Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
| − | + | Let <math>c=\frac{2\pi}{p}</math> and <math>n</math> be relatively prime to <math>p</math>. | |
| + | |||
| + | Then <math>\dfrac{\sin(nc)\sin(2nc)\ldots\sin(\frac{p-1}{2}nc}{\sin(c)\sin(2c)\ldots\sin(\frac{p-1}{2}c)}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:14, 16 November 2022
Contents
Problem
Let 
What is the value of
![\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]](http://latex.artofproblemsolving.com/5/c/b/5cb66360bc69b72e86eac3564c26bae7fd577bc9.png)
Solution
Plugging in 
, we get
![\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\]](http://latex.artofproblemsolving.com/f/f/8/ff84ea1918eb8039edc31877eb63ca11dfef1986.png)
Since 
and 
we get
![\[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.\]](http://latex.artofproblemsolving.com/e/7/e/e7e5d4583710682a595b107182f40d55b4cad8b7.png)
~kingofpineapplz ~Ziyao7294 (minor edit)
Solution 2
Let 
and 
be relatively prime to 
.
Then 
See Also
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
