The correct option is D cos(nπ)=1
Checking all be given options, putting n=0, we get
cos[(2n+1)π2]=cosπ2=0cos(2nπ)=cos0=1cos[(2n+1)π]=cosπ=−1cos(nπ)=cos0=1
Checking all be given options, putting n=1, we get
cos[(2n+1)π2]=cos3π2=0cos(2nπ)=cos2π=1cos[(2n+1)π]=cos3π=−1cos(nπ)=cosπ=−1
So, cos(nπ)=±1