Let f ( x ) = sin − 1 ⁡ ( 2 x 1 + x 2 ) − ∞ &lt; x &lt; ∞ .Show that,(a) f ( x ) = 2 tan − 1 ⁡ ( x ) for − 1 ≤ x ≤ 1 and(b) f ( x ) = π − 2 tan − 1 ⁡ ( x ) for x ≥ 1.

Question

Let  f  (  x  )  =      sin          −      1        ⁡      (                  2        x                    1        +                  x          2                      )          −  ∞  &amp;lt;  x  &amp;lt;  ∞  .Show that,(a)   f  (  x  )  =  2      tan          −      1        ⁡  (  x  ) for   −  1  ≤  x  ≤  1 and(b)   f  (  x  )  =  π  −  2      tan          −      1        ⁡  (  x  ) for   x  ≥  1.

Ryan Davies · Accepted Answer

sin          −      1        ⁡      (                  2        x                    1        +                  x          2                      )    =  θ      ⟺      sin  ⁡  (  θ  )  =            2      x              1      +              x        2                ∧    −      π    2    ≤  θ  ≤      π    2  Note that  cos  ⁡  (  θ  )  =      |                  1        −                  x          2                            1        +                  x          2                      |  When       |    x      |    ≤  1  tan  ⁡  (  θ      /    2  )  =            sin      ⁡      (      θ      )              1      +      cos      ⁡      (      θ      )        =                    2        x                    1        +                  x          2                            1      +                        1          −                      x            2                                    1          +                      x            2                                =  xTherefore,  θ  =  2      tan          −      1        ⁡  (  x  )When       |    x      |    &amp;gt;  1  tan  ⁡  (  θ      /    2  )  =            sin      ⁡      (      θ      )              1      +      cos      ⁡      (      θ      )        =                    2        x                    1        +                  x          2                            1      −                        1          −                      x            2                                    1          +                      x            2                                =  1      /    xTherefore,  θ  =  2      tan          −      1        ⁡  (  1      /    x  )Thus,      sin          −      1        ⁡      (                  2        x                    1        +                  x          2                      )    =      {                            2                      tan                          −              1                                ⁡          (          x          )                          if                       |                    x                      |                    ≤          1                                      2                      tan                          −              1                                ⁡          (          1                      /                    x          )                          if                       |                    x                      |                    &amp;gt;          1                        and      tan          −      1        ⁡  (  1      /    x  )  =      {                                        π            2                    −                      tan                          −              1                                ⁡          (          x          )                          if           x          &amp;gt;          0                                      −                      π            2                    −                      tan                          −              1                                ⁡          (          x          )                          if           x          &amp;lt;          0

Jared Lowe · Accepted Answer

Show that       f    ′    (  x  )  =      2          1      +              x        2            , hence f(x) and   2  arctan  ⁡  (  x  ) differ only by a constant. Then substitute   x  =  0 to show that this constant is 0.

Let f(x)=sin^(−1)((2x)/(1+x^2)) Show that f(x)=2 tan^(−1)(x)

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