f ( x ) = cos ( x ) ⋅ cos ( 2 x ) ⋅ cos ( 3 x ) ⋯ cos ( 9 9 9 x )
If f ( 1 9 9 9 2 π ) = 2 k 1 , find k .
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Let P : = f ( 2 π / 1 9 9 9 ) . It is well-known that k = 1 ∏ n − 1 cos ( k π / n ) = 2 n − 1 sin ( n π / 2 ) . In particular, k = 1 ∏ 1 9 9 8 cos ( k π / 1 9 9 9 ) = 2 1 9 9 8 1 . Thus, since k = 1 ∏ 1 9 9 8 cos ( k π / 1 9 9 9 ) = k = 1 ∏ 9 9 9 cos ( 2 k π / 1 9 9 9 ) k = 1 ∏ 9 9 9 cos ( ( 2 k − 1 ) π / 1 9 9 9 ) = P k = 1 ∏ 9 9 9 cos ( π − ( 2 k − 1 ) π / 1 9 9 9 ) = P k = 1 ∏ 9 9 9 cos ( ( 2 0 0 0 − 2 k ) π / 1 9 9 9 ) = P k = 1 ∏ 9 9 9 cos ( 2 k π / 1 9 9 9 ) = P 2 ,
we deduce that P = 1 / 2 1 9 9 8 . Therefore, k = 9 9 9 .
Sorry, this is my first time seeing that identity. Could you provide a link, proof, or name?
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Relevant wiki: Chebyshev Polynomials - Application to Polynomial Interpolation
Let I = f ( 1 9 9 9 2 π ) . Since I represents the product of all cosine of positive angles less than π , then I > 0 .
By recalling the identity cos ( π − x ) = − cos x , we can rewrite I as
I = = [ − cos ( 1 9 9 9 1 9 9 7 π ) ] [ − cos ( 1 9 9 9 1 9 9 5 π ) ] [ − cos ( 1 9 9 9 1 9 9 3 π ) ] ⋯ [ − cos ( 1 9 9 9 π ) ] ( − 1 ) 9 9 9 cos ( 1 9 9 9 π ) cos ( 1 9 9 9 3 π ) cos ( 1 9 9 9 5 π ) ⋯ cos ( 1 9 9 9 π )
Thus, I 2 = − n = 1 ∏ 1 9 9 8 cos ( 1 9 9 9 n π ) = n = 1 ∏ 1 9 9 9 cos ( 1 9 9 9 n π ) .
Note that { cos ( 1 9 9 9 n π ) } n = 1 n = 1 9 9 9 are roots of the equation cos ( 1 9 9 9 y ) = 0 .
By Chebyshev polynomials , we may rewrite cos ( 1 9 9 9 y ) as a polynomial of cos y , namely,
cos ( 1 9 9 9 y ) = 2 1 9 9 8 cos 1 9 9 9 ( y ) − cos y ( ⋯ ) − 1 .
By Vieta's formula , n = 1 ∏ 1 9 9 9 cos ( 1 9 9 9 n π ) can be expressed as the product of roots of the equation, 2 1 9 9 8 z 1 9 9 9 − ⋯ − 1 = 0 . That is,
I 2 = n = 1 ∏ 1 9 9 9 cos ( 1 9 9 9 n π ) = 2 1 9 9 8 1 ⇒ I = 2 9 9 9 1 ⇒ k = 9 9 9 .