MHB Prove A_n is an integer for all n in N

[Albert1]

$a,b\in N ,\, and \,\, a>b,\,\, sin \,\theta=\dfrac {2ab}{a^2+b^2}$

(where $0<\theta <\dfrac {\pi}{2}$)

$A_n=(a^2+b^2)^nsin \,n\theta$

prove :$A_n$ is an integer for all n $\in N$

 

[kaliprasad]

Re: prone A_n is an integer for all n in N

Spoiler

we have

$\sin \theta = \frac{2ab}{a^2+b^2}$

so as $\cos \theta \gt 0$

$\cos \theta = \frac{a^2- b^2}{a^2+b^2}$

hence $\sin 2\theta = 2 \ sin\theta \cos \theta $
= $\frac{2ab(a^2-b^2)}{(a^2+b^2)^2}$

$cos 2\theta = 2 \cos^2\theta -1 = 2 (\frac{a^2- b^2}{a^2+b^2})^2 - 1$
= $\frac{2(a^2- b^2)^2 - (a^2 + b^2 )^2}{(a^2+b^2)^2}$

so $\sin \theta,\cos \theta,\sin 2\theta,\cos 2\theta$ are of the form $\frac{I}{(a^2+b^2)^n}$ for n = 1 and 2

let

$\sin k\theta = \frac{A_{k}}{(a^2+b^2)^k}$

$\cos k\theta = \frac{B_{k}}{(a^2+b^2)^k}$

for k = 1 to n

then
$sin (n+1)\theta = \sin n\theta \cos \theta +\sin \theta \cos n \theta $
= $\frac{A_{n}}{(a^2+b^2)^n} \frac{B_{1}}{a^2+b^2} + \frac{A_{1}}{(a^2+b^2)} \frac{B_{n}}{(a^2+b^2)^n}$
= $\frac{A_{n}B_{1} + A_{1}B_{n}}{(a^2+b^2)^{n+1}}$

if the form is true for 1 to n then it is true for n+ 1 for $\sin$ But we depend on $\cos$ so we need to prove for $\cos$

$cos (n+1)\theta = \cos n\theta \cos \theta - \sin n\theta \sin \theta $
= $\frac{B_{n}}{(a^2+b^2)^n} \frac{B_{1}}{a^2+b^2} - \frac{A_{n}}{(a^2+b^2)^n} \frac{A_{1}}{(a^2+b^2}$
= $\frac{B_{n}B_{1} - A_{n}A_{1}}{(a^2+b^2)^{n+1}}$so if it is true for sin and cos for 1 and n then it is true for n+ 1
as we have proved for 2 and then induction step

so $\ sin n\theta = \frac{I}{(a^2+b^2)^n}$

or $\sin n \theta (a^2+b^2)^n$ is integer

hence proved