Prove: 1 sin ⁡ 2 x + 1 sin ⁡ 4 x + ⋯ + 1 sin ⁡ 2 n x = cot ⁡ x − cot ⁡ 2 n xwhere n ∈ N and x not a multiple of π 2 k for any k ∈ N

Question

Prove:       1          sin      ⁡      2      x        +      1          sin      ⁡      4      x        +  ⋯  +      1          sin      ⁡              2        n            x        =  cot  ⁡  x  −  cot  ⁡      2    n    xwhere   n  ∈      N   and x not a multiple of       π          2      k       for any   k  ∈      N

Jaylee Dodson · Accepted Answer

If we take the equation            1              sin        ⁡        2        x              +            1              sin        ⁡        4        x              +  ⋯  +            1              sin        ⁡                  2          n                x              =  cot  ⁡  x  −  cot  ⁡      2    n    xand replace x with 2x, we get            1              sin        ⁡        4        x              +            1              sin        ⁡        8        x              +  ⋯  +            1              sin        ⁡                  2                      n            +            1                          x              =  cot  ⁡  2  x  −  cot  ⁡      2          n      +      1        xAdding   1      /    sin  ⁡  2  x to both sides now gives            1              sin        ⁡        2        x              +            1              sin        ⁡        4        x              +  ⋯  +            1              sin        ⁡                  2                      n            +            1                          x              =            1              sin        ⁡        2        x              +  cot  ⁡  2  x  −  cot  ⁡      2          n      +      1        xso, to complete the inductive step, it suffices to prove            1              sin        ⁡        2        x              +  cot  ⁡  2  x  =  cot  ⁡  xBut this is the base case, just rewritten with the   cot  ⁡  2  x moved to the left hand side! So let&#039;s prove it:                                          1                          sin              ⁡              2              x                                      +        cot        ⁡        2        x                            =                                            1              +              cos              ⁡              2              x                                      sin              ⁡              2              x                                                                        =                                            1              +              (              2                              cos                2                            ⁡              x              −              1              )                                      2              sin              ⁡              x              cos              ⁡              x                                                                        =                                            cos              ⁡              x                                      sin              ⁡              x                                                                        =        cot        ⁡        x

Prove: 1 sin ⁡ 2 x </mrow> </mfrac> +

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