Proof that ∑ k = 0 N − 1 cos ⁡ ( 2 π N ( k + 1 2 ) ) = 0and I happened to notice that it equals 0 for all integer values of N &gt; 1. This it very easy to prove for N even, by noticing some symmetries in the angles that appear in the arguments of the cosines. This does not work for N odd. When trying to prove it for N odd, I split it as ∑ k = 0 N − 1 cos ⁡ ( 2 π N ( k + 1 2 ) ) = cos ⁡ ( π N ) ∑ k = 0 N − 1 cos ⁡ ( 2 π N k ) − sin ⁡ ( π N ) ∑ k = 0 N − 1 sin ⁡ ( 2 π N k ) and noticed (by checking it numerically) that the two terms are individually 0 for all integer N&gt;1 ∑ k = 0 N − 1 cos ⁡ ( 2 π N k ) = 0 = ∑ k = 0 N − 1 sin ⁡ ( 2 π N k ) .From here I don't know to proceed.

Question

Proof that       ∑          k      =      0              N      −      1        cos  ⁡      (                  2        π            N              (      k      +              1        2            )        )    =  0and I happened to notice that it equals 0 for all integer values of   N  &amp;gt;  1. This it very easy to prove for N even, by noticing some symmetries in the angles that appear in the arguments of the cosines. This does not work for N odd. When trying to prove it for N odd, I split it as      ∑          k      =      0              N      −      1        cos  ⁡      (                  2        π            N              (      k      +              1        2            )        )    =  cos  ⁡      (          π      N        )        ∑          k      =      0              N      −      1        cos  ⁡      (                  2        π            N        k    )    −  sin  ⁡      (          π      N        )        ∑          k      =      0              N      −      1        sin  ⁡      (                  2        π            N        k    )  and noticed (by checking it numerically) that the two terms are individually 0 for all integer N&amp;gt;1      ∑          k      =      0              N      −      1        cos  ⁡      (                  2        π            N        k    )    =  0  =      ∑          k      =      0              N      −      1        sin  ⁡      (                  2        π            N        k    )    .From here I don&#039;t know to proceed.

Kailee Abbott · Accepted Answer

HintYou face sums of sines and cosines of angles in arithmetic progression      ∑          k      =      0              N      −      p        cos  ⁡      (                  2        π            N        k    )    =      1    2        (    1    +    csc    ⁡          (              π        N            )        sin    ⁡          (                        π          −          2          π          p                N            )        )        ∑          k      =      0              N      −      p        sin  ⁡      (                  2        π            N        k    )    =      1    2    csc  ⁡      (          π      N        )        (    cos    ⁡          (                        π          (          N          +          1          −          2          p          )                N            )        +    cos    ⁡          (              π        N            )        )

Proof that sum_(k=0)^(N-1) cos((2pi)/N (k+1/2))=0

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