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 Trigonometric product (Posted on 2013-12-29)
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

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 re: computer solution ---second the motion | Comment 3 of 4 |
(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 :

f2(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.

 Posted by Ady TZIDON on 2013-12-29 18:03:09

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