Problem 365: (2021Fall AMC12B, 13)
Let $c = \frac{2\pi}{11}.$ What is the value of\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\] $\textbf{(A)}\ {-}1 \qquad$ $\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad$ $\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad$ $\textbf{(D)}\ \frac{10}{11} \qquad$ $\textbf{(E)}\ 1$
Solution
- Since $11c=2\pi$: $\sin6c=\sin((11-5)c)=-\sin5c$, $\sin9c=-\sin2c$, $\sin12c=\sin(c)$, $\sin15c=\sin(4c)$.
- Numerator (after canceling the common $\sin3c$ factor) $=\sin6c\sin9c\sin12c\sin15c=(-\sin5c)(-\sin2c)(\sin c)(\sin4c)=\sin c\sin2c\sin4c\sin5c$.
- This exactly matches the remaining denominator, so the ratio is $1$, answer (E).
Let $c=\frac{2\pi}{7}$. What is the value of $\dfrac{\sin3c\cdot\sin6c\cdot\sin9c}{\sin c\cdot\sin2c\cdot\sin3c}$?
