Finding value of ∏ k = 1 n − 1 cos ⁡ k π 2 n = n 2 n − 1 Try: let z 2 n = 1 ⇒ z = ( 1 ) 1 2 n = e k π n So z 2 n − 1 = ∏ k = 1 2 n ( z − e k π n ) = ( z 2 − 1 ) ∏ k = 1 n − 1 ( z − e k π n ) ⋅ ( z − e − k π n )So z 2 n − 1 = ( z 2 − 1 ) ∏ k = 1 n − 1 ( z 2 − 2 z cos ⁡ ( k π / n ) + 1Could some help me how to prove my original formula

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Finding value of       ∏          k      =      1              n      −      1        cos  ⁡            k      π              2      n        =            n              2              n        −        1            Try: let      z          2      n        =  1  ⇒  z  =  (  1      )                  1                  2          n                      =      e                            k          π                n            So      z          2      n        −  1  =      ∏          k      =      1              2      n        (  z  −      e                            k          π                n              )  =  (      z    2    −  1  )      ∏          k      =      1              n      −      1        (  z  −      e                            k          π                n              )  ⋅  (  z  −      e                            −          k          π                n              )So      z          2      n        −  1  =  (      z    2    −  1  )      ∏          k      =      1              n      −      1        (      z    2    −  2  z  cos  ⁡  (  k  π      /    n  )  +  1Could some help me how to prove my original formula

lodosr · Accepted Answer

You are pretty close to derive the answer yourself. Let S be the product at hand, we have                              S          2                                    =                  ∏                      k            =            1                                n            −            1                                    cos          2                ⁡                              k            π                                2            n                          =                  ∏                      k            =            1                                n            −            1                                    1          2                          (          1          +          cos          ⁡                                    k              π                        n                    )                =                  4                      1            −            n                                    ∏                      k            =            1                                n            −            1                                    (          2          +          2          cos          ⁡                                    k              π                        n                    )                                                  =                  4                      1            −            n                                    lim                      z            →            −            1                                    ∏                      k            =            1                                n            −            1                                    (          1          +                      z            2                    −          2          z          cos          ⁡                                    k              π                        n                    )                =                  4                      1            −            n                                    lim                      z            →            −            1                                                              z                              2                n                                      −            1                                              z              2                        −            1                          =                  n                      4                          n              −              1                                          Since S is clearly positive, taking square root give us   S  =            n              2              n        −        1

Finding value of <munderover> ∏ <mrow class="MJX-TeXAtom-ORD"> k

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