Solve: 1 sin ⁡ ( π / n ) = 1 sin ⁡ ( 2 π / n ) + 1 sin ⁡ ( 3 π / n )

Question

Solve:       1          sin      ⁡      (      π              /            n      )        =      1          sin      ⁡      (      2      π              /            n      )        +      1          sin      ⁡      (      3      π              /            n      )

Raven Hawkins · Accepted Answer

Using the formula  sin  ⁡  A  −  sin  ⁡  B  =  2  sin  ⁡      (                  A        −        B            2        )    cos  ⁡      (                  A        +        B            2        )  we have                    sin        ⁡                                            4              π                        n                                              =        sin        ⁡                                            3              π                        n                                              sin        ⁡                                            4              π                        n                          −        sin        ⁡                                            3              π                        n                                              =        0                            2        sin        ⁡                  (                      π                          2              n                                )                cos        ⁡                  (                                    7              π                                      2              n                                )                                    =        0            then                    sin        ⁡                  (                                    π                              2                n                                              )                            =                    0                                                  π                          2              n                                                  ∈                    {        k        π        :        k        ∈                              Z                                ≠            0                          }                            n                    ∈                              {                                    1                              2                k                                              :          k          ∈                                    Z                                      ≠              0                                }                         or                       cos        ⁡                  (                                                    7                π                                            2                n                                              )                            =                    0                                                  π                          2              n                                                  ∈                              {          (          2          k          +          1          )                      π            2                    :          k          ∈                      Z                    }                                    n                    ∈                              {                                    7                              2                k                +                1                                              :          k          ∈                      Z                    }                    Thus  n  ∈      {                  1                  2          k                      :    k    ∈                  Z                    ≠        0              }    ∪      {                  7                  2          k          +          1                      :    k    ∈          Z        }

Solve: 1/(sin(pi /n))=(1)/(sin(2 pi/n))+(1)/(sin(3 pi/n))

Answered question

Answer & Explanation

New Questions in Trigonometry