# Solution to puzzle 157: Trigonometric product

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 2.

For 0 x 2, the quantities sin(x) cos(x/2), sin(x/2) cos(x/4), ... are all non-negative.
Further, all of the quantities are positive except for sin(x) cos(x/2), which equals 0 when x = 0, , or 2.

Let P(n) be the partial product of the first n bracketed terms.  The infinite product is equal to the limit as n tends to infinity of P(n), if the limit exists.
In each bracketed term we apply the identity sin(2t) = 2 sin(t) cos(t):

 P(n) = [2 sin(x/2) cos2(x/2)]1/2 · [2 sin(x/4) cos2(x/4)]1/4 · ... · [2 sin(x/2n) cos2(x/2n)]1/2n = 21 − 1/2n [sin(x/2) cos(x/4)]1/2 · [sin(x/4) cos(x/8)]1/4 · [2 sin(x/2n−1) cos2(x/2n)]1/2n · |cos(x/2)| · [sin(x/2n)]1/2n

Note that [cos2(x/2)]1/2 = |cos(x/2)|, as we must take the positive square root.
Note also that, for n > 1, sin(x/2n) is positive, and thus [sin(x/2n)]1/2n is real.

The terms of the form [sin(x/2k−1) cos(x/2k)]1/2k−1 are the squares of the corresponding terms in the original expression for P(n).

Hence, if sin(x) cos(x/2) is non-zero (i.e., x is not a multiple of ), we can write

 P(n) = 21 − 1/2n |cos(x/2)| [sin(x/2n)]1/2n [P(n)/(sin(x) cos(x/2))1/2]2 = 21 − 1/2n [sin(x/2n)]1/2n P(n)2/|sin(x)|

If P(n) is non-zero, we can divide by P(n) and rearrange

P(n) = |sin(x)|/[21 − 1/2n (sin(x/2n))1/2n].

Now, lim n [21 − 1/2n] = 2, and lim n [sin(x/2n)1/2n] = lim n [(x/2n)1/2n] = 1.

Thus lim n [P(n)] = |sin(x)|/2.
This holds unless P(n) = 0 for some n.  But this can occur only for x = 0, , or 2, in which case the result still holds.

Thus [sin(x) cos(x/2)]1/2 · [sin(x/2) cos(x/4)]1/4 · [sin(x/4) cos(x/8)]1/8 · ... = ½|sin(x)|, for 0 x 2.

Source: M500 Magazine, Issue 211