Sejam p=\frac{5 \pi}{12}, q=\frac{7 \pi}{12} e r=\frac{\pi}{12}. O valor da \operatorname{expressão} \operatorname{sen}(p)+\operatorname{sen}(q)+\operatorname{sen}(r)-\operatorname{sen}(p+q+r) é
(A) \sqrt{2}
(B) 2 \sqrt{3}
(C) 3 \sqrt{3}
(D) \sqrt{6}
(E) 2 \sqrt{6}

Passo 1

O enunciado dá os ângulos \,p=\frac{5\pi}{12}\,,\;q=\frac{7\pi}{12}\,,\;r=\frac{\pi}{12}\, e pede o valor da expressão
\sin(p)+\sin(q)+\sin(r)-\sin(p+q+r).
Lista clara do que é pedido:
  • Calcular numericamente \sin(p)+\sin(q)+\sin(r)-\sin(p+q+r).
  • Usar as relações trigonométricas e valores conhecidos de senos e cossenos de ângulos especiais.

Passo 2

Calculo primeiro \,p+q+r. Somando os numeradores:
p+q+r=\frac{5\pi}{12}+\frac{7\pi}{12}+\frac{\pi}{12}=\frac{13\pi}{12}.
Agora avalio \sin(p+q+r)=\sin\!\left(\frac{13\pi}{12}\right). Observo que
\frac{13\pi}{12}=\pi+\frac{\pi}{12},
e uso a identidade \sin(\pi+x)=-\sin(x). Portanto
\sin\!\left(\frac{13\pi}{12}\right)=-\sin\!\left(\frac{\pi}{12}\right)=-\sin(r).

Passo 3

Substituo esse resultado na expressão original:
\sin(p)+\sin(q)+\sin(r)-\sin(p+q+r) =\sin\!\left(\tfrac{5\pi}{12}\right)+\sin\!\left(\tfrac{7\pi}{12}\right)+\sin\!\left(\tfrac{\pi}{12}\right)-\big(-\sin\!\left(\tfrac{\pi}{12}\right)\big).
Isso simplifica para
\sin\!\left(\tfrac{5\pi}{12}\right)+\sin\!\left(\tfrac{7\pi}{12}\right)+2\sin\!\left(\tfrac{\pi}{12}\right).

Passo 4

Agora uso a fórmula soma-para-produto para \sin A+\sin B:
\sin A+\sin B = 2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right).
Aplicando com A=\tfrac{5\pi}{12} e B=\tfrac{7\pi}{12}:
\sin\!\left(\tfrac{5\pi}{12}\right)+\sin\!\left(\tfrac{7\pi}{12}\right) =2\sin\!\left(\frac{\tfrac{5\pi}{12}+\tfrac{7\pi}{12}}{2}\right)\cos\!\left(\frac{\tfrac{5\pi}{12}-\tfrac{7\pi}{12}}{2}\right).
Calculo os meios:
\frac{\tfrac{5\pi}{12}+\tfrac{7\pi}{12}}{2}=\frac{\tfrac{12\pi}{12}}{2}=\frac{\pi}{2}, \qquad \frac{\tfrac{5\pi}{12}-\tfrac{7\pi}{12}}{2}=\frac{-\tfrac{2\pi}{12}}{2}=-\frac{\pi}{12}.
Como \sin\!\left(\tfrac{\pi}{2}\right)=1 e \cos é par, obtemos
\sin\!\left(\tfrac{5\pi}{12}\right)+\sin\!\left(\tfrac{7\pi}{12}\right)=2\cos\!\left(\tfrac{\pi}{12}\right).
Logo a expressão se torna
2\cos\!\left(\tfrac{\pi}{12}\right)+2\sin\!\left(\tfrac{\pi}{12}\right)=2\big(\cos\!\left(\tfrac{\pi}{12}\right)+\sin\!\left(\tfrac{\pi}{12}\right)\big).

Passo 5

Preciso agora dos valores de \sin\!\left(\tfrac{\pi}{12}\right) e \cos\!\left(\tfrac{\pi}{12}\right). Note que \tfrac{\pi}{12}=15^\circ e posso usar a fórmula de diferença de ângulos \sin(a-b)=\sin a\cos b-\cos a\sin b com a=\tfrac{\pi}{4} e b=\tfrac{\pi}{6} (pois \tfrac{\pi}{4}-\tfrac{\pi}{6}=\tfrac{\pi}{12}):
Calculo \sin\!\left(\tfrac{\pi}{12}\right)=\sin\!\left(\tfrac{\pi}{4}-\tfrac{\pi}{6}\right):
\sin\!\left(\tfrac{\pi}{12}\right)=\sin\!\left(\tfrac{\pi}{4}\right)\cos\!\left(\tfrac{\pi}{6}\right)-\cos\!\left(\tfrac{\pi}{4}\right)\sin\!\left(\tfrac{\pi}{6}\right).
Usando \sin\!\left(\tfrac{\pi}{4}\right)=\cos\!\left(\tfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}, \cos\!\left(\tfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}, \sin\!\left(\tfrac{\pi}{6}\right)=\dfrac{1}{2}, vem
\sin\!\left(\tfrac{\pi}{12}\right)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2} =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}.
Calculo \cos\!\left(\tfrac{\pi}{12}\right)=\cos\!\left(\tfrac{\pi}{4}-\tfrac{\pi}{6}\right) usando \cos(a-b)=\cos a\cos b+\sin a\sin b:
\cos\!\left(\tfrac{\pi}{12}\right)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2} =\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}.

Passo 6

Somando \cos\!\left(\tfrac{\pi}{12}\right)+\sin\!\left(\tfrac{\pi}{12}\right):
\cos\!\left(\tfrac{\pi}{12}\right)+\sin\!\left(\tfrac{\pi}{12}\right) =\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4} =\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}.
Multiplicando por 2 (lembre da fatoração anterior), a expressão original vale
2\cdot\frac{\sqrt{6}}{2}=\sqrt{6}.

Resposta

\sqrt{6}

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