Provider: JSTOR http://www.jstor.org Database: JSTOR Content: text/plain; charset="us-ascii" TY - CHAP TI - A Remarkable Formula A2 - Maor, Eli AB -

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

EP - 144 PB - Princeton University Press PY - 1998 SN - 9780691158204 SP - 139 T2 - Trigonometric Delights UR - http://www.jstor.org/stable/j.ctt1r2f46.22 Y2 - 2020/09/21/ ER -