RADII

Algebra Level pending

A = 2 n = 3 ( 1 tan 4 ( π 2 n ) ) A=2\prod_{n=3}^{\infty} \left(1-\tan^4\left(\frac{\pi}{2^n}\right)\right)

If R R is the radius of a circle whose area is A A , find the radius of a circle whose area is R R .

1 8 \frac{1}{8} sin R \sin R log 2 ( sec R ) \log_2 (\sec R) 1 4 \frac{1}{4}

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1 solution

Chew-Seong Cheong
Jul 16, 2020

A = 2 k = 3 ( 1 tan 4 π 2 n ) = 2 k = 3 ( 1 tan 2 π 2 k ) ( 1 + tan 2 π 2 k ) = 2 k = 3 1 tan 2 π 2 k 1 + tan 2 π 2 k ( 1 + tan 2 π 2 k ) 2 = 2 k = 3 cos π 2 n 1 sec 4 π 2 k = 2 k = 3 cos π 2 k 1 cos 4 π 2 k = 2 cos π 4 k = 3 1 cos 3 π 2 k = k = 3 2 cos 3 π 2 k See proof k = 2 n cos π 2 k = 1 2 n 1 sin π 2 n = lim n 2 cos 3 π 4 ( k = 2 n 1 cos π 2 k ) 3 = 1 2 ( lim n 2 n 1 sin π 2 n ) 3 = 1 2 ( lim n π sin π 2 n 2 π 2 n ) 3 = π 3 16 \begin{aligned} A & = 2 \prod_{k=3}^\infty \left(1-\tan^4 \frac \pi {2^n} \right) = 2 \prod_{k=3}^\infty \left(1-\tan^2 \frac \pi {2^k} \right) \left(1+\tan^2 \frac \pi {2^k} \right) \\ & = 2 \prod_{k=3}^\infty \frac {1-\tan^2 \frac \pi {2^k}}{1+\tan^2 \frac \pi {2^k}} \left(1+\tan^2 \frac \pi {2^k} \right)^2 = 2 \prod_{k=3}^\infty \cos \frac \pi{2^{n-1}} \sec^4 \frac \pi {2^k} \\ & = 2 \prod_{k=3}^\infty \frac {\cos \frac \pi{2^{k-1}}}{\cos^4 \frac \pi {2^k}} = 2\cos \frac \pi 4 \prod_{k=3}^\infty \frac 1{\cos^3 \frac \pi {2^k}} = \prod_{k=3}^\infty \frac {\sqrt 2}{\cos^3 \frac \pi {2^k}} & \small \blue{\text{See proof }\prod_{k=2}^n \cos \frac \pi{2^k} = \frac 1{2^{n-1}\sin \frac \pi{2^n}}} \\ & = \lim_{n \to \infty} \sqrt 2 \cos^3 \frac \pi 4 \left( \prod_{k=2}^n \frac 1{\cos \frac \pi{2^k}} \right)^3 = \frac 12 \left(\lim_{n \to \infty} 2^{n-1} \sin \frac \pi {2^n} \right)^3 \\ & = \frac 12 \left(\lim_{n \to \infty} \frac { \pi \sin \frac \pi {2^n}}{2 \cdot \frac \pi{2^n}} \right)^3 = \frac {\pi^3}{16} \end{aligned}

Therefore A = π 2 16 = π R 2 A = \frac {\pi^2}{16} = \pi R^2 , R = π 4 \implies R = \frac \pi 4 and the radius of the circle with area R R is r = 1 2 = log 2 ( 2 ) = log 2 sec π 4 = log 2 sec R r = \frac 12 = \log_2 (\sqrt 2) = \log_2 \sec \frac \pi 4 = \boxed{\log_2 \sec R} .


Proof: Let us prove by induction the claim P ( n ) = k = 2 n cos π 2 k = 1 2 n 1 sin π 2 n \displaystyle P(n) = \prod_{k=2}^n \cos \frac \pi{2^k} = \frac 1{2^{n-1} \sin \frac \pi {2^n}} for n 2 n \ge 2 . For n = 2 n=2 , P ( 2 ) = cos π 4 = 1 2 = 1 2 sin π 4 P(2) = \cos \dfrac \pi 4 = \dfrac 1{\sqrt 2} = \dfrac 1{2 \sin \frac \pi 4} . The claim is true for n = 2 n=2 . Assuming that the claim is true for n n , then:

P ( n + 1 ) = P ( n ) cos π 2 n + 1 = cos π 2 n + 1 2 n 1 sin π 2 n = cos π 2 n + 1 2 n sin π 2 n + 1 cos π 2 n + 1 = 1 2 n sin π 2 n + 1 \begin{aligned} P(n+1) & = P(n) \cos \frac \pi{2^{n+1}} = \frac {\cos \frac \pi{2^{n+1}}}{2^{n-1}\sin \frac \pi {2^n}} = \frac {\cos \frac \pi{2^{n+1}}}{2^n \sin \frac \pi {2^{n+1}}\cos \frac \pi{2^{n+1}}} = \frac 1{2^n \sin \frac \pi {2^{n+1}}} \end{aligned}

The claim is also true of n + 1 n+1 and therefore is true for all n 2 n \ge 2 .

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