Difference between revisions of "2008 AIME II Problems/Problem 8"

Revision as of 22:34, 4 July 2013

Problem

Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that $2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]$ is an integer.

Solution

By the product-to-sum identities, we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$ . Therefore, this reduces to a telescoping series: $\begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}$

Thus, we need $\sin \left(\frac{n(n+1)\pi}{2008}\right)$ to be an integer; this can be only $\{-1,0,1\}$ , which occur when $2 \cdot \frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1$ . It easily follows that $n = \boxed{251}$ is the smallest such integer.

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	[[Category:Intermediate Trigonometry Problems]]		[[Category:Intermediate Trigonometry Problems]]
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Difference between revisions of "2008 AIME II Problems/Problem 8"

Revision as of 22:34, 4 July 2013

Problem

Solution

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