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Show that ##\sin(\arctan x) < x < \tan(\arcsin x)##

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  1. Aug 17, 2016 #1

    Rectifier

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    Gold Member

    The problem
    For ##0 < x < 1 ## . Show that
    $$ \sin(\arctan x) < x < \tan(\arcsin x) $$

    The attempt

    I know that ## \sin x < x < \tan x ## is true for ## 0 < x < \ \pi / 2 ##

    220px-Sinxoverx.png
    x - is by definition the length DA (in radians)

    I draw a right triangle with sides x and 1 and with hypotenuse ## \sqrt{x^2+1} ##

    ## \sin v = \frac{x}{\sqrt{x^2+1}} ##

    ## \tan v = \frac{x}{1} \Rightarrow v = \arctan \ x ##

    ## \sin (\arctan \ x) = \frac{x}{\sqrt{x^2+1}} ##


    I am trying to write an expression for ## \tan(\arcsin(x)) ##

    ## \sin v = \frac{x}{\sqrt{x^2+1}} \Rightarrow v = \arcsin \left( \frac{x}{\sqrt{x^2+1}} \right)##

    ## \tan v = x \Rightarrow \tan \left( \arcsin \left( \frac{x}{\sqrt{x^2+1}} \right) \right) = x##

    But I fail. Please Help me write an expression for ## \tan(\arcsin(x)) ##.
     
    Last edited: Aug 17, 2016
  2. jcsd
  3. Aug 17, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    I don't think this is necessary. You can compare arctan(x) and arcsin(x) with x.
     
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