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 Identity)
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 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
~ERiccc
Solution 7(single formula)
We use
for
vladimir.shelomovskii@gmail.com, vvsss
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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