By chance I found: 2 sin ⁡ ( 2 m + 1 ) x − sin ⁡ ( 2 m − 1 ) x = sin ⁡ ( x ) ( 1 + 2 cos ⁡ ( 2 m x ) + 2 ∑ k = 1 m cos ⁡ ( 2 k x ) )

Question

By chance I found:  2  sin  ⁡  (  2  m  +  1  )  x  −  sin  ⁡  (  2  m  −  1  )  x  =  sin  ⁡  (  x  )      (    1    +    2    cos    ⁡    (    2    m    x    )    +    2          ∑              k        =        1                    m              cos    ⁡    (    2    k    x    )    )

Immanuel Glenn · Accepted Answer

You may write, for any real number   x  ∉      Z    π                              ∑                      k            =            1                                m                          cos        ⁡        (        2        k        x        )                            =        ℜ                  ∑                      k            =            1                                m                                    e                      2            i            k            x                                                            =        ℜ                  (                      e                          2              i              x                                                                          e                                  2                  i                  m                  x                                            −              1                                                      e                                  2                  i                  x                                            −              1                                )                                                  =        ℜ                  (                      e                          2              i              x                                                                          e                                  i                  m                  x                                                            (                                  e                                      i                    m                    x                                                  −                                  e                                      −                    i                    m                    x                                                  )                                                                    e                                  i                  x                                                            (                                  e                                      i                    x                                                  −                                  e                                      −                    i                    x                                                  )                                              )                                                  =        ℜ                  (                      e                          i              (              m              +              1              )              x                                                          sin              ⁡              (              m              x              )                                      sin              ⁡              (              x              )                                )                                                  =        ℜ                  (                      (            cos            ⁡            (            (            m            +            1            )            x            )            +            i            sin            ⁡            (            (            m            +            1            )            x            )            )                                              sin              ⁡              (              m              x              )                                      sin              ⁡              (              x              )                                )                                                  =                              cos            ⁡            (            (            m            +            1            )            x            )                                sin            ⁡            (            x            )                          sin        ⁡        (        m        x        )        .            Thus                          sin        ⁡        (        x        )                  (          1          +          2          cos          ⁡          (          2          m          x          )          +          2                      ∑                          k              =              1                                      m                                cos          ⁡          (          2          k          x          )          )                                                  =        sin        ⁡        (        x        )        +        2        sin        ⁡        (        x        )        cos        ⁡        (        2        m        x        )        +        2        cos        ⁡        (        (        m        +        1        )        x        )        sin        ⁡        (        m        x        )            and, using       2    sin    ⁡    a    cos    ⁡    b    =    sin    ⁡    (    a    +    b    )    +    sin    ⁡    (    a    −    b    )  , we get                          sin        ⁡        (        x        )                  (          1          +          2          cos          ⁡          (          2          m          x          )          +          2                      ∑                          k              =              1                                      m                                cos          ⁡          (          2          k          x          )          )                                                  =        sin        ⁡        (        x        )        +        sin        ⁡        (        (        2        m        +        1        )        x        )        −        sin        ⁡        (        (        2        m        −        1        )        x        )        +        sin        ⁡        (        (        2        m        +        1        )        x        )        −        sin        ⁡        (        x        )                                          =        2        sin        ⁡        (        (        2        m        +        1        )        x        )        −        sin        ⁡        (        (        2        m        −        1        )        x        )            as announced.

By chance I found: 2 sin ⁡ ( 2 m + 1 ) x −

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