Respuesta :

LammettHash
Provided that [tex]n[/tex] is an integer, the statement is true.

[tex](-1)^{2n}=((-1)^2)^n=(1)^n=1[/tex]

Meanwhile, Euler's formula gives us

[tex](-1)^{2n}=(e^{i\pi})^{2n}=e^{i(2n\pi)}=\cos(2n\pi)+i\sin(2n\pi)[/tex]

and we know that [tex]\sin2n\pi=0[/tex] for all integers [tex]n[/tex], while [tex]\cos2n\pi=1[/tex].

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