A last year Putnam question maximum $\sum{\cos 3x}$

Visualising the solution

You have asked for help in visualising the solution. I think you will find it useful to have in mind the picture of $y=x^3$ for $-1\le x\le1$.

Now consider the arrangement of the 10 numbers in the maximum position. (We have a continuous function on a compact set and so the maximum is attained.)

First suppose that there is a number, $s$, smaller in magnitude than the least negative number $l$. Increasing $l$ whilst decreasing $s$ by the same amount would increase the sum of cubes and therefore cannot occur.

So, all the negative numbers are equal, to $l$ say, and all the positive numbers are greater than $|l|$.

Now suppose that a positive number was not $1$. Then increasing it to $1$ whilst reducing one of the $l$s would increase the sum of cubes and therefore cannot occur.

Hence we need only consider the case where we have $m$ $1$s and $10-m$ numbers equal to $-\frac{m}{10-m}$.