I would like to show that sin − 1 is an increasing function. In other words, I want to show that x 1 ≤ x 2 → sin − 1 ⁡ ( x 1 ) ≤ sin − 1 ⁡ ( x 2 ) .I fail to see how to start this. Also, if possible, I would like to show this without using other functions (e.g. sin) known to be increasing/decreasing.

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I would like to show that       sin          −      1       is an increasing function. In other words, I want to show that      x          1        ≤      x          2        →      sin          −      1        ⁡  (      x          1        )  ≤      sin          −      1        ⁡  (      x          2        )  .I fail to see how to start this. Also, if possible, I would like to show this without using other functions (e.g. sin) known to be increasing/decreasing.

Elias Flores · Accepted Answer

HintIf f is increasing, then       f          −      1       is increasing.Indeed, let   u  ≤  v and x,y such that   u  =  f  (  x  ) and   v  =  f  (  y  ). Then, since f is increasing,  u  ≤  v    ⟹    f  (  x  )  ≤  f  (  y  )    ⟹    x  ≤  y    ⟹        f          −      1        (  u  )  ≤      f          −      1        (  v  )  .PS: Use   arcsin instead of       sin          −      1

Janet Forbes · Accepted Answer

HINT:      d          d      x        (      sin          −      1        ⁡  x  )  =      1          1      −              x        2              &amp;gt;  0        ∀        −  1  &amp;lt;  x  &amp;lt;  1

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