Verified Answer :

LammettHash

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.

boffeemadrid

Answer:

True

Step-by-step 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.

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