Difference between revisions of "2005 AIME II Problems/Problem 9"
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== Problem == | == Problem == | ||
− | + | For how many positive integers <math> n </math> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>? | |
− | For how many positive integers <math> n </math> less than or equal to 1000 is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>? | ||
== Solution == | == Solution == | ||
+ | We know by [[De Moivre's Theorem]] that <math>(\cos t + i \sin t)^n = \cos nt + i \sin nt</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. | ||
− | + | Recall the [[trigonometric identities]] <math>\cos \frac{\pi}2 - u = \sin u</math> and <math>\sin \frac{\pi}2 - u = \cos u</math> hold for all real <math>u</math>. If our original equation holds for all <math>t</math>, it must certainly hold for <math>t = \frac{\pi}2 - u</math>. Thus, the question is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</math> we have that <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)</math> holds for all real <math>u</math>. | |
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− | we | + | <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu</math>. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all <math>n</math> such that <math>\cos n u = \sin n\left(\frac\pi2 - u\right)</math> and <math>\sin nu = \cos n\left(\frac\pi2 - u\right)</math> hold for all real <math>u</math>. |
− | + | <math>\sin x = \cos y</math> if and only if either <math>x + y = \frac \pi 2 + 2\pi \cdot k</math> or <math>x - y = \frac\pi2 + 2\pi\cdot k</math> for some integer <math>k</math>. So from the equality of the real parts we need either <math>nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n = 1 + 4k</math>, or we need <math>-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n</math> will depend on <math>u</math> and so the equation will not hold for all real values of <math>u</math>. Checking <math>n = 1 + 4k</math> in the equation for the imaginary parts, we see that it works there as well, so exactly those values of <math>n</math> congruent to <math>1 \pmod 4</math> work. There are <math>\boxed{250}</math> of them in the given range. | |
== See also == | == See also == |
Revision as of 16:45, 25 July 2008
Problem
For how many positive integers less than or equal to
is
true for all real
?
Solution
We know by De Moivre's Theorem that for all real numbers
and all integers
. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities and
hold for all real
. If our original equation holds for all
, it must certainly hold for
. Thus, the question is equivalent to asking for how many positive integers
we have that
holds for all real
.
. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all
such that
and
hold for all real
.
if and only if either
or
for some integer
. So from the equality of the real parts we need either
, in which case
, or we need
, in which case
will depend on
and so the equation will not hold for all real values of
. Checking
in the equation for the imaginary parts, we see that it works there as well, so exactly those values of
congruent to
work. There are
of them in the given range.
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |