x , y &amp; z are real number such that sin ⁡ x + sin ⁡ y + sin ⁡ z sin ⁡ ( x + y + z ) = cos ⁡ x + cos ⁡ y + cos ⁡ z cos ⁡ ( x + y + z ) = 2find the value ∑ c y c sin ⁡ x sin ⁡ y

Question

x , y &amp;amp; z are real number such that            sin      ⁡              x            +      sin      ⁡              y            +      sin      ⁡              z                    sin      ⁡              (        x        +        y        +        z        )              =            cos      ⁡              x            +      cos      ⁡              y            +      cos      ⁡              z                    cos      ⁡              (        x        +        y        +        z        )              =  2find the value      ∑          c      y      c        sin  ⁡  x  sin  ⁡  y

Conor Frederick · Accepted Answer

I used the Blue&#039;s beautiful idea.Let       e          i      x        =  a,       e          i      y        =  b and       e          i      z        =  c.Hence,   sin  ⁡      x    =            a      −              1        a                    2      i        =                    a        2            −      1              2      a      i       ,   cos  ⁡      x    =                    a        2            +      1              2      a       ,   sin  ⁡      y    =                    b        2            −      1              2      b      i       ,   sin  ⁡      x    =                    c        2            −      1              2      c      i       and   cos  ⁡      x    =                    c        2            +      1              2      c      .Thus,       ∑          c      y      c        sin  ⁡      x    =  2  sin  ⁡  (  x  +  y  +  z  ) gives       ∑          c      y      c        (      a    2    b  c  −  a  b  )  =  2  (      a    2        b    2        c    2    −  1  ) and      ∑          c      y      c        cos  ⁡      x    =  2  cos  ⁡  (  x  +  y  +  z  ) gives       ∑          c      y      c        (      a    2    b  c  +  a  b  )  =  2  (      a    2        b    2        c    2    +  1  ) or ab+ac+bc=2 and a+b+c=2abc.Thus,  −      ∑          c      y      c        sin  ⁡      x    sin  ⁡      y    =      ∑          c      y      c                  (              a        2            −      1      )      (              b        2            −      1      )              4      a      b        =                    ∑                  c          y          c                    c      (              a        2            −      1      )      (              b        2            −      1      )              4      a      b      c        =  =            a      b      c      (      a      b      +      a      c      +      b      c      )      −              ∑                  c          y          c                    (              a        2            b      +              a        2            c      )      +      a      +      b      +      c              4      a      b      c        =            a      b      c      (      a      b      +      a      c      +      b      c      )      −      (      a      +      b      +      c      )      (      a      b      +      a      c      +      b      c      )      +      3      a      b      c      +      a      +      b      +      c              4      a      b      c        =      1    2    −  1  +      3    4    +      1    2    =      3    4

x , y & z are real number such that sin ⁡ <mrow

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