There exist constants b, and d such that ( sin ⁡ x ) 7 = a sin ⁡ 7 x + b sin ⁡ 5 x + c sin ⁡ 3 x + d sin ⁡ x for all angles x. Find d

Question

There exist constants   b, and   d such that  (  sin  ⁡  x      )    7    =  a  sin  ⁡  7  x  +  b  sin  ⁡  5  x  +  c  sin  ⁡  3  x  +  d  sin  ⁡  x for all angles   x. Find   d

thatuglygirlyu · Accepted Answer

Evaluate the equation for different   x:                                            π          6                                    →        −        a        +        b        +        2        c                    +        d                            =                  1          64                                              π          4                                    →        −        a        −        b        +                 c                    +        d                            =                  8          64                                              π          3                                    →        +        a        −        b                    +        d                            =                  27          64                                              π          2                                    →        −        a        +        b        −                 c                    +        d                            =                  64          64                    (after normalization of the coefficient of   d)Elimination by the combination   (  1  )  +  3  ×  (  3  )  +  2  ×  (  4  ) yields  6  d  =      210    64    .The evaluation for   x  =            π      4       was not even necessary.

Jorden Pace · Accepted Answer

sin          7                ⁡        x                                          =                  sin          6                ⁡        x        ⋅        sin        ⁡        x                                          =                  1          16                ⋅        16                  sin          6                ⁡        x        ⋅        sin        ⁡        x                                          =                  1          16                ⋅        (        4                  sin          3                ⁡        x                  )          2                ⋅        sin        ⁡        x                                          =                  1          16                ⋅        (        3        sin        ⁡        x        −        sin        ⁡        3        x                  )          2                ⋅        sin        ⁡        x                                          =                  1          16                ⋅        (        9                  sin          2                ⁡        x        −        6        sin        ⁡        x        sin        ⁡        3        x        +                  sin          2                ⁡        3        x        )        ⋅        sin        ⁡        x                                          =                  1          16                ⋅        [                  9          2                (        1        −        cos        ⁡        2        x        )        −        3        (        cos        ⁡        2        x        −        cos        ⁡        4        x        )        +                  1          2                (        1        −        cos        ⁡        6        x        )        ]        ⋅        sin        ⁡        x                                          =                  9          32                (        sin        ⁡        x        −        sin        ⁡        x        ⋅        cos        ⁡        2        x        )        −                  3          16                (        sin        ⁡        x        ⋅        cos        ⁡        2        x        −        sin        ⁡        x        ⋅        cos        ⁡        4        x        )        +                  1          32                (        sin        ⁡        x        −        sin        ⁡        x        ⋅        cos        ⁡        6        x        )                                          =                  9          32                [        sin        ⁡        x        −                  1          2                (        sin        ⁡        3        x        −        sin        ⁡        x        )        ]        −                  3          32                [        (        sin        ⁡        3        x        −        sin        ⁡        x        )        −        (        sin        ⁡        5        x        −        sin        ⁡        3        x        )        ]        +                  1          32                [        sin        ⁡        x        −                  1          2                (        sin        ⁡        7        x        −        sin        ⁡        5        x        )        ]            The coefficient of   sin  ⁡  x in the above expression is            9      32        +          9      64        +          3      32        +          1      32        =            35      64

There exist constants b , and d such that ( sin ⁡ x ) 7

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