MHB Is There a Rational Multiple for the Product of Sines from 1 to 90 Degrees?

[anemone]

Here is this week's POTW:

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Show that there is a rational number $k$ such that $\sin 1^\circ \sin 2^\circ \cdots \sin 89^\circ \sin 90^\circ =k\sqrt{10}$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. Opalg

Solution from Olinguito:

Spoiler

Observe that
$$2\sin x^\circ\sin(60-x)^\circ$$
$=\ \cos(2x-60)^\circ-\cos60^\circ$

$=\ \cos(2x-60)^\circ-\dfrac12$

and so
$$4\sin x^\circ\sin(60-x)^\circ\sin(60+x)^\circ$$
$=\ 2\left[\cos(2x-60)^\circ-\dfrac12\right]\sin(60+x)^\circ$

$=\ 2\cos(2x-60)^\circ\sin(60+x)^\circ-\sin(60+x)^\circ$

$=\ \sin(3x)^\circ+\sin(120-x)^\circ-\sin(60+x)^\circ$

$=\ \sin(3x)^\circ+\sin(180-[120-x])^\circ-\sin(60+x)^\circ$

i.e.
$$\boxed{\sin x^\circ\sin(60-x)^\circ\sin(60+x)^\circ\ =\ \frac{\sin(3x)^\circ}4}.$$

So:
$$\sin1^\circ\sin2^\circ\cdots\sin90^\circ$$
$=\ (\sin1^\circ\sin59^\circ\sin61^\circ)(\sin2^\circ\sin58^\circ\sin62^\circ)\cdots(\sin29^\circ\sin31^\circ\sin89^\circ)\sin30^\circ\sin60^\circ\sin90^\circ$

$=\ \dfrac{\sin3^\circ}4\cdot\dfrac{\sin6^\circ}4\cdots\dfrac{\sin87^\circ}4\cdot\dfrac12\cdot\dfrac{\sqrt3}2\cdot1$

$=\ \dfrac{\sqrt3}{4^{30}}\sin3^\circ\sin6^\circ\cdots\sin87^\circ$

$=\ \dfrac{\sqrt3}{4^{30}}(\sin3^\circ\sin57^\circ\sin63^\circ)(\sin6^\circ\sin54^\circ\sin66^\circ)\cdots(\sin27^\circ\sin33^\circ\sin87^\circ)\sin30^\circ\sin60^\circ$

$=\ \dfrac{\sqrt3}{4^{30}}\cdot\dfrac{\sin9^\circ}4\cdot\dfrac{\sin18^\circ}4\cdots\dfrac{\sin81^\circ}4\cdot\dfrac12\cdot\dfrac{\sqrt3}2$

$=\ \dfrac3{4^{40}}\sin9^\circ\sin18^\circ\cdots\sin81^\circ$

$=\ \dfrac3{4^{40}}(\sin9^\circ\sin81^\circ)(\sin18^\circ\sin72^\circ)(\sin27^\circ\sin63^\circ)(\sin36^\circ\sin54^\circ)\sin45^\circ$

$=\ \dfrac3{4^{40}}(\sin9^\circ\cos9^\circ)(\sin18^\circ\cos18^\circ)(\sin27^\circ\cos27^\circ)(\sin36^\circ\cos36^\circ)\cdot\dfrac{\sqrt2}2$

$=\ \dfrac{3\sqrt2}{2^{81}}\cdot\dfrac{\sin18^\circ}2\dfrac{\sin36^\circ}2\cdot\dfrac{\sin54^\circ}2\cdot\dfrac{\sin72^\circ}2$

$=\ \dfrac{3\sqrt2}{2^{85}}\sin18^\circ\cos18^\circ\sin36\cos36^\circ$

$=\ \dfrac{3\sqrt2}{2^{87}}\sin36^\circ\sin72^\circ$

$=\ \dfrac{3\sqrt2}{2^{86}}\sin^236^\circ\cos36^\circ$

$=\ \dfrac{3\sqrt2}{2^{86}}(\cos36^\circ-\cos^336^\circ).$

It remains to work out $\cos36^\circ$. We have
$$\sin36^\circ\ =\ \cos54^\circ$$
$\implies\ 2\sin18^\circ\cos18^\circ\ =\ 4\cos^318^\circ-3\cos18^\circ$

$\implies\ 2\sin18^\circ\ =\ 4\cos^218^\circ-3\ =\ 4-4\sin^218^\circ-3\ =\ 1-4\sin^218^\circ$

$\implies\ 4\sin^218^\circ+2\sin18^\circ-1\ =\ 0$

$\implies\ \sin18^\circ\ (>0) =\ \dfrac{-2+\sqrt{20}}8\ =\ \dfrac{-1+\sqrt5}4.$

So $\cos36^\circ\ =\ 1-2\sin^218^\circ\ =\ \dfrac{1+\sqrt5}4$

$\implies\ \cos^336^\circ\ =\ \dfrac{2+\sqrt5}8$

$\implies \cos36^\circ-\cos^336^\circ\ =\ \dfrac{\sqrt5}8$.

Hence
$$\sin1^\circ\sin2^\circ\cdots\sin90^\circ$$
$=\ \dfrac{3\sqrt2}{2^{86}}\cdot\dfrac{\sqrt5}8\ =\ \boxed{k\sqrt{10}\ \text{where}\ k=\dfrac3{2^{89}}\in\mathbb Q}$.

Alternate solution from Opalg:

Spoiler

This proof uses Chebyshev polynomials. By equation (14) in Multiple-Angle Formulas -- from Wolfram MathWorld, $\sin(180\theta) = -\cos\theta\, U_{179}(\sin \theta)$, where $U_{179}(x)$ is a Chebyshev polynomial of the second kind. And by equation (17) in Chebyshev Polynomial of the Second Kind -- from Wolfram MathWorld, $$U_{179}(x) = \sum_0^{89}(-1)^n{179-n\choose n}(2x)^{179-2n}.$$ It follows that $$\sin(180\theta) = -\cos\theta \sum_0^{89}(-1)^n{179-n\choose n}(2\sin\theta)^{179-2n}. \qquad(*)$$ For every integer $k$, $\sin(180k^\circ) = 0$. If $-89\leqslant k \leqslant 89$ it is also true that $\cos k^\circ \ne0$. It follows from (*) that for $-89\leqslant k \leqslant 89$, $\sin k^\circ$ satisfies the equation $$ \sum_0^{89}(-1)^n{179-n\choose n}(2x)^{179-2n} = 0. \qquad(**)$$ But that is an equation of degree 179, and the 179 solutions $\sin k^\circ \ (-89\leqslant k \leqslant 89)$ are all distinct. So they comprise all the solutions. One of the solutions is $x=0$ (corresponding to $k=0$), and the others are $\pm\sin k^\circ \ (1\leqslant k \leqslant 89)$ (because $\sin(-\theta) = -\sin\theta$). Divide (**) by $x$ to eliminate the solution $x=0$, and it follows that the solutions of $$ \sum_0^{89}(-1)^n{179-n\choose n}2^{179-2n}x^{178-2n} = 0$$ are $\pm\sin k^\circ \ (1\leqslant k \leqslant 89)$. Write the equation as $$ 2^{179}x^{178} - \ldots - 2{90\choose 89} = 0, \\ 2^{178}x^{178} - \ldots - 90 = 0, $$ to see that the product of the roots is $$(\sin 1^\circ \sin 2^\circ \cdots \sin 89^\circ)^2 = \dfrac{90}{2^{178}}.$$ Now take the square root and toss in the fact that $\sin90^\circ = 1$, to get $$\sin 1^\circ \sin 2^\circ \cdots \sin 89^\circ \sin 90^\circ = \frac3{2^{89}}\sqrt{10}.$$