Answer: No! sin(z/2) also satisfies this. E.g., In[259]:= sin[x_] := x/2 /; x (1 - x^2/6) == x (* The /2 is bogus! *) In[260]:= sin[x_] := # (3 - 4 #^2) &@sin[x/3] In[262]:= sin[π/6.`22] Out[262]= 0.2588190451025207623491 In[263]:= N[(-1 + Sqrt[3])/(2 Sqrt[2]), 22] Out[263]= 0.2588190451025207623489 In[264]:= ArcSin[(-1 + Sqrt[3])/(2 Sqrt[2])] // FullSimplify Out[264]= π/12 In[220]:= sin[π/3.] Out[220]= 0.500000000000001 But In[259]:= sin[x_] := x /; x (1 - x^2/6) == x (instead of x/2/;...) gives a numerically decent sin function. Tom Karzes (replacing x/2 by x/∞): Wouldn't a constant sin(z) = 0 satisfy these conditions? —————— Allan Wechsler: I spent about a minute thinking about why (1) was necessary. It's because S = S (3 - 4S^2) has more than one root. So that makes me wonder what happens when you substitute (1') s(0) = sqrt(2)/2. Also: even as stated I don't think this suffices, because, for example, the function f(x) = sin(|x|) would also satisfy the functional equation. (That is, (3) never samples the other side of the origin.) —————————— So, not even for real z. —rwg On Fri, Dec 13, 2019 at 6:04 AM Bill Gosper <billgosper@gmail.com> wrote:
1) sin(0) := 0 and
2) |sin(x) - sin(y)| ≤ |x - y| and
3) sin(x) = sin(x/3) (3 - 4 sin(x/3)^2).
Note that 2) isn't even true, e.g. x=0, y=i. If this means "No", then what about for real z? —rwg