MathDB
Trigonometric identity

Source: Russian TST 2022, Day 7 P1

March 21, 2023
algebratrigonometry

Problem Statement

Let aa{} and bb{} be positive integers. Prove that for any real number xx{} j=0a(aj)(2cos((2ja)x))b=j=0b(bj)(2cos((2jb)x))a.\sum_{j=0}^a\binom{a}{j}\big(2\cos((2j-a)x)\big)^b=\sum_{j=0}^b\binom{b}{j}\big(2\cos((2j-b)x)\big)^a.