Com conhecimento das $\text{Fórmulas de Werner}$:\begin{matrix} z &=&[ \ \cos{(2x)}\sin{(2x)} \ ]^2\sin{(x)} &=& {{\left[ \dfrac{\sin{(4x) + \sin{(0)}}}{2} \right]^2}} \sin{(x)} &=& 2^{-2}\sin{(4x)} \sin{(4x)} \sin{(x)} \end{matrix}Continuando, \begin{matrix} z &=& 2^{-2}\sin{(4x)} {{\left[ \dfrac{\cos{(5x) \ - \ \cos{(3x)}}}{-2} \right]}} &=& 2^{-3} [\sin{(4x)}\cos{(3x)} - \sin{(4x)}\cos{(5x)} ] \end{matrix} \begin{matrix} z &=& 2^{-3} {{\left[ \dfrac{\sin{(7x) \ + \ \sin{(x)}}}{2} - \dfrac{\sin{(9x) \ - \ \sin{(x)}}}{2} \right]}} \end{matrix}Por fim, \begin{matrix} z &=& 2^{-4} \ [\ 2\sin{(x)} + \sin{(7x)} - \sin{(9x)} \ ] \end{matrix} \begin{matrix} Letra \ (B) \end{matrix}

Outra maneira de se resolver essa questão seria por complexos, com conhecimento das equações: $z^{n} + z^{-n}= \ce2{cos}(nx)$ e $z^{n} - z^{-n}= 2i\ce{sen}(nx)$ Assim, $[ \ce{cos}(2x).\ce{se}n(2x) ]^2 . \ce{sen}(x)= \left( \frac{z^{2} + z^{-2}}{2} \right)^2 . \left( \frac{z^{2} - z^{-2}}{2i} \right)^2 . \left( \frac{z - z^{-1}}{2i} \right)$ = $\left(\frac{z^{4} + z^{-4} +2}{4} \right) . \left( \frac{z^{4} + z^{-4} -2}{-4} \right) . \left( \frac{z - z^{-1}}{2i} \right)$ = $\left(\frac{z^{8} + z^{-8} -2}{-16} \right) . \left( \frac{z - z^{-1}}{2i} \right)$ Continuando, $\left(\frac{z^{8} + z^{-8} -2}{-16} \right) . \left( \frac{z - z^{-1}}{2i} \right)= \left(\frac{(z^{9} + z^{-9}) -(z^7-z^{-7})-2(z-z^{-1})}{-32i} \right) = 2^{−4}[ 2sen(x)+sen(7x)−sen(9x) ]$ $Letra (B)$