Let x =2sin(theta), -pi/2< theta

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Let x =2sin(theta), -pi/2< theta

Mathematics
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\[\frac{ x }{ \sqrt{4-x^2} }\]
\[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\]
@UsukiDoll not sure where to go with this one! \[\frac{ 2\sin(\theta) }{ 2-2\sin(\theta) }\] ??

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hmm try factoring the 2 out of the denominator
er..how do I do this xP
is it 2(1-sintheta) ?
in the denominator of course.
\[\frac{ 2\sin(\theta) }{ 2(1-\sin(\theta)) }\] then cancel the 2
and we're left with (sin(theta))/(1-sin(theta))
and then it turns into a mess... geez...
haha aii
I'm wondering if that's it... It's been a while. but I know that the domain is restricted counter clockwise 90 degrees to clockwise 90 degrees.
what if we ^2 the whole thing?
wait...
we can add 180 later.. right? to get it into the correct domain.
\[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\] \[\frac{ 2\sin(\theta) }{ \sqrt{4-4\sin^2(\theta))} }\] \[\[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]\]
\[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]
\[\cos^2x+\sin^2x=1 \] \[\cos^2x = 1-\sin^2x\] \[\frac{ 2\sin(\theta) }{ \sqrt{4(\cos^2(\theta))} }\]
\[\frac{2\sin(\theta)}{2\cos(\theta)}\]
\[\Large\rm \sqrt{4-4\sin^2x}\ne 2-2\sin x\]You silly billy Miriam -_-
I got tangent theta in return?!
I saw something weird when I saw that I mean can't we yank the 4 out and use a trig identity
Woah how'd you get tangent out of there? because the sin and cos are the same here?
\[\sqrt{4-4\sin^2x} \rightarrow \sqrt{4(1-\sin^2x)}\]
\[\sqrt{4\cos^2x} \rightarrow 2cosx \]
\[\frac{2sinx}{2cosx} \rightarrow \frac{sinx}{cosx} \rightarrow tanx\]
oou..I see!
you've skipped a step and that trickled down later on
\[(2\sin(\theta))^2 \rightarrow (2\sin(\theta))(2\sin(\theta))\]
\[4\sin^2(\theta)\]
you've neglected to take the 2sin(theta) to the second power... that's why everything became nonsense
Sorry, my internets really bad. yeah I was jumping around too much without checking o.0
Sowriee

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