Question

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

Answer 1

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.

Answer 2

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.