Question

trigonometrie-cls a-9-a (60a)

Answer 1

Notăm:

$\it \dfrac{\pi}{7}=a$.

Membrul drept al egalității devine:

$\it E =cos2a+cos4a+cos6a\Big|_{\cdot\dfrac{sina}{sina}} \Rightarrow E=\dfrac{sinacos2a+sinacos4a+sinacos6a}{sina}= \\ \\ \\ =\dfrac{\dfrac{1}{2}\Big[sin(a-2a)+sin(a+2a)+sin(a-4a)+sin(a+4a)+sin(a-6a)+sin(a+6a)\Big]}{sinx}=\\ \\ \\ =\dfrac{-sina+sin3a-sin3a+sin5a-sin5a+sin7a}{2sina}=\dfrac{-sina+sin7a}{2sina}=\\ \\ \\ =-\dfrac{sina}{2sina}+\dfrac{sin7a}{2sina}=-\dfrac{1}{2} +\dfrac{sin7a}{2sina}$

Revenim asupra notației și obținem:

$\it E = -\dfrac{1}{2}+\dfrac{sin7\cdot\dfrac{\pi}{7}}{2sin\dfrac{\pi}{7}}=-\dfrac{1}{2}+\dfrac{sin\pi}{2sin\dfrac{\pi}{7}}=-\dfrac{1}{2}+0=-\dfrac{1}{2}$