
(%i11) asin(sqrt(3)*sqrt(2)/3) + asin(sqrt(3)/3) - x + y;
(%o11) y-x+%pi/2
(%i12) asin(sqrt(3)*sqrt(2)/3) + asin(sqrt(3)/3) - asin(x) + u;
(%o12) -asin(x)+u+%pi/2
(%i13) asin(sqrt(3)*sqrt(2)/3) + asin(sqrt(3)/3) + asin(sqrt(1 - u^2)) +
asin (u);
(%o13) %pi
This is just what I got from some half-hearted hacking; I'm sure there
are serious limitations.
Good luck, have fun, & hope this helps.
Robert Dodier
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