Sinus product

**dhiab** · August 15th 2009, 09:00 AM

Calculate : $p = \sin 1^\circ \times \sin 2^\circ \times \sin 3^\circ \times ..... \times \sin 89^\circ \times \sin 90^\circ <br />$

**Opalg** · August 16th 2009, 11:44 AM

Originally Posted by dhiab

Calculate : $p = \sin 1^\circ \times \sin 2^\circ \times \sin 3^\circ \times ..... \times \sin 89^\circ \times \sin 90^\circ <br />$

Outline solution: For x = 1,2,3,...,90, $\sin(180x)=0$ (where x is measured in degrees). The formula for $\sin(180x)$ is of the form $\sin(180x) = \cos x\sin x(180 - \ldots - 2^{179}\sin^{178}x)$ . The part in parentheses is a polynomial of degree 178 in $\sin x$ , with all but the first and last terms omitted.

If we put that expression equal to 0 then the factor $\cos x=0$ corresponds to x=90 and x=–90, and the factor $\sin x=0$ corresponds to x=0. If we write s for $\sin x$ then the 178 roots of the polynomial $180 - \ldots - 2^{179}s^{178}$ are $\sin x$ for $x=\pm1,\pm2,\ldots,\pm89$ . But the product of the roots of a polynomial of even degree is the constant term divided by the coefficient of the leading term. Therefore $-\sin^21^\circ\sin^22^\circ\cdots\sin^289^\circ = \frac{-180}{2^{179}}$ , and hence $\sin1^\circ\sin2^\circ\cdots\sin89^\circ = \frac{\sqrt{90}}{2^{89}}$ (and of course $\sin90^\circ=1$ , so that does not contribute to the product).

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