@inbook{10.2307/j.ctt1r2f46.22,
ISBN = {9780691158204},
URL = {http://www.jstor.org/stable/j.ctt1r2f46.22},
abstract = {We are not quite done yet with the function$\left( {\sin x} \right)/x$. Browsing one day through a handbook of mathematical formulas, I came across the following equation:$\frac{{\sin x}}{x} = \cos \frac{x}{4}.\cos \frac{x}{4}.\cos \frac{x}{8}...$.As I had never seen this formula before, I expected the proof to be rather difficult. To my surprise, it turned out to be extremely simple:$\sin x = 2\sin x/2{\rm{ }}\cdot{\rm{ }}\cos x/2$$\begin{array}{l}= 4\sin x/4 \cdot \cos x/4 \cdot \cos x/2 \\ = 8\sin x/8 \cdot \cos x/8 \cdot \cos x/4 \cdot \cos x/2 \\ = ... \\\end{array}$After repeating this process n times, we get$\sin x = {2^n}\sin x/{2^n} \cdot \cos x/{2^n} \cdot ... \cdot \cos x/2.$Let us multiply and divide the first term of this product by x (assuming,of course,that$x \ne 0$and rewrite it as$x \cdot [(\sin x/{2^n})/(x/{2^n})];$we then have$\sin x = x \cdot \left[ {\frac{{\sin x/{2^n}}}{{x/{2^n}}}} \right] \cdot \cos x/2 \cdot \cos x/4 \cdot .......... \cdot \cos x/{2^n}$Note that we have reversed the order of the},
bookauthor = {Eli Maor},
booktitle = {Trigonometric Delights},
pages = {139--144},
publisher = {Princeton University Press},
title = {A Remarkable Formula},
year = {1998}
}