Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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==Solution== | ==Solution== | ||
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+ | Part 1: solving for <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math> | ||
+ | |||
+ | <math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math> | ||
+ | |||
+ | Multiply by <math>8 \sin{2\pi}7</math> | ||
+ | |||
+ | <math>c 8 \sin{2\pi}7 = -8 \sin{2\pi}7 \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math> | ||
+ | |||
+ | Then use sine addition formula backwards: | ||
+ | |||
+ | <math>2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7 | ||
+ | |||
+ | </math>c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7<math> | ||
+ | |||
+ | </math>c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7<math> | ||
+ | |||
+ | </math>c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7<math> | ||
+ | |||
+ | </math>c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7<math> | ||
+ | |||
+ | </math>c = -\frac{1}8$ | ||
+ | |||
+ | |||
+ | This is in progress - I am working on solutions for a and b. | ||
+ | |||
+ | ~Tucker | ||
+ | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:23, 11 February 2021
Problem
Suppose that the roots of the polynomial are
and
, where angles are in radians. What is
?
Solution
Part 1: solving for
Multiply by
Then use sine addition formula backwards:
c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$$ (Error compiling LaTeX. Unknown error_msg)c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7$$ (Error compiling LaTeX. Unknown error_msg)c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7$$ (Error compiling LaTeX. Unknown error_msg)c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7$$ (Error compiling LaTeX. Unknown error_msg)c = -\frac{1}8$
This is in progress - I am working on solutions for a and b.
~Tucker
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.