Solving without squaringIf sec ⁡ θ − csc ⁡ θ = 4 3 , then prove that θ = 1 2 arcsin ⁡ 3 4 I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.The closest thing I reached is sin ⁡ θ = 3 ( 1 − 2 sin 2 ⁡ ( θ / 2 ) ) 8 sin 2 ⁡ ( θ / 2 ) − 1 I also obtained sin ⁡ 2 θ = − 3 2 ( 1 − tan ⁡ θ / 2 ) 2 1 + tan 2 ⁡ ( θ / 2 )

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Solving without squaringIf   sec  ⁡  θ  −  csc  ⁡  θ  =            4      3      , then prove that   θ  =            1      2        arcsin  ⁡                    3        4            I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.The closest thing I reached is   sin  ⁡  θ  =                    3        (        1        −        2                  sin          2                ⁡        (        θ                  /                2        )        )                    8                  sin          2                ⁡        (        θ                  /                2        )        −        1            I also obtained   sin  ⁡  2  θ  =                    −        3            2                          (        1        −        tan        ⁡        θ                  /                2                  )          2                            1        +                  tan          2                ⁡        (        θ                  /                2        )

embraci4i · Accepted Answer

Let   sin  ⁡      θ    −  cos  ⁡      θ    =  tThus,   sin  ⁡      θ    cos  ⁡      θ    =            1      −              t        2              2   and we have            2      t              1      −              t        2              =      4    3  or  2      t    2    +  3  t  −  2  =  0and since by C-S      |    sin  ⁡      θ    −  cos  ⁡      θ        |    ≤      (          1      2        +          1      2        )    (          sin      2        ⁡    θ    +          cos      2        ⁡    θ    )    =      2    ,we obtain   t  =      1    2    .Thus,  sin  ⁡  2  θ  =  2  ⋅            1      −              1        4              2    =      3    4  and we are done!

Jean Farrell · Accepted Answer

⟹    3  (  sin  ⁡  θ  −  cos  ⁡  θ  )  =  4  sin  ⁡  θ  cos  ⁡  θ  =  2  sin  ⁡  2  θ    ⟺    3      2    sin  ⁡      (    θ    −                  π        4              )    =  2  cos  ⁡  2      (    θ    −                  π        4              )    =  2      (    1    −    2          sin      2        ⁡          (      θ      −                        π          4                    )        )      ⟺    3      2    sin  ⁡      (    θ    −                  π        4              )    =  2  cos  ⁡  2      (    θ    −                  π        4              )    =  2      (    1    −    2          sin      2        ⁡          (      θ      −                        π          4                    )        )      ⟺    4      sin    2    ⁡      (    θ    −                  π        4              )    +  3      2    sin  ⁡      (    θ    −                  π        4              )    −  2  =  0  sin  ⁡      (    θ    −                  π        4              )    =                    −        3                  2                ±        5                  2                    8        =  −      2     or             1              2                  2                    As   sin  ⁡      (    θ    −                  π        4              )    ≥  −  1  ,  sin  ⁡      (    θ    −                  π        4              )    =            1              2                  2                        ⟹    sin  ⁡  2  θ  =  cos  ⁡  2      (    θ    −                  π        4              )    =  1  −  2      sin    2    ⁡      (    θ    −                  π        4              )    =  ?

Solving without squaring If sec theta - csc theta = 4/3, then prove that theta= 1/2 arcsin (3/4)

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