Morrie's law is a special trigonometric identity. Its name is due to the physicist Richard Feynman, who used to refer to the identity under that name. Feynman picked the name from that of Morrie Jacobs, a boy whom he knew during his childhood, and afterwards remembered it for all of his life.
Identity and generalisation
cos ( 20 ∘ ) ⋅ cos ( 40 ∘ ) ⋅ cos ( 80 ∘ ) = 1 8 . {\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
2 n ⋅ ∏ k = 0 n − 1 cos ( 2 k α ) = sin ( 2 n α ) sin ( α ) {\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
sin ( 160 ∘ ) sin ( 20 ∘ ) = sin ( 180 ∘ − 20 ∘ ) sin ( 20 ∘ ) = 1 , {\displaystyle {\frac {\sin(160^{\circ })}{\sin(20^{\circ })}}={\frac {\sin(180^{\circ }-20^{\circ })}{\sin(20^{\circ })}}=1,}
since
sin ( 180 ∘ − x ) = sin ( x ) . {\displaystyle \sin(180^{\circ }-x)=\sin(x).}
Similar identities A similar identity for the sine function also holds:
sin ( 20 ∘ ) ⋅ sin ( 40 ∘ ) ⋅ sin ( 80 ∘ ) = 3 8 . {\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:
tan ( 20 ∘ ) ⋅ tan ( 40 ∘ ) ⋅ tan ( 80 ∘ ) = 3 = tan ( 60 ∘ ) . {\displaystyle \tan(20^{\circ })\cdot \tan(40^{\circ })\cdot \tan(80^{\circ })={\sqrt {3}}=\tan(60^{\circ }).}
Proof
Geometric proof of Morrie's law
Consider a regular nonagon A B C D E F G H I {\displaystyle ABCDEFGHI} with side length 1 {\displaystyle 1} and let M {\displaystyle M} be the midpoint of A B {\displaystyle AB} , L {\displaystyle L} the midpoint B F {\displaystyle BF} and J {\displaystyle J} the midpoint of B D {\displaystyle BD} . The inner angles of the nonagon equal 140 ∘ {\displaystyle 140^{\circ }} and furthermore γ = ∠ F B M = 80 ∘ {\displaystyle \gamma =\angle FBM=80^{\circ }} , β = ∠ D B F = 40 ∘ {\displaystyle \beta =\angle DBF=40^{\circ }} and α = ∠ C B D = 20 ∘ {\displaystyle \alpha =\angle CBD=20^{\circ }} (see graphic). Applying the cosinus definition in the right angle triangles △ B F M {\displaystyle \triangle BFM} , △ B D L {\displaystyle \triangle BDL} and △ B C J {\displaystyle \triangle BCJ} then yields the proof for Morrie's law:
1 = | A B | = 2 ⋅ | M B | = 2 ⋅ | B F | ⋅ cos ( γ ) = 2 2 | B L | cos ( γ ) = 2 2 ⋅ | B D | ⋅ cos ( γ ) ⋅ cos ( β ) = 2 3 ⋅ | B J | ⋅ cos ( γ ) ⋅ cos ( β ) = 2 3 ⋅ | B C | ⋅ cos ( γ ) ⋅ cos ( β ) ⋅ cos ( α ) = 2 3 ⋅ 1 ⋅ cos ( γ ) ⋅ cos ( β ) ⋅ cos ( α ) = 8 ⋅ cos ( 80 ∘ ) ⋅ cos ( 40 ∘ ) ⋅ cos ( 20 ∘ ) {\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 the generalised identity Recall the double angle formula for the sine function
sin ( 2 α ) = 2 sin ( α ) cos ( α ) . {\displaystyle \sin(2\alpha )=2\sin(\alpha )\cos(\alpha ).}
Solve for cos ( α ) {\displaystyle \cos(\alpha )}
cos ( α ) = sin ( 2 α ) 2 sin ( α ) . {\displaystyle \cos(\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}.}
It follows that:
cos ( 2 α ) = sin ( 4 α ) 2 sin ( 2 α ) cos ( 4 α ) = sin ( 8 α ) 2 sin ( 4 α ) ⋮ cos ( 2 n − 1 α ) = sin ( 2 n α ) 2 sin ( 2 n − 1 α ) . {\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:
cos ( α ) cos ( 2 α ) cos ( 4 α ) ⋯ cos ( 2 n − 1 α ) = sin ( 2 α ) 2 sin ( α ) ⋅ sin ( 4 α ) 2 sin ( 2 α ) ⋅ sin ( 8 α ) 2 sin ( 4 α ) ⋯ sin ( 2 n α ) 2 sin ( 2 n − 1 α ) . {\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,
∏ k = 0 n − 1 cos ( 2 k α ) = sin ( 2 n α ) 2 n sin ( α ) , {\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.
See also Viète's formula, same identity taking α = 2 − n x {\displaystyle \alpha =2^{-n}x} on Morrie's law List of trigonometric identities
References
Further reading Van Brummelen, Glen (23 January 2020). Trigonometry: A Very Short Introduction. Oxford University Press. pp. 79–83. ISBN 978-0-19-254546-6. Anderson, Ernest C. (1998). "Morrie's Law and Experimental Mathematics". Journal of Recreational Mathematics. 29 (2): 85–88.
External links Weisstein, Eric W. "Morrie's Law". MathWorld.