, 2015

Solution:

Answer: $-9 + 3 \sqrt{13}$

Let $f(n) = \cos a_{n}$ and $m = 1$. The second ("quadratic interpolation") condition on $f(m), f(m+1), f(m+2), f(m+3)$ is equivalent to having a vanishing third finite difference
$$f(m+3) - 3 f(m+2) + 3 f(m+1) - f(m) = 0$$
This is equivalent to
$$\begin{aligned} f(m+3) - f(m) &= 3 [f(m+2) - f(m+1)] \\ \Longleftrightarrow \cos(a_{m+3}) - \cos(a_{m}) &= 3 (\cos(a_{m+2}) - \cos(a_{m+1})) \\ &= -6 \sin\left(\frac{a_{m+2} + a_{m+1}}{2}\right) \sin\left(\frac{a_{m+2} - a_{m+1}}{2}\right) \\ &= -6 \sin\left(\frac{a_{m+3}}{2}\right) \sin\left(\frac{a_{m}}{2}\right) . \end{aligned}$$
Set $x = \sin\left(\frac{a_{m+3}}{2}\right)$ and $y = \sin\left(\frac{a_{m}}{2}\right)$. Then the above rearranges to
$$(1 - 2x^{2}) - (1 - 2y^{2}) = -6 x y \Longleftrightarrow x^{2} - y^{2} = 3 x y$$
Solving gives $y = x \frac{-3 \pm \sqrt{13}}{2}$. The expression we are trying to maximize is $2(x^{2} - y^{2}) = 6 x y$, so we want $x, y$ to have the same sign; thus $y = x \frac{-3 + \sqrt{13}}{2}$.
Then $|y| \leq |x|$, so since $|x|, |y| \leq 1$, to maximize $6 x y$ we can simply set $x = 1$, for a maximal value of $6 \cdot \frac{-3 + \sqrt{13}}{2} = -9 + 3 \sqrt{13}$.