Contents

Morrie's law

${\displaystyle \cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {1}{8}}.}$

It is a special case of the more general identity

${\displaystyle 2^{n}\cdot \prod _{k=0}^{n-1}\cos(2^{k}\alpha )={\frac {\sin(2^{n}\alpha )}{\sin(\alpha )}}}$

with n = 3 and α = 20° and the fact that

${\displaystyle {\frac {\sin(160^{\circ })}{\sin(20^{\circ })}}={\frac {\sin(180^{\circ }-20^{\circ })}{\sin(20^{\circ })}}=1,}$

since

${\displaystyle \sin(180^{\circ }-x)=\sin($x).}

A similar identity for the sine function also holds:

${\displaystyle \sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })={\frac {\sqrt {3}}{8}}.}$

Moreover, dividing the second identity by the first, the following identity is evident:

${\displaystyle \tan(20^{\circ })\cdot \tan(40^{\circ })\cdot \tan(80^{\circ })={\sqrt {3}}=\tan(60^{\circ }).}$

Consider a regular nonagon ${\displaystyle ABCDEFGHI}$ with side length ${\displaystyle 1}$ and let ${\displaystyle M}$ be the midpoint of ${\displaystyle AB}$, ${\displaystyle L}$ the midpoint ${\displaystyle BF}$ and ${\displaystyle J}$ the midpoint of ${\displaystyle BD}$. The inner angles of the nonagon equal ${\displaystyle 140^{\circ }}$ and furthermore ${\displaystyle \gamma =\angle FBM=80^{\circ }}$, ${\displaystyle \beta =\angle DBF=40^{\circ }}$ and ${\displaystyle \alpha =\angle CBD=20^{\circ }}$ (see graphic). Applying the cosinus definition in the right angle triangles ${\displaystyle \triangle BFM}$, ${\displaystyle \triangle BDL}$ and ${\displaystyle \triangle BCJ}$ then yields the proof for Morrie's law:^[2]

${\displaystyle {\begin{aligned}1&=|AB|\\&=2\cdot |MB|\\&=2\cdot |BF|\cdot \cos(\gamma )\\&=2^{2}|BL|\cos(\gamma )\\&=2^{2}\cdot |BD|\cdot \cos(\gamma )\cdot \cos(\beta )\\&=2^{3}\cdot |BJ|\cdot \cos(\gamma )\cdot \cos(\beta )\\&=2^{3}\cdot |BC|\cdot \cos(\gamma )\cdot \cos(\beta )\cdot \cos(\alpha )\\&=2^{3}\cdot 1\cdot \cos(\gamma )\cdot \cos(\beta )\cdot \cos(\alpha )\\&=8\cdot \cos(80^{\circ })\cdot \cos(40^{\circ })\cdot \cos(20^{\circ })\end{aligned}}}$

Recall the double angle formula for the sine function

${\displaystyle \sin(2\alpha )=2\sin(\alpha )\cos(\alpha ).}$

Solve for ${\displaystyle \cos(\alpha )}$

${\displaystyle \cos(\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}.}$

It follows that:

${\displaystyle {\begin{aligned}\cos(2\alpha )&={\frac {\sin(4\alpha )}{2\sin(2\alpha )}}\\[6pt]\cos(4\alpha )&={\frac {\sin(8\alpha )}{2\sin(4\alpha )}}\\&\,\,\,\vdots \\\cos \left(2^{n-1}\alpha \right)&={\frac {\sin \left(2^{n}\alpha \right)}{2\sin \left(2^{n-1}\alpha \right)}}.\end{aligned}}}$

Multiplying all of these expressions together yields:

${\displaystyle \cos(\alpha )\cos(2\alpha )\cos(4\alpha )\cdots \cos \left(2^{n-1}\alpha \right)={\frac {\sin(2\alpha )}{2\sin(\alpha )}}\cdot {\frac {\sin(4\alpha )}{2\sin(2\alpha )}}\cdot {\frac {\sin(8\alpha )}{2\sin(4\alpha )}}\cdots {\frac {\sin \left(2^{n}\alpha \right)}{2\sin \left(2^{n-1}\alpha \right)}}.}$

The intermediate numerators and denominators cancel leaving only the first denominator, a power of 2 and the final numerator. Note that there are n terms in both sides of the expression. Thus,

${\displaystyle \prod _{k=0}^{n-1}\cos \left(2^{k}\alpha \right)={\frac {\sin \left(2^{n}\alpha \right)}{2^{n}\sin(\alpha )}},}$

which is equivalent to the generalization of Morrie's law.

• Glen Van Brummelen: Trigonometry: A Very Short Introduction. Oxford University Press, 2020, ISBN 9780192545466, pp. 79–83
• Ernest C. Anderson: Morrie's Law and Experimental Mathematics. In: Journal of recreational mathematics, 1998

• Weisstein, Eric W. "Morrie's Law". MathWorld.