It's not true for any [tex]t[/tex], but it is for [tex]t=\pi[/tex].
Start with [tex]n=0[/tex] (which is even). Then
[tex]\cos(0\pi)=\cos0=1[/tex]
The cosine function is [tex]2\pi[tex]-periodic, meaning that for any [tex]x[/tex] we have
[tex]\cos(x+2\pi)=\cos x[/tex]
Now, [tex]2\pi=0+2\pi[/tex], so
[tex]\cos(2\pi)=\cos(0+2\pi)=\cos0=1[/tex]
This is the basis for an argument via induction to show that [tex]\cos(n\pi)=1[/tex] whenever [tex]n[/tex] is even. Also, when [tex]n[/tex] is even we have [tex](-1)^n=1[/tex].
In a similar way, starting with the fact that
[tex]\cos(1\pi)=\cos\pi=-1[/tex]
and that [tex]\cos x[/tex] is periodic, we have
[tex]\cos(3\pi)=\cos(\pi+2\pi)=\cos\pi=-1[/tex]
and so for odd [tex]n[/tex] we have [tex]\cos(n\pi)=-1=(-1)^n[/tex].
