Evaluate the following where k is an integer sinpi22kpi Than

Solution

sin[(/2) + (2k)]

= cos(2k)       since sin(/2 + ) = cos

k is an integer ==> 2k is always even

= 1 (since cos0 = cos(2) = cos(4) = cos(8) --- = 1)

Hence sin[(/2) + (2k)] = 1