Suppose sin 4 ( α ) a + cos 4 ( α ) b = 1 a + b for some a , b ≠ 0Why does sin 8 ( α ) a 3 + cos 8 ( α ) b 3 = 1 ( a + b ) 3

Question

Suppose                                          sin                    4                    (      α      )        a    +                                          cos                    4                    (      α      )        b    =      1          a      +      b      for some   a  ,  b  ≠  0Why does                                          sin                    8                    (      α      )                                a          3                      +                                          cos                    8                    (      α      )                                b          3                      =      1                                        (            a            +            b            )                    3

Braedon Rivas · Accepted Answer

There is a very direct way to solve this. First, use the given equation to find, for example,       sin    2    ⁡  α:  x  =      sin    2    ⁡  α      1    a        x    2    +      1    b    (  1  −  x      )    2    =      1          a      +      b      Solving this easy quadratic, we get a single root:  x  =      a          a      +      b        =      sin    2    ⁡  αIt follows:  1  −  x  =      b          a      +      b        =      cos    2    ⁡  αNow substitute these values into the second equation and see that it holds

Cory Patrick · Accepted Answer

The general expression is:                    sin                  4          n                    ⁡      θ              a              2        n        −        1              +                    cos                  4          n                    ⁡      θ              b              2        n        −        1              =      1          (      a      +      b              )                  2          n          −          1                         ,        n  ∈      N  From the given relation, it can be written                    (        a        +        b        )                  (                                                    sin                4                            ⁡              θ                        a                    +                                                    cos                4                            ⁡              θ                        b                    )                                    =                              (                          sin              2                        ⁡            θ            +                          cos              2                        ⁡            θ            )                    2                                            ⟹                          sin          4                ⁡        θ        +                  cos          4                ⁡        θ        +                  b          a                          sin          4                ⁡        θ        +                  a          b                          cos          4                ⁡        θ                            =                  sin          4                ⁡        θ        +        2                  sin          2                ⁡        θ                  cos          2                ⁡        θ        +                  cos          4                ⁡        θ                                    ⟹                                                                                                                                         (                                          b                a                                                    sin              2                        ⁡            θ            −                                          a                b                                                    cos              2                        ⁡            θ            )                    2                                    =        0                                    ⟹                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             sin              2                        ⁡            θ                    a                                    =                                            cos              2                        ⁡            θ                    b                                            ⟹                                                                                                                                                                                                                                                                                                                                                    sin              2                        ⁡            θ                    a                =                                            cos              2                        ⁡            θ                    b                                    =                                            sin              2                        ⁡            θ            +                          cos              2                        ⁡            θ                                a            +            b                              From this, it can be concluded that      sin    2    ⁡  θ  =      a          a      +      b        ;               cos    2    ⁡  θ  =      b          a      +      b          .It&#039;s a matter of substituting the values to prove the general and the desired expression.

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