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Mathematics 6 Online
OpenStudy (anonymous):

what is the value of cos(2npi)??

OpenStudy (aum):

cos(0) = 1 When you make 1 full 2pi rotation, you end up at the same place. So cos(2*n*pi) = cos(0) = 1.

OpenStudy (anonymous):

cos(npi)=??

OpenStudy (aum):

Depends on the value of n. If n is even, cos(n*pi) = cos(0) = 1 If n is odd, cos(n*pi) = cos(pi) = -1

OpenStudy (anonymous):

sin(npi)=?? sin(2npi)=??

OpenStudy (aum):

sin(npi). Put n = 0, 1, 2, 3, .... sin(0) = 0, sin(pi) = 0, sin(2pi) = 0, sin(3pi) = 0. So sin(npi) = 0 for all integer values of n. sin(2npi). Put n = 0, 1, 2, .... sin(0) = 0, sin(2pi) = 0, sin(4pi) = 0 So sin(2npi) = 0 for all integer values of n.

OpenStudy (anonymous):

thanks a lot....

OpenStudy (aum):

you are welcome.

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