Mostre que

$\frac{1}{2} + \cos{x} + \cos{2x} + \ldots + \cos{nx} = \frac{\sin{\frac{(2n+1)x}{2}}}{2 \sin{\frac{x}{2}}}$


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ITA IIIT 28/04/2022 18:31
$-$ Pela expressão do enunciado: \begin{matrix} \sin{\frac{x}{2}} \ + \ \underbrace{2\sin{\frac{x}{2}} \cos{x} \ + \ 2\sin{\frac{x}{2}} \cos{2x} \ + \ ... \ + \ 2\sin{\frac{x}{2}} \cos{nx}}_{\large{\delta}} &=& \sin{\frac{(2n+1)x}{2}} \end{matrix}$-$ Com conhecimento das $\text{Fórmulas de Werner}$, analisemos $\delta$: \begin{matrix} 2\sin{\frac{x}{2}} \cos{x} &=& \sin{(x + \frac{x}{2})} & -& \sin{(x - \frac{x}{2})} \\ 2\sin{\frac{x}{2}} \cos{2x} &=& \sin{(2x + \frac{x}{2})} & -& \sin{(2x - \frac{x}{2})} \\ 2\sin{\frac{x}{2}} \cos{3x} &=& \sin{(3x + \frac{x}{2})} & -& \sin{(3x - \frac{x}{2})} \\ \vdots && \vdots && \vdots \\ 2\sin{\frac{x}{2}} \cos{nx} &=& \sin{(nx + \frac{x}{2})} & -& \sin{(nx - \frac{x}{2})} \\ \hline \\ \delta &=& \sin{(nx + \frac{x}{2})} & -& \sin{(x - \frac{x}{2})} \end{matrix}Conclui-se que, \begin{matrix} \delta &=& \sin{\frac{(2n+1)x}{2}} &-& \sin{\frac{x}{2}} \end{matrix}Então, \begin{matrix} \sin{\frac{x}{2}} + \delta = \sin{\frac{(2n+1)x}{2}} \end{matrix}Isto é, \begin{matrix} \frac{1}{2} \ + \ \cos{x} \ + \ \cos{2x} \ + \ ... \ + \ \cos{nx}&=& \large{\frac{\sin{\frac{(2n+1)x}{2}}}{2\sin{\frac{x}{2}}}} & \tiny{ \blacksquare} \end{matrix}
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