Question

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Why is cosnπ = (–1)^n true, when n could be any integer?

Answer 1

Let's think about this systematically. If n can be an integer, then n can be 1, 2, 3, 0, -1, -2, -3, etc. Look at what happens with \(\cos(n\pi)\) with these values. No matter which integer value of n it is, \(n\pi\) will have a factor of \(\pi\) in it. The value of cosine when the angle is at a value like 0, \(\pi\), \(2\pi\), \(3\pi\), \(-\pi\), \(-2\pi\), etc. are either 1 or -1. Cosine would evaluate to 1 when n is an even integer (because then the angle would have factors of \(2\pi\)) and -1 when n is an odd number. Likewise, on the right side, \((-1)^n\) evaluates to -1 when n is odd and to 1 when n is even. Because the right side equals the left at the same integer values of n, they are equal.

Answer 2

Does this make sense or should I try and clarify further?

Answer 3

no i understand. thank you @Calcmathlete

Answer 4

np :)