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anonymous
 one year ago
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Why is cosnπ = (–1)^n true, when n could be any integer?
anonymous
 one year ago
I will fan and medal! Why is cosnπ = (–1)^n true, when n could be any integer?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let'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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense or should I try and clarify further?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i understand. thank you @Calcmathlete
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