Let d n be the determinant of the n × n matrix whose entries, from left to right and then from top to bottom, are cos 1 , cos 2 , … , cos n 2 . (For example, d 3 = ∣ ∣ ∣ ∣ ∣ ∣ cos 1 cos 4 cos 7 cos 2 cos 5 cos 8 cos 3 cos 6 cos 9 ∣ ∣ ∣ ∣ ∣ ∣ . The argument of cos is always in radians, not degrees.) Evaluate lim n → ∞ d n .
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It turns out that d n = 0 for n ≥ 3 , so that the limit is 0 as well.
The reason, in a nutshell, is that the vectors ( cos 1 , cos 2 , . . . , cos n ) and ( sin 1 , sin 2 , . . . , sin n ) span the row space of A n , so that r a n k A n ≤ 2 and A n fails to be invertible for n ≥ 3 .
We will illustrate this observation in the case of the second row of the given matrix A 3 ; it is a simple application of the addition theorem for the cosine:
( cos 4 , cos 5 , cos 6 ) = ( cos ( 3 + 1 ) , cos ( 3 + 2 ) , cos ( 3 + 3 ) ) = cos 3 ( cos 1 , cos 2 , cos 3 ) − sin 3 ( sin 1 , sin 2 , sin 3 )
PS: This works out the same way if were to measure angles in degrees, of course.