f ( 2 π 7 ) + f ( 4 π 7 ) + f ( 6 π 7 ) = 1let f ( x ) = 1 1 + 2 cos ⁡ x prove that : f ( 2 π 7 ) + f ( 4 π 7 ) + f ( 6 π 7 ) = 1My Try : f ( 2 π 7 ) = 1 1 + 2 cos ⁡ ( 2 π 7 ) f ( 2 π 7 ) = 1 1 + 2 cos ⁡ ( 4 π 7 ) f ( 2 π 7 ) = 1 1 + 2 cos ⁡ ( 6 π 7 ) L = 1 1 + 2 cos ⁡ ( 6 π 7 ) + 1 1 + 2 cos ⁡ ( 4 π 7 ) + 1 1 + 2 cos ⁡ ( 2 π 7 ) what now ?

Question

f  (            2      π        7    )  +  f  (            4      π        7    )  +  f  (            6      π        7    )  =  1let  f  (  x  )  =      1          1      +      2      cos      ⁡      x      prove that :  f  (            2      π        7    )  +  f  (            4      π        7    )  +  f  (            6      π        7    )  =  1My Try :  f  (            2      π        7    )  =      1          1      +      2      cos      ⁡      (                        2          π                7            )        f  (            2      π        7    )  =      1          1      +      2      cos      ⁡      (                        4          π                7            )        f  (            2      π        7    )  =      1          1      +      2      cos      ⁡      (                        6          π                7            )        L  =      1          1      +      2      cos      ⁡      (                        6          π                7            )        +      1          1      +      2      cos      ⁡      (                        4          π                7            )        +      1          1      +      2      cos      ⁡      (                        2          π                7            )      what now ?

Clarissa Adkins · Accepted Answer

Use factor       z    7    −  1 into linear and quadratic factors and prove that   cos  ⁡  (  π      /    7  )  ⋅  cos  ⁡  (  2  π      /    7  )  ⋅  cos  ⁡  (  3  π      /    7  )  =  1      /    8or if   7  x  =  2  m  π where   m is any integer  sin  ⁡  4  x  =  sin  ⁡  (  2  m  π  −  3  x  )  =  −  sin  ⁡  3  x    ⟹    4  sin  ⁡  x  cos  ⁡  x  cos  ⁡  2  x  =  sin  ⁡  x  (  4      sin    2    ⁡  x  −  3  )    ⟹    4  sin  ⁡  x  cos  ⁡  x  (  2      cos    2    ⁡  x  −  1  )  =  sin  ⁡  x  {  4  (  1  −      cos    2    ⁡  x  )  −  1  }So, the roots of   4  cos  ⁡  x  (  2      cos    2    ⁡  x  −  1  )  =  4  (  1  −      cos    2    ⁡  x  )  −  1    ⟺    8      cos    3    ⁡  x  +  4      cos    2    ⁡  x  −  4  cos  ⁡  x  −  3  =  0              (  1  )are   7  x  =  2  m  π where   m  ≡  ±  1  ,  ±  2  ,  ±  3    (  mod    7  )Now if   u  =            1              1        +        2        cos        ⁡        x                ⟺    cos  ⁡  x  =  ?Replace the values of   cos  ⁡  x in   (  1  )

Levi Rasmussen · Accepted Answer

Let   2  cos  ⁡            2      π        7    =  a,   2  cos  ⁡            4      π        7    =  b and   2  cos  ⁡            6      π        7    =  cHence,  a  +  b  +  c  =            2      sin      ⁡              π        7            cos      ⁡                        2          π                7            +      2      sin      ⁡              π        7            cos      ⁡                        4          π                7            +      2      sin      ⁡              π        7            cos      ⁡                        6          π                7                    sin      ⁡              π        7              =  =            sin      ⁡                        3          π                7            −      sin      ⁡              π        7            +      sin      ⁡                        5          π                7            −      sin      ⁡                        3          π                7            +      sin      ⁡                        7          π                7            −      sin      ⁡                        5          π                7                    sin      ⁡              π        7              =  −  1  ;  a  b  +  a  c  +  b  c  =  4      (    cos    ⁡                  2        π            7        cos    ⁡                  4        π            7        +    cos    ⁡                  2        π            7        cos    ⁡                  6        π            7        +    cos    ⁡                  4        π            7        cos    ⁡                  6        π            7        )    =  =  2      (    cos    ⁡                  6        π            7        +    cos    ⁡                  2        π            7        +    cos    ⁡                  6        π            7        +    cos    ⁡                  4        π            7        +    cos    ⁡                  4        π            7        +    cos    ⁡                  2        π            7        )    =  −  2and  a  b  c  =  8  cos  ⁡            2      π        7    cos  ⁡            4      π        7    cos  ⁡            6      π        7    =            8      sin      ⁡                        2          π                7            cos      ⁡                        2          π                7            cos      ⁡                        4          π                7            cos      ⁡                        8          π                7                    sin      ⁡                        2          π                7              =            sin      ⁡                        16          π                7                    sin      ⁡                        2          π                7              =  1.Id est,  f      (                  2        π            7        )    +  f      (                  4        π            7        )    +  f      (                  6        π            7        )    =      ∑          c      y      c            1          1      +      a        =                    ∑                  c          y          c                    (      a      +      1      )      (      b      +      1      )                      ∏                  c          y          c                    (      1      +      a      )        =  =            a      b      +      a      c      +      b      c      +      2      (      a      +      b      +      c      )      +      3              1      +      a      +      b      +      c      +      a      b      +      a      c      +      b      c      +      a      b      c        =            −      2      −      2      +      3              1      −      1      −      2      +      1        =  1.Done!

let f(x)=1/(1+2cosx) prove that : f((2 pi)/(7))+f((4 pi)/(7))+f((6pi)/(7})=1

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