Please, note that we have:

the Taylor sequence of cos(z), is:

cos (z)= sum(n=0 to infinity) [((-1)^n)/(2n)!]*z^(2n)

now if we set sqrt[z-(pi/2)] in place of z, we get (where pi=3.14159...):

cos (sqrt(z-pi/2))= sum(n=0 to infinity) [((-1)^n)/(2n)!]*(z-pi/2)^(n)

so setting z=-1+(pi/2), we get:

cos(-1)=cos(1)=sum(n=0 to infinity) [1/(2n)!]

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