2024 AMC 12A Problems/Problem 23
Contents
[hide]Problem
What is the value of
Solution 1 (Trigonometric Identities)
First, notice that
Here, we make use of the fact that
Hence,
Note that
Hence,
Therefore, the answer is .
~tsun26
Solution 2 (Another Indentity)
First, notice that
Here, we make use of the fact that
Hence,
Therefore, the answer is .
Solution 3 (Complex Numbers)
Let . Then, Expanding by using a binomial expansion, Divide by and notice we can set where . Then, define so that
Notice that we can have because we are only considering the real parts. We only have this when , meaning . This means that we have as unique roots (we get them from ) and by using the fact that , we get Since we have a monic polynomial, by the Fundamental Theorem of Algebra, Looking at the term in the expansion for and using vietas gives us Since and Therefore
Solution 4(Just do it)
Look at the options, A is too small and E is too big, you have BCD left, you can make a guess but can also estimate the values. After factorising in solution 1, tan(3pi/16)^2 is around 0.45,tan(5pi/16)^2 is around 2.24 and so on. Therefore, the answer is .
Solution 5(transform)
set x = , 7x = - x , set C7 = , C5 = , C3 = , C= , S2 = , S6 =
First, notice that
Solution 6 (Half angle formula twice)
So from the question we have:
Using
Using
Using
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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