Proving ∑ i = 1 n i sin ⁡ i x = sin ⁡ ( ( n + 1 ) x ) 4 sin 2 ⁡ x 2 − ( n + 1 ) cos ⁡ ( ( 2 n + 1 ) x 2 ) 2 sin ⁡ x 2 by induction

Question

Proving       ∑          i      =      1        n    i  sin  ⁡      i    x    =            sin      ⁡              (        (        n        +        1        )        x        )                    4              sin        2            ⁡                        x          2                      −            (      n      +      1      )      cos      ⁡              (        (        2        n        +        1        )                  x          2                )                    2      sin      ⁡                        x          2                     by induction

arhaitategr · Accepted Answer

Base case   n  =  1, we need to prove:  sin  ⁡  x  =            sin      ⁡              (        2        x        )                    4              sin        2            ⁡                        x          2                      −            2      cos      ⁡              (                              3            x                    2                )                    2      sin      ⁡                        x          2                      .We have            2      cos      ⁡              (                              3            x                    2                )                    2      sin      ⁡                        x          2                      =            2      cos      ⁡              (                              3            x                    2                )            sin      ⁡              x        2                    2              sin        2            ⁡                        x          2                      =            sin      ⁡      2      x      −      sin      ⁡      x              2              sin        2            ⁡                        x          2                      .So we have  R  H  S  =            sin      ⁡              (        2        x        )                    4              sin        2            ⁡                        x          2                      −            sin      ⁡      2      x      −      sin      ⁡      x              2              sin        2            ⁡                        x          2                      =            2      sin      ⁡      x      −      sin      ⁡      2      x              2      (      1      −      cos      ⁡      x      )        =            2      sin      ⁡      x      −      2      sin      ⁡      x      cos      ⁡      x              2      (      1      −      cos      ⁡      x      )        =  sin  ⁡  x  =  L  H  S  .Induction step, we need to prove:                    n        sin        ⁡                  n          x                                    =                              sin            ⁡                          (              (              n              +              1              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    −                              (            n            +            1            )            cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                    −                              sin            ⁡                          (              (              n              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    +                              (            n            )            cos            ⁡                          (              (              2              n              −              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                                      =                              sin            ⁡                          (              (              n              +              1              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    −                              sin            ⁡                          (              (              n              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                                      +        n                  (                                    cos              ⁡                              (                (                2                n                −                1                )                                  x                  2                                )                                                    2              sin              ⁡                                                x                  2                                                              −                                    cos              ⁡                              (                (                2                n                +                1                )                                  x                  2                                )                                                    2              sin              ⁡                                                x                  2                                                              )                .            We know that                                                sin            ⁡                          (              (              n              +              1              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    −                              sin            ⁡                          (              (              n              )              x              )                                            4                          sin              2                        ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                        =                                          sin            ⁡                          (              (              n              +              1              )              x              )                        −            sin            ⁡            (            n            x            )                                4                          sin              2                        ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                        =                                          2            sin            ⁡                          x              2                        cos            ⁡                                          2                n                +                1                            2                        x                                4                          sin              2                        ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                        =                                          cos            ⁡                                          2                n                +                1                            2                        x                                2            sin            ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                    =        0.            Also we have                                                cos            ⁡                          (              (              2              n              −              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                    −                              cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                                        =                                          cos            ⁡                          (              (              2              n              −              1              )                              x                2                            )                        −            cos            ⁡                          (              (              2              n              +              1              )                              x                2                            )                                            2            sin            ⁡                                          x                2                                                    =                              2            sin            ⁡            n            x            sin            ⁡                          x              2                                            2            sin            ⁡                                          x                2                                                    =        sin        ⁡        n        x        .            Therefore, we have  R  H  S  =            sin      ⁡              (        (        n        +        1        )        x        )                    4              sin        2            ⁡                        x          2                      −            sin      ⁡              (        (        n        )        x        )                    4              sin        2            ⁡                        x          2                      −            cos      ⁡              (        (        2        n        +        1        )                  x          2                )                    2      sin      ⁡                        x          2                        +  n      (                  cos        ⁡                  (          (          2          n          −          1          )                      x            2                    )                            2        sin        ⁡                              x            2                                −                  cos        ⁡                  (          (          2          n          +          1          )                      x            2                    )                            2        sin        ⁡                              x            2                                )    =  0  +  n  sin  ⁡  n  x  =  n  sin  ⁡  n  x  =  L  H  S  .So, we are done.

Proving <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> i =

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