$S_2 = \large{\frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \cdots \infty}$ $=$ $\pi \cdot (\underbrace{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \infty}_{1})$ $\implies$ $S_2 = \pi$ $S_3 = \large{\frac{\pi}{3} + \frac{\pi}{9} + \frac{\pi}{27} + \cdots \infty}$ $=$ $\pi \cdot (\underbrace{\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \infty}_{\large{\frac{1}{2}}})$ $\implies$ $S_3 = \Large{\frac{\pi}{2}}$ $S = S_2 + S_3$ $=$ $\Large{\frac{3\pi}{2}}$ $\implies$ $\boxed{\cos \left(S-x \right) = -\sin (x)}$ Alternativa $\mathbb{(A)}$

Podemos dividir S em duas somas $S_1$ e $S_2$, tal que: $$S_1=\frac{π}{2} + \frac{π}{4} +...+ \frac{π}{2^n}$$ e $$S_2=\frac{π}{3} + \frac{π}{9} +...+ \frac{π}{3^n}$$, onde $$S=S_1+S_2$$ $S_1=\frac{a_1}{1-q}$, onde $a_1=\frac{π}{2}$ e $q=\frac{1}{2}$, onde q é a razão da P.G $S_1=\frac{\frac{π}{2}}{1-\frac{1}{2}}=π$ $S_2=\frac{\frac{π}{3}}{1-\frac{1}{3}}=\frac{π}{2}$ $S=π + \frac{π}{2}=\frac{3π}{2}$ Calculemos $cos(S-x)$: $$cos(S-x)=cos(S)cos(x) + sen(S)sen(x)=cos(\frac{3π}{2})cos(x) + sen(\frac{3π}{2})sen(x)=0.cos(x)-1.sen(x)=-sen(x)$$