Putnam 1957, 17th Problem I have spent the better part of today trying to prove the following question from Putnam 1957

If $\cos B\neq \cos A$, and $k>1$ is any natural number, prove that $$\left|\frac{\cos(kB)\cos(A)-\cos(kA)\cos(B)}{\cos(B)-\cos(A)}\right|<k^2-1$$

I tried using the mean value theorem for the function $f(x)=\cos[k(\cos^{-1}(x))]\cos[A+B-\cos^{-1}(x)]$. The rationale behind using this function is that if $q=\cos B$ and $p=\cos A$, then $\left|\frac{f(q)-f(p)}{q-p}\right|=\left|\frac{\cos(kB)\cos(A)-\cos(kA)\cos(B)}{\cos(B)-\cos(A)}\right|$. Now I would just have to prove that the derivative of this function at all points is bounded above by $k^2-1$.
I also tried using the Taylor expansion for all the cosines. Although it seemed promising at first, I got a lot of cross terms I didn't know how to get rid of.
Other methods tried- complex numbers, trig identities, etc 
I eventually did find a solution here. However, I would be interested in alternate/cleaner proofs of this assertion. How would you approach this question?

The given inequality is equivalent to 
$$|\cos(kB)\cos A-\cos(kA)\cos B|<(k^2-1)|\cos B-\cos A|$$
Using the formula $2\cos x\cos y=\cos(x-y)+\cos(x+y)$ we can rewrite the last inequality as
$$|\cos(kB-A)+\cos(kB+A)-\cos(kA-B)-\cos(kA+B)|<2(k^2-1)|\cos B-\cos A|$$
which is equivalent to 
$$|[\cos(kB-A)-\cos(kA-B)]+[\cos(kB+A)-\cos(kA+B)]|<2(k^2-1)|\cos B-\cos A|$$
Now we use formula $2\sin x\sin y=\cos(x-y)-\cos(x+y)$ to both sides of the last inequality to get the equivalent form
$$\Big|2\sin\Big((k-1)\frac{A+B}{2}\Big)\sin\Big((k+1)\frac{A-B}{2}\Big)+2\sin\Big((k+1)\frac{A+B}{2}\Big)\sin\Big((k-1)\frac{A-B}{2}\Big)\Big|\\ <4(k^2-1)\sin\Big(\frac{A+B}{2}\Big)\sin\Big(\frac{A-B}{2}\Big)$$
Since $\cos A\neq \cos B$ then necessarily $A\neq B$ which in turn implies $(A-B)/2\neq 0$, moreover by assumption $k>1$ we have $|\sin(kx)<k|\sin x|$ (proof by induction as in the link).  Now by triangle inequality we have
$$\Big|2\sin\Big((k-1)\frac{A+B}{2}\Big)\sin\Big((k+1)\frac{A-B}{2}\Big)+2\sin\Big((k+1)\frac{A+B}{2}\Big)\sin\Big((k-1)\frac{A-B}{2}\Big)\Big|\\\leqslant 4\Big|\sin\Big((k-1)\frac{A+B}{2}\Big)\sin\Big((k+1)\frac{A-B}{2}\Big)\Big|\\<4(k-1)(k+1)\Big|\sin\Big(\frac{A+B}{2}\Big)\sin\Big(\frac{A-B}{2}\Big)\Big|\\=4(k^2-1)\Big|\sin\Big(\frac{A+B}{2}\Big)\sin\Big(\frac{A-B}{2}\Big)\Big|$$
as desired.
