34.1k views
2 votes
Is the following statement valid according to Euler's formula?
Cos(2npi) = (-1)^2n

1 Answer

3 votes
Provided that
n is an integer, the statement is true.


(-1)^(2n)=((-1)^2)^n=(1)^n=1

Meanwhile, Euler's formula gives us


(-1)^(2n)=(e^(i\pi))^(2n)=e^(i(2n\pi))=\cos(2n\pi)+i\sin(2n\pi)

and we know that
\sin2n\pi=0 for all integers
n, while
\cos2n\pi=1.
User James Dunmore
by
8.2k points