# 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 1000 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 == | ||

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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>. | 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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== See Also == | == See Also == | ||

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*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||

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+ | [[Category:Intermediate Complex Numbers Problems]] |

## Revision as of 16:53, 7 September 2006

## Problem

For how many positive integers less than or equal to 1000 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 250 of them in the given range.