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Lemma
Trigonometric Identities Difficulty 4

TMUA practice: Trig Identities, problem 1

A TMUA-calibre trig identities problem at difficulty 4 of 5, with the full worked solution.

The question
Let c=2π11.c = \frac{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?\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}?
A −1{-}1
B −115{-}\frac{\sqrt{11}}{5}
C 115\frac{\sqrt{11}}{5}
D 1011\frac{10}{11}
E 11

The correct answer is highlighted. E

Worked solution

Plugging in cc , we get

sin⁡3c⋅sin⁡6c⋅sin⁡9c⋅sin⁡12c⋅sin⁡15csin⁡c⋅sin⁡2c⋅sin⁡3c⋅sin⁡4c⋅sin⁡5c=sin⁡6π11⋅sin⁡12π11⋅sin⁡18π11⋅sin⁡24π11⋅sin⁡30π11sin⁡2π11⋅sin⁡4π11⋅sin⁡6π11⋅sin⁡8π11⋅sin⁡10π11.\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}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.

Since sin⁡(x+2π)=sin⁡(x),\sin(x+2\pi)=\sin(x), sin⁡(2π−x)=sin⁡(−x),\sin(2\pi-x)=\sin(-x), and sin⁡(−x)=−sin⁡(x),\sin(-x)=-\sin(x), we get

sin⁡6π11⋅sin⁡12π11⋅sin⁡18π11⋅sin⁡24π11⋅sin⁡30π11sin⁡2π11⋅sin⁡4π11⋅sin⁡6π11⋅sin⁡8π11⋅sin⁡10π11=sin⁡6π11⋅sin⁡−10π11⋅sin⁡−4π11⋅sin⁡2π11⋅sin⁡8π11sin⁡2π11⋅sin⁡4π11⋅sin⁡6π11⋅sin⁡8π11⋅sin⁡10π11=(E) 1.\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.