AMC 12B 2021 Fall (Problem 13)Let c=2π11c=\dfrac{2\pi}{11}c=112π. What is the value of sin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c\dfrac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}sinc⋅sin2c⋅sin3c⋅sin4c⋅sin5csin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15c?(A) −1\text{(A)}\;-1(A)−1(B) −115\text{(B)}\;-\dfrac{\sqrt{11}}{5}(B)−511(C) 115\text{(C)}\;\dfrac{\sqrt{11}}{5}(C)511(D) 1011\text{(D)}\;\dfrac{10}{11}(D)1110(E) 1\text{(E)}\;1(E)1Related TopicsToolkit 37 — Trigonometric TransformationsHints (7)Hint 1c=2π11⇒11c=2π.c=\frac{2\pi}{11}\Rightarrow 11c=2\pi.c=112π⇒11c=2π.Hint 2Use 37. Trigonometry Transformations.Hint 3sin6c=sin(2π−6c)=−sin(11c−6c)=−sin5c.\sin 6c=\sin(2\pi-6c)=-\sin(11c-6c)=-\sin 5c.sin6c=sin(2π−6c)=−sin(11c−6c)=−sin5c.Hint 4sin9c=sin(11c−2c)=−sin2c.\sin 9c=\sin(11c-2c)=-\sin 2c.sin9c=sin(11c−2c)=−sin2c.Hint 5sin12c=sin(11c+c)=sinc.\sin 12c=\sin(11c+c)=\sin c.sin12c=sin(11c+c)=sinc.Hint 6sin15c=sin(11c+4c)=sin4c.\sin 15c=\sin(11c+4c)=\sin 4c.sin15c=sin(11c+4c)=sin4c.Hint 7sin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c=sin3c⋅(−sin5c)⋅(−sin2c)⋅sinc⋅sin4csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c=1.\dfrac{\sin3c\cdot\sin6c\cdot\sin9c\cdot\sin12c\cdot\sin15c}{\sin c\cdot\sin2c\cdot\sin3c\cdot\sin4c\cdot\sin5c} = \dfrac{\sin3c\cdot(-\sin5c)\cdot(-\sin2c)\cdot\sin c\cdot\sin4c}{\sin c\cdot\sin2c\cdot\sin3c\cdot\sin4c\cdot\sin5c}=1.sinc⋅sin2c⋅sin3c⋅sin4c⋅sin5csin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15c=sinc⋅sin2c⋅sin3c⋅sin4c⋅sin5csin3c⋅(−sin5c)⋅(−sin2c)⋅sinc⋅sin4c=1.Final Answer(E) 1