AMC 12B 2021 Fall (Problem 13)Let c=2π11c=\frac{2\pi}{11}c=112π. What is the value of sin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c\frac{\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)}\;-\frac{\sqrt{11}}{5}(B)−511(C) 115\text{(C)}\;\frac{\sqrt{11}}{5}(C)511(D) 1011\text{(D)}\;\frac{10}{11}(D)1110(E) 1\text{(E)}\;1(E)1Related TopicsCoreToolkit 37 — Trigonometric TransformationsHints (6)Hint 1c=2π11c=\frac{2\pi}{11}c=112π⟹11c=2π.\Longrightarrow 11c=2\pi.⟹11c=2π.Hint 2sin6c=−sin(−6c)\sin 6c=-\sin(-6c)sin6c=−sin(−6c)=−sin(2π−6c)=-\sin(2\pi-6c)=−sin(2π−6c)=−sin(11c−6c)=-\sin(11c-6c)=−sin(11c−6c)=−sin5c.=-\sin 5c.=−sin5c.Hint 3sin9c=sin(11c−2c)\sin 9c=\sin(11c-2c)sin9c=sin(11c−2c)=sin(2π−2c)=\sin(2\pi-2c)=sin(2π−2c)=sin(−2c)=\sin(-2c)=sin(−2c)=−sin2c.=-\sin 2c.=−sin2c.Hint 4sin12c=sin(11c+c)\sin 12c=\sin(11c+c)sin12c=sin(11c+c)=sin(2π+c)=\sin(2\pi+c)=sin(2π+c)=sinc.=\sin c.=sinc.Hint 5sin15c=sin(11c+4c)\sin 15c=\sin(11c+4c)sin15c=sin(11c+4c)=sin(2π+4c)=\sin(2\pi+4c)=sin(2π+4c)=sin4c.=\sin 4c.=sin4c.Hint 6By Hints 2, 3, 4, and 5,sin3c⋅sin6c⋅sin9c⋅sin12c⋅sin15csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c\frac{\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=sin3c⋅(−sin5c)⋅(−sin2c)⋅sinc⋅sin4csinc⋅sin2c⋅sin3c⋅sin4c⋅sin5c=\frac{\sin 3c\cdot(-\sin 5c)\cdot(-\sin 2c)\cdot\sin c\cdot\sin 4c}{\sin c\cdot\sin 2c\cdot\sin 3c\cdot\sin 4c\cdot\sin 5c}=sinc⋅sin2c⋅sin3c⋅sin4c⋅sin5csin3c⋅(−sin5c)⋅(−sin2c)⋅sinc⋅sin4c=1.=1.=1.Final Answer(E) 111Related Problems (2)AMC 12A Spring 2021 (Problem 19)AMC 12B 2024 (Problem 11)