On the board after the class, there was a note left:

"Calculate $t(0)-t(\pi / 5)+t\left((\pi / 5)-t(3 \pi / 5)+\ldots+t\left({ }^{8 \pi} / 5\right)-t(9 \pi / 5)\right.$, where $t(x)=\cos 5 x+* \cos 4 x+* \cos 3 x+* \cos 2 x+$ $*^{\cos x}+* "$.

Seeing it, a math student told a friend that he could calculate this sum even without knowing the values of the coefficients erased from the board (denoted by * in our notation). Is he mistaken?
Let $t_{k}(x)=\cos k x$ for $k=0,1,2,3,4$. If we consider a similar sum for $t_{5}$ instead of $t$, we will get 10 as the result. We will check that for all other $t_{k}$ the sums are equal to 0. For $k=0$ this is obvious. For $k=1,2,3,4$ we arrive at the same equality

$\cos 0+\cos (2 \pi / 5)+\cos ((\pi / 5) \cos (6 \pi / 5)+\cos ((\pi / 5)=\cos (\pi / 5)+\cos (3 \pi / 5)+\cos (5 \pi / 5)+\cos (7 \pi / 5)+\cos ((\pi / 5)$. The expression on both the right and left sides is equal to 0. This follows from the fact that the sum of vectors going from the center of a regular pentagon to its vertices is 0 (see problem $\underline{55373}$ a or problem $\underline{85241}$ ).

## Answer

It is correct; the sum is 10.