The Cosine Product!

Geometry Level 1

f ( x ) = cos ( x ) cos ( 2 x ) cos ( 3 x ) cos ( 999 x ) \large f(x) = \cos(x) \cdot \cos(2x) \cdot \cos(3x)\cdots \cos(999x)

If f ( 2 π 1999 ) = 1 2 k f \left(\dfrac{2\pi }{1999}\right) = \dfrac{1}{2^{k}} , find k k .


The answer is 999.

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2 solutions

Pi Han Goh
Aug 10, 2016

Relevant wiki: Chebyshev Polynomials - Application to Polynomial Interpolation

Let I = f ( 2 π 1999 ) I = f\left( \dfrac{2\pi}{1999} \right) . Since I I represents the product of all cosine of positive angles less than π \pi , then I > 0 I > 0 .

By recalling the identity cos ( π x ) = cos x \cos(\pi - x) = - \cos x , we can rewrite I I as

I = [ cos ( 1997 π 1999 ) ] [ cos ( 1995 π 1999 ) ] [ cos ( 1993 π 1999 ) ] [ cos ( π 1999 ) ] = ( 1 ) 999 cos ( π 1999 ) cos ( 3 π 1999 ) cos ( 5 π 1999 ) cos ( π 1999 ) \begin{aligned} I& =& \left [ - \cos \left( \dfrac{1997\pi}{1999} \right) \right ] \left [ - \cos \left( \dfrac{1995\pi}{1999} \right) \right ] \left [ - \cos \left( \dfrac{1993\pi}{1999} \right) \right ] \cdots \left [ - \cos \left( \dfrac{\pi}{1999} \right) \right ] \\ &=& (-1)^{999} \cos \left( \dfrac{\pi}{1999} \right) \cos \left( \dfrac{3\pi}{1999} \right) \cos \left( \dfrac{5\pi}{1999} \right) \cdots \cos \left( \dfrac{\pi}{1999} \right) \end{aligned}

Thus, I 2 = n = 1 1998 cos ( n π 1999 ) = n = 1 1999 cos ( n π 1999 ) \displaystyle I^2 = - \prod_{n=1}^{1998} \cos \left( \dfrac{n\pi}{1999} \right) = \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right) .

Note that { cos ( n π 1999 ) } n = 1 n = 1999 \left \{ \cos \left( \dfrac{n\pi}{1999}\right) \right \}_{n=1}^{n=1999} are roots of the equation cos ( 1999 y ) = 0 \cos(1999y) = 0 .

By Chebyshev polynomials , we may rewrite cos ( 1999 y ) \cos(1999y) as a polynomial of cos y \cos y , namely,

cos ( 1999 y ) = 2 1998 cos 1999 ( y ) cos y ( ) 1 . \cos(1999y) = 2^{1998} \cos^{1999} (y ) - \cos y (\cdots) - 1 \; .

By Vieta's formula , n = 1 1999 cos ( n π 1999 ) \displaystyle \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right) can be expressed as the product of roots of the equation, 2 1998 z 1999 1 = 0 2^{1998} z^{1999} - \cdots - 1 = 0 . That is,

I 2 = n = 1 1999 cos ( n π 1999 ) = 1 2 1998 I = 1 2 999 k = 999 . I^2 = \prod_{n=1}^{1999} \cos \left( \dfrac{n\pi}{1999} \right) = \dfrac{ 1}{2^{1998}} \quad \Rightarrow \quad I = \dfrac1{2^{999}} \quad \Rightarrow \quad k = \boxed{999} \; .

Diego G
Aug 16, 2016

Let P : = f ( 2 π / 1999 ) P:=f(2\pi/1999) . It is well-known that k = 1 n 1 cos ( k π / n ) = sin ( n π / 2 ) 2 n 1 . \prod_{k=1}^{n-1}\cos(k\pi/n)=\frac{\sin(n\pi/2)}{2^{n-1}}. In particular, k = 1 1998 cos ( k π / 1999 ) = 1 2 1998 . \prod_{k=1}^{1998}\cos(k\pi/1999)=\frac1{2^{1998}}. Thus, since k = 1 1998 cos ( k π / 1999 ) = k = 1 999 cos ( 2 k π / 1999 ) k = 1 999 cos ( ( 2 k 1 ) π / 1999 ) = P k = 1 999 cos ( π ( 2 k 1 ) π / 1999 ) = P k = 1 999 cos ( ( 2000 2 k ) π / 1999 ) = P k = 1 999 cos ( 2 k π / 1999 ) = P 2 , \begin{aligned} \prod_{k=1}^{1998}\cos(k\pi/1999)&=\prod_{k=1}^{999}\cos(2k\pi/1999)\prod_{k=1}^{999}\cos((2k-1)\pi/1999) =P \prod_{k=1}^{999}\cos(\pi-(2k-1)\pi/1999)\\ &=P \prod_{k=1}^{999}\cos((2000-2k)\pi/1999) =P \prod_{k=1}^{999}\cos(2k\pi/1999) =P^2, \end{aligned}

we deduce that P = 1 / 2 1998 P=\sqrt{1/2^{1998}} . Therefore, k = 999 k=999 .

Sorry, this is my first time seeing that identity. Could you provide a link, proof, or name?

Eric Kim - 4 years, 6 months ago

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