Domain of y=arcsin(tanx) (1 Viewer)

Kimyia · Aug 7, 2012

Can anyone please show me how you would work out the domain of y=arcsin(tanx). I'm not sure how to go about working it out. Thank you!

SpiralFlex · Aug 7, 2012

$We know that the arc of sine must hav a domain in the range of [-1, 1]. Now instead of our normal arcsin x, we know have a new function forming a composite in tan x. But we can apply the same technique.$

$$So this means,$\; -1 \leq \tan x \leq 1$

$Graphically where is tangent x in between the y values of -1 and 1? There are many solutions, in fact an infinite amount. One solution is a a quarter of pi, and is minus a quarter of pi. We need a way to generalise the amount of solutions. So x will lie inbetween the aforementioned.$

$You can see by the tangent graph and graphing y=-1 and y=1, the solutions will lie,$

$\therefore -\frac{\pi}{4}+n\pi \leq x \leq \frac{\pi}{4}+n\pi$

$Where n is an integer.$

Off study English kimmy. GL for tomorrow!

Kimyia · Aug 7, 2012

Thank you Spi! That makes sense now

asianese · Aug 7, 2012

Well the domain of tanx is all x except x=/=k pi/2 , k integral.

Now sin^-1 is only defined for -1 to 1. So you can only put in values of 'u' where u=tanx from -1 to 1. Where does -1 <= tanx <= 1? Well from -pi/4 < x < pi/4. and then the left branch of that, -5pi/4 <= x <= -3pi/4 and then the branch to the right, 3pi/4 <=x <= 5pi/4 etc.

asianese · Aug 7, 2012

LOL too slow for the spi.

Kimyia · Aug 7, 2012

its alright, thank you for your help!

Domain of y=arcsin(tanx) (1 Viewer)

Kimyia

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SpiralFlex

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Kimyia

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asianese

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asianese

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Kimyia

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