Problem
ALG-B1-M06-P024 A rational linear combination
For some \(x\) and \(y\), the numbers \(A=\sin x+\cos y\) and \(B=\cos x-\sin y\) are positive rational numbers. Prove that there exist positive integers \(m\) and \(n\) such that \(m\sin x+n\cos x\) is a positive integer.
Express \(\cos y\) and \(\sin y\) through \(A,B,\sin x,\cos x\), then use \(\sin^2y+\cos^2y=1\).
We have \(\cos y=A-\sin x\) and \(\sin y=\cos x-B\). Square and add: \[(A-\sin x)^2+(\cos x-B)^2=1.\]
After expanding and using \(\sin^2x+\cos^2x=1\), we get \(A^2+B^2-2A\sin x-2B\cos x=0\), that is \(2A\sin x+2B\cos x=A^2+B^2\).
Let \(N\) be a common denominator of the rational numbers \(A\) and \(B\). Then \(m=2AN^2\) and \(n=2BN^2\) are positive integers, and \((A^2+B^2)N^2\) is a positive integer. Multiplying the equality by \(N^2\) gives the result.
Source analysis: the objects are rational trigonometric sums; the first approach is to try to determine the sines and cosines themselves; the hidden observation is that quadratic identities give a linear combination; the key ideas are expressing the \(y\)-part and clearing denominators. The signs and structure of the sums are changed.