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

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== Problem ==
 
== Problem ==
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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>?
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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 ==
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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. 
  
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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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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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<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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==Solution 2==
 
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We can rewrite <math>(\sin t + i \cos t)^n</math> as <math>(\cos(90-t) + i\sin(90-t))^n</math> which, by De Moivre's Theorem is equal to <math>\cos{n(90-t)} + i\sin{n(90-t)}</math>, but we know that is is equal to <math>\sin nt + i \cos nt</math>, now if we replace <math>\sin nt + i \cos nt</math> with <math>\cos{90-nt} + i\sin{90-nt}</math>. This gives us the equation:
 
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<math>\cos{90n - nt} + i\sin{90 - nt} = \cos{90-nt} + i\sin{90-nt}</math>
 
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Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get
 
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<math>90n - nt = 90 - nt</math>
 
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but <math>90</math> in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is:
 
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<math>90n = 90 + 360x</math> for <math>x</math>  being an integer.
 
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<math>n = 4x + 1</math>
 
  
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we know that <math>n \le 1000</math>
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<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>.
  
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so we want all numbers that are less than or equal to <math>1000</math> and also <math>\equiv 1 (\mbox{mod} 4)</math> and there are <math>\fbox{250}</math> such numbers
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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\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 17:45, 25 July 2008

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 $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$.

$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$. 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 $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ 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 $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, 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 $\boxed{250}$ 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
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