### 17th Putnam 1957

Problem A3

A and B are real numbers such that cos A ≠ cos B. Show that for any integer n > 1, |cos nA cos B - cos A cos nB| < (n2 - 1) |cos A - cos B|.

Solution

It is not clear how to use induction, expanding cos (n+1)A = cos nA cos A - sin nA sin A gives a profusion of sines and it is not clear how to get rid of them. So some other approach is needed.

An ingenious manipulation puts the expression entirely in terms of sines. Put x = (A + B)/2, y = (A - B)/2. Then A = x + y, B = x - y. Now 2 cos nA cos B = cos(nA + B) + cos(nA - B). Similarly, 2 cos A cos nB = cos(nB + A) + cos(nB - A). Switch to x and y and then go back to products of cosines. We have cos(nA + B) - cos(nB + A) = cos( (n+1)x + (n-1)y) - cos( (n+1)x - (n-1)y ) = -2 sin(n+1)x sin(n-1)y. Similarly, cos(nA - B) - cos(nB - A) = - 2 sin(n-1)x sin(n+1)y. Hence |cos nA cos B - cos A cos nB| = |sin(n+1)x sin(n-1)y + sin(n-1)x sin(n+1)y|. Obviously |cos A - cos B| = |2 sin x sin y|, so we have to prove that |sin(n+1)x sin(n-1)y + sin(n-1)x sin(n+1)y| < 2(n2 - 1) |sin x sin y|.

If would evidently be sufficient to show that |sin mx| < m |sin x|. Unfortunately, that is not true. We have equality if sin x = 0 or if m = 1. However, we can easily prove by induction that we have strict inequality in all other cases.

For m = 2, we have sin 2x = 2 sin x cos x, which establishes the result since |cos x| < 1 for sin x non-zero. Suppose it is true for m. Then sin(m+1)x = sin mx cos x + sin x cos mx, so |sin(m+1)x| <= |sin mx| |cos x| + |sin x| |cos mx| <= |sin mx| + |sin x| < (m+1) |sin x|, so it is true for m+1.

Now if sin x = 0, then A + B is a multiple of 2π, so cos A = cos B. Similarly if sin y = 0, then A - B is a multiple of 2π, so cos A = cos B. But we are told that cos A ≠ cos B, so sin x and sin y are both non-zero. Also we are given that n > 1, so we have strict inequality on |sin(n+1)x| > (n+1) |sin x| and |sin(n+1)y| > (n+1) |sin y|, whilst |sin(n-1)x| ≥ (n-1) |sin x| and |sin(n-1)y| ≥ (n-1) |sin y|. Thus we get the required strict inequality.