Morrie's Law

Morrie's law (dt. Act Morrie ) is a special trigonometric identity . Its name goes back to the physicist Richard Feynman , who called it that because he was shown it by a boy named Morrie Jacobs during his childhood.

Identity and generalization

Morrie's law reads:

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

It is a special case of the following more general trigonometric identity:

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

For and then one obtains Morrie's law if one observes that ${\displaystyle n=3}$ ${\displaystyle \alpha =20^{\circ }}$

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

applies because of

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

More similar identities

There is a similar identity for the sine function:

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

A corresponding identity for the tangent function can be obtained by dividing the two previous identities into one another:

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

proofs

Geometric proof

Regular nine-corner with as the center of its circumference . The following applies to the angles: ${\displaystyle ABCDEFGHI}$ ${\displaystyle O}$
${\displaystyle {\begin{aligned}40^{\circ }&={\frac {360^{\circ }}{9}}\\70^{\circ }&={\frac {180^{\circ }-40^{\circ }}{2}}\\\alpha &=180^{\circ }-90^{\circ }-70^{\circ }=20^{\circ }\\\beta &=180^{\circ }-90^{\circ }-(70^{\circ }-\alpha )=40^{\circ }\\\gamma &=140^{\circ }-\beta -\alpha =80^{\circ }\end{aligned}}}$

One considers a regular hexagon with side length , furthermore denotes the center of , the center of and the center of . The interior angles of the hexagon are , in addition , and (see drawing). Now one applies the definition of the cosine in the right triangle to the triangles , and one after the other , and thus a proof of identity is obtained: ${\displaystyle ABCDEFGHI}$ ${\displaystyle 1}$ ${\displaystyle M}$ ${\displaystyle AB}$ ${\displaystyle L}$ ${\displaystyle BF}$ ${\displaystyle J}$ ${\displaystyle BD}$ ${\displaystyle 140^{\circ }}$ ${\displaystyle \gamma =\angle FBM=80^{\circ }}$ ${\displaystyle \beta =\angle DBF=40^{\circ }}$ ${\displaystyle \alpha =\angle CBD=20^{\circ }}$ ${\displaystyle \triangle BFM}$ ${\displaystyle \triangle BDL}$ ${\displaystyle \triangle BCJ}$

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

Algebraic proof of general identity

The following formula applies to doubling the angle of the sine function:

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

Resolved according to you get: ${\displaystyle \cos(\alpha )}$

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

Accordingly, it follows:

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

If you multiply all right and left sides together, you get:

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

The right-hand side is a telescope product , that is, apart from the last sine term in the numerator and the first sine term in the denominator, all sine terms are reduced and the equation to be proven is obtained

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

literature

• 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

Web links

• Eric W. Weisstein : Morrie's law . In: MathWorld (English).

Individual evidence

1. ↑ WA Beyer, JD Louck, and D. Zeilberger , A Generalization of a Curiosity that Feynman Remembered All His Life , Math. Mag. 69, 1996, pp. 43-44, ( JSTOR )
2. ^ Samuel G. Moreno, Esther M. García-Caballero: A Geometric Proof of Morrie's Law . In: American Mathemtical Montly , Volume 122, No. 2 (February 2015), p. 168 ( JSTOR )