For n odd and positive integer, proof that sin ⁡ ( n x ) sin ⁡ ( x ) = ( − 4 ) ( n − 1 ) / 2 ∏ 1 ≤ j ≤ ( n − 1 ) / 2 ( sin 2 ⁡ ( x ) − sin 2 ⁡ ( 2 π j n ) )It says this is elementary, and suggest first proving that sin ⁡ ( n x ) sin ⁡ ( x ) is a polynomial of degree ( n − 1 ) / 2 in sin 2 ⁡ ( x ), then remark that it has sin 2 ⁡ ( 2 π j n ) as roots, then comparing coefficients of e i ( n − 1 ) / 2 to get ( − 4 ) ( n − 1 ) / 2 .

Question

For n odd and positive integer, proof that             sin      ⁡      (      n      x      )              sin      ⁡      (      x      )        =  (  −  4      )          (      n      −      1      )              /            2            ∏          1      ≤      j      ≤      (      n      −      1      )              /            2        (        sin    2    ⁡  (  x  )  −      sin    2    ⁡  (            2      π      j        n    )    )It says this is elementary, and suggest first proving that             sin      ⁡      (      n      x      )              sin      ⁡      (      x      )       is a polynomial of degree   (  n  −  1  )      /    2 in       sin    2    ⁡  (  x  ), then remark that it has       sin    2    ⁡  (            2      π      j        n    ) as roots, then comparing coefficients of       e          i      (      n      −      1      )              /            2       to get   (  −  4      )          (      n      −      1      )              /            2      .

Shyla Larson · Accepted Answer

Step 1On the left you have an expression of the form                    z        n            −              z                  −          n                            z      −              z                  −          1                      =      z          n      −      1        +      z          n      −      3        +      z          n      −      5        +  .  .  .  +      z          1      −      n      which divides perfectly and has the unit roots of degree 2n except   ±  1 as its roots. Because of symmetry, for every root       λ   one has also                     λ            ¯      ,   −      λ  ,   −                    λ            ¯       as roots. Because of n odd the special case       λ    =  i does not occur, so that all roots can be grouped into these quadrupels.Instead of pairing conjugate roots one can here also pair mirror roots       λ   and   −                    λ            ¯       relative to the imaginary axis resulting in                                                        z              n                        −                          z                              −                n                                                          z            −                          z                              −                1                                                                        =                  ∏                      k            =            1                                (            n            −            1            )                          /                        2                          (        z        −                  e                      i            π                        k                          /                        n                          )        (        1        +                  z                      −            1                                    e                      −            i            π                        k                          /                        n                          )                (        z        −                  e                      −            i            π                        k                          /                        n                          )        (        1        +                  z                      −            1                                    e                      i            π                        k                          /                        n                          )                                          =                  ∏                      k            =            1                                (            n            −            1            )                          /                        2                          (        z        −                  z                      −            1                          −                  e                      i            π                        k                          /                        n                          +                  e                      −            i            π                        k                          /                        n                          )        (        z        −                  z                      −            1                          +                  e                      i            π                        k                          /                        n                          −                  e                      −            i            π                        k                          /                        n                          )                                          =                  ∏                      k            =            1                                (            n            −            1            )                          /                        2                          (        z        −                  z                      −            1                          −        2        i        sin        ⁡        (        π                k                  /                n        )        )        (        z        −                  z                      −            1                          +        2        i        sin        ⁡        (        π                k                  /                n        )        )            Step 2Replacing   z  =      e          i      x      , collecting exponentials into sine terms and applying a binomial formula gives a formula similar to the stated result.Check that also using       λ    =      e          i      2      π            k              /            n       gives one representant per root quadruple, here it is also important that n is odd. This will finally result in the stated formula.

For n odd and positive integer, proof that (sin(nx))/(sin(x))=(-4)^{(n-1)/2} prod_{1 <= j <= (n-1)/2}(\, sin^2(x)-sin^2((2pi j)/(n))\,)

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