Calculate : $\displaystyle p = \sin 1^\circ \times \sin 2^\circ \times \sin 3^\circ \times ..... \times \sin 89^\circ \times \sin 90^\circ
$
Outline solution: For x = 1,2,3,...,90, $\displaystyle \sin(180x)=0$ (where x is measured in degrees). The formula for $\displaystyle \sin(180x)$ is of the form $\displaystyle \sin(180x) = \cos x\sin x(180 - \ldots - 2^{179}\sin^{178}x)$. The part in parentheses is a polynomial of degree 178 in $\displaystyle \sin x$, with all but the first and last terms omitted.
If we put that expression equal to 0 then the factor $\displaystyle \cos x=0$ corresponds to x=90 and x=–90, and the factor $\displaystyle \sin x=0$ corresponds to x=0. If we write s for $\displaystyle \sin x$ then the 178 roots of the polynomial $\displaystyle 180 - \ldots - 2^{179}s^{178}$ are $\displaystyle \sin x$ for $\displaystyle x=\pm1,\pm2,\ldots,\pm89$. 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 $\displaystyle -\sin^21^\circ\sin^22^\circ\cdots\sin^289^\circ = \frac{-180}{2^{179}}$, and hence $\displaystyle \sin1^\circ\sin2^\circ\cdots\sin89^\circ = \frac{\sqrt{90}}{2^{89}}$ (and of course $\displaystyle \sin90^\circ=1$, so that does not contribute to the product).