- #1
Rectifier
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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 ##
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)) ##.
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 ##
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)) ##.
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