Assinale a opção que indica a soma dos elementos de $A\cup B$, sendo:$$A=\left\{x_k=\sin^2\left(\frac{k^2\pi}{24}\right): k = 1,2\right\}$$ $$B=\left\{y_k=\sin^2\left(\frac{(3k+5)\pi}{24}\right): k = 1,2\right\}$$


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ITA IIIT 20/12/2021 14:28
$-$ Do enunciado, podemos escrever: \begin{matrix} A &=& \{ & \sin^2{(\frac{\pi}{24})} & , & \sin^2{(\frac{\pi}{6})} & \} \\ B &=& \{ & \sin^2{(\frac{\pi}{3})} & , & \sin^2{(\frac{11\pi}{24})} & \} \end{matrix} \begin{matrix} \color{gray}{\fbox{ $\begin{matrix} \sin{\alpha} = \cos{(\frac{\pi}{2}- \alpha)} \\ \sin^2{\alpha} + \cos^2{\alpha}= 1 \end{matrix}$}} \end{matrix} \begin{matrix} A &=& \{ & \sin^2{(\frac{\pi}{24})} & , & \frac{3}{4} & \} \\ B &=& \{ & \frac{1}{4} & , & \cos^2{(\frac{\pi}{24})} & \} \end{matrix} \begin{matrix} \color{gray}{\fbox{$ \sin^2{(\frac{11\pi}{24})} = \cos^2{(\frac{\pi}{2} - \frac{11\pi}{24})} = \cos^2{(\frac{\pi}{24})} $}} \end{matrix} \begin{matrix} A \cup B &=& \{ & \sin^2{(\frac{\pi}{24})} & , & \frac{3}{4} &,& \frac{1}{4} &,& \cos^2{(\frac{\pi}{24})} & \} \\ \end{matrix} Assim, a soma: \begin{matrix} [\sin^2{(\frac{\pi}{24})} + \frac{3}{4} ] &+& [ \frac{1}{4} + \cos^2{(\frac{\pi}{24})} ] \\ \end{matrix} \begin{matrix} \fbox{$1 + \frac{3}{4} + \frac{1}{4} = 2$} \\ \\ \\ Letra \ (C) \end{matrix}
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