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2021 Fall AMC 12B Problem 13

Problem 13 of 25IntermediateGeometryProblem-Solving Techniques

Let c=2π11.c = \dfrac{2\pi}{11}. What is the value of sin⁡3c⋅sin⁡6c⋅sin⁡9c⋅sin⁡12c⋅sin⁡15csin⁡c⋅sin⁡2c⋅sin⁡3c⋅sin⁡4c⋅sin⁡5c?\small \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}?

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Solution

Write each angle as kc=2πk11.kc = \dfrac{2\pi k}{11}. Reducing modulo 2π,2\pi, sin⁡12c=sin⁡c\sin 12c = \sin c and sin⁡15c=sin⁡4c.\sin 15c = \sin 4c. So the numerator is sin⁡3c⋅sin⁡6c⋅sin⁡9c\sin 3c \cdot \sin 6c \cdot \sin 9c ⋅sin⁡c⋅sin⁡4c.\cdot \sin c \cdot \sin 4c. Cancelling the common factors sin⁡c,\sin c, sin⁡3c,\sin 3c, sin⁡4c\sin 4c leaves sin⁡6c⋅sin⁡9csin⁡2c⋅sin⁡5c.\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}. Now sin⁡9c=sin⁡(2π−2c)=−sin⁡2c\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c and sin⁡6c=sin⁡(2π−5c)=−sin⁡5c,\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c, so the ratio equals (−sin⁡5c)(−sin⁡2c)sin⁡2c⋅sin⁡5c=1.\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1. Thus, the correct answer is E.
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