A geometry problem by neelesh vij

Geometry Level 3

cos ( x ) cos ( 2 x ) + cos ( 2 x ) cos ( 3 x ) + + cos ( ( n 1 ) x ) cos ( n x ) + cos ( n x ) cos ( x ) = n + sin 2 ( n x ) \cos(x) \cos(2x) + \cos(2x) \cos(3x) + \cdots + \cos((n-1)x) \cos(nx) + \cos(nx) \cos(x) = n + \sin^2(nx)

Find the number of solutions for the equation above, where x [ 0 , 2 π ] x \in [0,2 \pi] for a given value of n n where, n 2 n \geq 2 .


The answer is 2.

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1 solution

Neelesh Vij
Feb 20, 2017

Notice that the LHS has a maximum value of n n as cos ( A ) × cos ( B ) \cos(A) \times \cos(B) has max value of 1 1 and RHS has minimum value of n n

So x = 2 k π x = 2k \pi where k is a whole number.

Please do not let your notation do double duty. Is the n n in your solution (both the first and second lines) equal to the n n in the equation?

Calvin Lin Staff - 4 years, 3 months ago

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thanks for pointing out. i have done the necessary changes.

neelesh vij - 4 years, 3 months ago

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