7.55 If three real numbers $x<y<z$, satisfy $y-x=z-y$, then $x, y, z$ are called an "arithmetic array". For $\triangle A B C$, let $B C=a, C A=b, A B=c$, and the opposite angles of $a$, $b, c$ are denoted as $A, B, C$ respectively. Given that $a, b, c$ form an "arithmetic array", then it must be true that
(A) $A, B, C$ form an "arithmetic array".
(B) $\cos A, \cos B, \cos C$ form an "arithmetic array".
(C) $\sin A, \sin B, \sin C$ form an "arithmetic array".
(D) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ form an "arithmetic array".
(China Beijing High School Mathematics Competition, 1990)
[Solution] By the Law of Sines, we have $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$, $\therefore a=2 R \sin A, \quad b=2 R \sin B, \quad c=2 R \sin C$,
Given that $2 b=a+c$,
then $4 R \sin B=2 R \sin A+2 R \sin C$,
which simplifies to $2 \sin B=\sin A+\sin C$.
Therefore, the correct choice is (C).