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

Question

asked Aug 21, 2018 67.1k views

2 Answers

Cander · Answer 1 · 2018-08-23T12:01:37+0000

This is because when

is odd, i.e.
n=2k+1

for integers

, you have

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

while for even

, i.e. when
n=2k

, you have

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

Meanwhile,
(-1)^n=-1

is

is odd, while
(-1)^n=1

if

is even.

Jim Nutt · Answer 2 · 2018-08-27T07:11:08+0000

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,
for when n is odd and
when n is even.

Hence, the given statement is true.