Difference between revisions of "2005 AIME II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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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>(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = \sin n(\frac\pi2 -u) + i\cos n(\frac\pi2 - u)</math> holds for all real <math>u</math>.
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<math>(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = (\cos u + i \sin u)^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(\frac\pi2 - u)</math> and <math>\sin nu = \cos n(\frac\pi2 - u)</math> hold for all real <math>u</math>.
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<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(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n = 1 + 4k</math>, or we need <math>-nu + n(\frac\pi2 - u) = \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 250 of them in the given range.
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== See Also ==
 
== See Also ==
 
*[[2005 AIME II Problems]]
 
*[[2005 AIME II Problems]]

Revision as of 09:39, 21 July 2006

Problem

For how many positive integers $n$ less than or equal to 1000 is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$?

Solution

We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$. 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 $\cos \frac{\pi}2 - u = \sin u$ and $\sin \frac{\pi}2 - u = \cos u$ hold for all real $u$. If our original equation holds for all $t$, it must certainly hold for $t = \frac{\pi}2 - u$. Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = \sin n(\frac\pi2 -u) + i\cos n(\frac\pi2 - u)$ holds for all real $u$.

$(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = (\cos u + i \sin u)^n = \cos nu + i\sin nu$. 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 $n$ such that $\cos n u = \sin n(\frac\pi2 - u)$ and $\sin nu = \cos n(\frac\pi2 - u)$ hold for all real $u$.

$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$. So from the equality of the real parts we need either $nu + n(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k$, in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$. Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work. There are 250 of them in the given range.

See Also

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