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

2 Answers

1 vote
This is because when
n is odd, i.e.
n=2k+1 for integers
k, you have


\cos(\pm\pi)=\cos(\pm3\pi)=\cdots=\cos(\pm(2k+1)\pi)=-1

while for even
n, i.e. when
n=2k, you have


\cos0=\cos(\pm2\pi)=\cos(\pm4\pi)=\cdots=\cos(\pm2k\pi)=1

Meanwhile,
(-1)^n=-1 is
n is odd, while
(-1)^n=1 if
n is even.
User Cander
by
7.8k points
4 votes

Answer:

True

Explanation:

The value of
cosn{\pi}=(-1)^n is true as:

when n is odd that is of the form
n=2m+1 where m is any integer, thus we have


cos(\pm\pi)=cos(\pm3\pi)=...=cos(\pm(2m+1)\pi=-1

and when n is even that is of the form
n=2m, where m is any integer, thus we have


cos0=cos(\pm2\pi)=....=cos(\pm2m\pi)=1

Thus,
(-1)^n=-1 for when n is odd and
(-1)^n=1 when n is even.

Hence, the given statement is true.

User Jim Nutt
by
8.8k points

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