Compute the infinite product

[sin(x) cos(x/2)]^1/2 · [sin(x/2) cos(x/4)]^1/4 · [sin(x/4) cos(x/8)]^1/8 · ... ,

where 0 ≤ x ≤ 2π

(In reply to

computer solution by Charlie)

**I
have tried to solve it analytically and therefore compiled a table similar to
yours,**

**but
did not find a formula as requested. **

**After seeing
your surmise, based on the fact that the infinite
product column and the "sin(x)/2 column" match it is relatively easy to show the correctness
of your conclusion.<o:p></o:p>**

**If
f(x)=(sin(x))/2 then f(x/2)= (sin(x/2))/2 **

**Squaring
the infinite expression we get : **

** f**^{2}(x)= sin(x)*cos(x/2)*(sin(x/2))/2 which can be rewritten
as

**=1/4*( sin(x)*2*cos(x/2)*(sin(x/2))==(sin(x))**^{2}/4

**.So f(x) as
calculated from the LHS equals the
RHS i.e. (sin(x/2))/2 **

**The above as
presented does not constitute deriving an unknown formula, but turning a
conjecture into fact. **

**If anyone
wants to translate it into a "formal" proof, go ahead. **

** **