2021 Fall AMC 12B Problem 13

Write each angle as $kc = \dfrac{2\pi k}{11}.$ Reducing modulo $2\pi,$ $\sin 12c = \sin c$ and $\sin 15c = \sin 4c.$ So the numerator is $\sin 3c \cdot \sin 6c \cdot \sin 9c$ $\cdot \sin c \cdot \sin 4c.$ Cancelling the common factors $\sin c,$ $\sin 3c,$ $\sin 4c$ leaves $\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}.$ Now $\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c$ and $\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c,$ so the ratio equals $\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1.$ Thus, the correct answer is E.