Prove that: sec 2 ⁡ 20 ∘ + sec 2 ⁡ 40 ∘ + sec 2 ⁡ 80 ∘ = 3 6

Question

Prove that:       sec    2    ⁡      20    ∘    +      sec    2    ⁡      40    ∘    +      sec    2    ⁡      80    ∘    =      3    6

Harper Heath · Accepted Answer

Starting like dxiv,let   a  =  cos  ⁡      20    ∘    ,  b  =  −  cos  ⁡      40    ∘    ,  c  =  −  cos  ⁡      80    ∘    ,As   cos  ⁡  (  3  ⋅      20    ∘    )  =            1      2        cos  ⁡  (  3  ⋅      40    ∘    )  =  −            1      2        cos  ⁡  (  3  ⋅      80    ∘    )  =  −            1      2      As   cos  ⁡  3  x  =  cos  ⁡      60    ∘    ,  3  x  =      360    ∘    m  ±      60    ∘   where m is any integer.    ⟹    x  =      120    ∘    m  +      20    ∘   where   m  ≡  −  1  ,  0  ,  1    (  mod    3  )Now as   cos  ⁡  3  x  =  4      cos    3    ⁡  x  −  3  cos  ⁡  xThe roots of   4      t    3    −  3  t  −            1      2        =  0 are a,b,cand we need to find             1              a        2              +            1              b        2              +            1              c        2              =                              a          2                          b          2                +                  b          2                          c          2                +                  c          2                          a          2                            (        a        b        c                  )          2                      =                    (        a        b        +        b        c        +        c        a                  )          2                −        2        a        b        c        (        a        +        b        +        c        )                    (        a        b        c                  )          2                    By Vieta&#039;s formula  a  +  b  +  c  =            0      4        a  b  +  b  c  +  c  a  =  −            3      4        a  b  c  =  −            1              2        ⋅        4

Prove that: sec 2 </msup> ⁡ 20 ∘<!-- ∘ --

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