Question

Why is cos(nt) = (-1)^n true, when n could be any integer?

Answer 1

It's not true for any
`t`, but it is for
`t=\pi`.

Start with
`n=0` (which is even). Then

`\cos(0\pi)=\cos0=1`

The cosine function is
`2\pi[tex]-periodic, meaning that for any [tex]x` we have

`\cos(x+2\pi)=\cos x`

Now,
`2\pi=0+2\pi`, so

`\cos(2\pi)=\cos(0+2\pi)=\cos0=1`

This is the basis for an argument via induction to show that
`\cos(n\pi)=1` whenever
`n` is even. Also, when
`n` is even we have
`(-1)^n=1`.

In a similar way, starting with the fact that

`\cos(1\pi)=\cos\pi=-1`

and that
`\cos x` is periodic, we have

`\cos(3\pi)=\cos(\pi+2\pi)=\cos\pi=-1`

and so for odd
`n` we have
`\cos(n\pi)=-1=(-1)^n`.