Question

Problema din atasament. Daca se poate cu rezolvare completa sau ideile, multumesc.

Answer 1

$\displaystyle f(x) = \sum\limits_{k=1}^{1009}\cos(2k-1)x \\ \\ 2f(x)\sin x = \sum\limits_{k=1}^{1009}2\sin x\cos(2k-1)x \\ \\ \\2f(x)\sin x = \sum\limits_{k=1}^{1009}\Bigg[\sin\Big[x+(2k-1)x\Big]+\sin\Big[x-(2k-1)x\Big]\Bigg] \\ \\\\ 2f(x)\sin x = \sum\limits_{k=1}^{1009}\Big[\sin 2kx-\sin (2k-2)x\Big]\\ \\\\ 2f(x)\sin x = \sin 2x+\sin 4x+...+\sin 2018x -\\ \\ - \sin 0-\sin 2x-\sin 4x-...-\sin 2016x \\ \\\\ 2f(x)\sin x = \sin 2018x$

$\Rightarrow \ f(x) = \dfrac{\sin 2018x}{2\sin x}\\ \\ \Rightarrow f\Big(\dfrac{\pi}{2019}\Big) = \dfrac{\sin \dfrac{2018\pi}{2019}}{2\sin \dfrac{\pi}{2019}}= \dfrac{\sin \Big(\pi -\dfrac{\pi}{2019}\Big)}{2\sin \dfrac{\pi}{2019}} = \\ \\ = \dfrac{\sin \dfrac{\pi}{2019}}{2\sin \dfrac{\pi}{2019}} = \boxed{\dfrac{1}{2}}$