Prove that 1 2 + ∑ k = 1 n cos ⁡ ( k x ) = sin ⁡ ( n + 1 2 ) x 2 sin ⁡ ( 1 2 x ) for any x ∈ R

Question

Prove that       1    2    +      ∑          k      =      1        n    cos  ⁡  (  k  x  )  =            sin      ⁡      (      n      +              1        2            )      x              2      sin      ⁡      (              1        2            x      )       for any   x  ∈      R

Travis Fernandez · Accepted Answer

Consider, when   x  ≠  m  .2  π  ,  m  ∈      Z        ∑          k      =      1              n            e          i      k      x        =                    e                  i          x                    (              e                  i          n          x                    −      1      )                      e                  i          x                    −      1      (being the sum of a geometric series whose ratio is   ≠  1)  =                    e                  i                      1            2                    x                            e                  i          n          x                    −              e                  i                      1            2                    x                                    e                  i                      1            2                    x                    −              e                  −          i                      1            2                    x                      =                    e                  i          x          (          n          +                      1            2                    )                    −              e                  i                      1            2                    x                            2      i      sin      ⁡              x        2            Considering the real part of both sides,      ∑          k      =      1              n        cos  ⁡  (  k  x  )  =            sin      ⁡      (      n      +              1        2            )      x      −      sin      ⁡      (              x        2            )              2      sin      ⁡      (              x        2            )        ,and the result follows immediately.For   x  =  m  .2  π, the proof follows the same lines as you have already done.

Prove that 1 2 </mfrac> + <munderover> ∑ <mrow class="MJX

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