i have to prove this cos ⁡ 3 x sin ⁡ 2 x sin ⁡ 4 x + cos ⁡ 5 x sin ⁡ 4 x sin ⁡ 6 x + cos ⁡ 7 x sin ⁡ 6 x sin ⁡ 8 x + cos ⁡ 9 x sin ⁡ 8 x sin ⁡ 10 x = 1 2 csc ⁡ x ( csc ⁡ 2 x − csc ⁡ 10 x )

Question

i have to prove this            cos      ⁡      3      x              sin      ⁡      2      x      sin      ⁡      4      x        +            cos      ⁡      5      x              sin      ⁡      4      x      sin      ⁡      6      x        +            cos      ⁡      7      x              sin      ⁡      6      x      sin      ⁡      8      x        +            cos      ⁡      9      x              sin      ⁡      8      x      sin      ⁡      10      x        =      1    2    csc  ⁡  x  (  csc  ⁡  2  x  −  csc  ⁡  10  x  )

candydulce168nlid · Accepted Answer

The right side looks like a result of a telescopic sum, so we&#039;ll try to prove            cos      ⁡      (      n      x      )              sin      ⁡      (      (      n      −      1      )      x      )      sin      ⁡      (      (      n      +      1      )      x      )        =      1    2    csc  ⁡  (  x  )  (  csc  ⁡  (  (  n  −  1  )  x  )  −  csc  ⁡  (  (  n  +  1  )  x  )  )Why this? It may not be true, but if it is, we have                                                cos            ⁡            3            x                                sin            ⁡            2            x            sin            ⁡            4            x                          +                              cos            ⁡            5            x                                sin            ⁡            4            x            sin            ⁡            6            x                          +                              cos            ⁡            7            x                                sin            ⁡            6            x            sin            ⁡            8            x                          +                              cos            ⁡            9            x                                sin            ⁡            8            x            sin            ⁡            10            x                                              =                              1          2                csc        ⁡        x                  (          csc          ⁡          2          x          −          csc          ⁡          4          x          +          csc          ⁡          4          x          −          csc          ⁡          6          x          +          csc          ⁡          6          x          −          csc          ⁡          8          x          +          csc          ⁡          8          x          −          csc          ⁡          10          x          )                                    =                              1          2                csc        ⁡        x        (        csc        ⁡        2        x        −        csc        ⁡        10        x        )            So, we&#039;ll start operating the right side. We have                                    1          2                csc        ⁡        (        x        )        (        csc        ⁡        (        (        n        −        1        )        x        )        −        csc        ⁡        (        (        n        +        1        )        x        )        )                            =                              1                      2            sin            ⁡            (            x            )                                    (                      1                          sin              ⁡              (              (              n              −              1              )              x              )                                −                      1                          sin              ⁡              (              (              n              +              1              )              x              )                                )                                    =                              1                      2            sin            ⁡            (            x            )                          ⋅                              sin            ⁡            (            (            n            +            1            )            x            )            −            sin            ⁡            (            (            n            −            1            )            x            )                                sin            ⁡            (            (            n            −            1            )            x            )            sin            ⁡            (            (            n            +            1            )            x            )                                              =                                          sin            ⁡            (            (            n            +            1            )            x            )            −            sin            ⁡            (            (            n            −            1            )            x            )                                2            sin            ⁡            (            x            )            sin            ⁡            (            (            n            −            1            )            x            )            sin            ⁡            (            (            n            +            1            )            x            )                                              =                                          (            sin            ⁡            (            n            x            )            cos            ⁡            (            x            )            +            sin            ⁡            (            x            )            cos            ⁡            (            n            x            )            )            −            (            sin            ⁡            (            n            x            )            cos            ⁡            (            x            )            −            sin            ⁡            (            x            )            cos            ⁡            (            n            x            )            )                                2            sin            ⁡            (            x            )            sin            ⁡            (            (            n            −            1            )            x            )            sin            ⁡            (            (            n            +            1            )            x            )                                              =                                          2            sin            ⁡            (            x            )            cos            ⁡            (            n            x            )                                2            sin            ⁡            (            x            )            sin            ⁡            (            (            n            −            1            )            x            )            sin            ⁡            (            (            n            +            1            )            x            )                                              =                                          cos            ⁡            (            n            x            )                                sin            ⁡            (            (            n            −            1            )            x            )            sin            ⁡            (            (            n            +            1            )            x            )                              And we&#039;re ready

Alissa Hutchinson · Accepted Answer

Hint: Observe                                          cos            ⁡            3            x                                sin            ⁡            2            x            sin            ⁡            4            x                          =                  1                      2            sin            ⁡            x                                    (                      1                          sin              ⁡              2              x                                −                      1                          sin              ⁡              4              x                                )                    since                              1                      sin            ⁡            2            x                          −                  1                      sin            ⁡            4            x                          =                                                   sin            ⁡            4            x            −            sin            ⁡            2            x                                sin            ⁡            2            x            sin            ⁡            4            x                                              =                                                   sin            ⁡            3            x            cos            ⁡            x            +            sin            ⁡            x            cos            ⁡            3            x            −            sin            ⁡            3            x            cos            ⁡            x            +            sin            ⁡            x            cos            ⁡            3            x                                sin            ⁡            2            x            sin            ⁡            4            x                                              =                                                   2            cos            ⁡            3            x            sin            ⁡            x                                sin            ⁡            2            x            sin            ⁡            4            x                          .            Telescoping sum.

i have to prove this cos ⁡ 3 x </mrow>

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Answer & Explanation

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