[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) cos^{2}(x/2)]^{1/2} · [2 sin(x/4) cos^{2}(x/4)]^{1/4} · ... · [2 sin(x/2^{n}) cos^{2}(x/2^{n})]^{1/2n} |

= 2^{1 − 1/2n} [sin(x/2) cos(x/4)]^{1/2} · [sin(x/4) cos(x/8)]^{1/4} · [2 sin(x/2^{n−1}) cos^{2}(x/2^{n})]^{1/2n} · |cos(x/2)| · [sin(x/2^{n})]^{1/2n} |

Note that [cos^{2}(x/2)]^{1/2} = |cos(x/2)|, as we must take the positive square root.

Note also that, for n > 1, sin(x/2^{n}) is positive, and thus [sin(x/2^{n})]^{1/2n} is real.

The terms of the form [sin(x/2^{k−1}) cos(x/2^{k})]^{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) | = 2^{1 − 1/2n} |cos(x/2)| [sin(x/2^{n})]^{1/2n} [P(n)/(sin(x) cos(x/2))^{1/2}]^{2} |

= 2^{1 − 1/2n} [sin(x/2^{n})]^{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)|/[2^{1 − 1/2n} (sin(x/2^{n}))^{1/2n}].

Now, lim n [2^{1 − 1/2n}] = 2, and lim n [sin(x/2^{n})^{1/2n}] = lim n [(x/2^{n})^{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