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Question

Which of the following option is incorrect?
(nZ)

A
cos[(2n+1)π2]=0
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B
cos(2nπ)=1
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C
cos[(2n+1)π]=1
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D
cos(nπ)=1
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Solution

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

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