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Babynini
 one year ago
Let x =2sin(theta), pi/2< theta<pi/2
Simplify the expression
x/(sqroot(4x^2)
Babynini
 one year ago
Let x =2sin(theta), pi/2< theta<pi/2 Simplify the expression x/(sqroot(4x^2)

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x }{ \sqrt{4x^2} }\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2\sin(\theta) }{ \sqrt{4(2\sin(\theta))^2} }\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll not sure where to go with this one! \[\frac{ 2\sin(\theta) }{ 22\sin(\theta) }\] ??

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2hmm try factoring the 2 out of the denominator

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0er..how do I do this xP

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0is it 2(1sintheta) ?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0in the denominator of course.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ 2\sin(\theta) }{ 2(1\sin(\theta)) }\] then cancel the 2

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0and we're left with (sin(theta))/(1sin(theta))

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2and then it turns into a mess... geez...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'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.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0what if we ^2 the whole thing?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0we can add 180 later.. right? to get it into the correct domain.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ 2\sin(\theta) }{ \sqrt{4(2\sin(\theta))^2} }\] \[\frac{ 2\sin(\theta) }{ \sqrt{44\sin^2(\theta))} }\] \[\[\frac{ 2\sin(\theta) }{ \sqrt{4(1\sin^2(\theta))} }\]\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ 2\sin(\theta) }{ \sqrt{4(1\sin^2(\theta))} }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos^2x+\sin^2x=1 \] \[\cos^2x = 1\sin^2x\] \[\frac{ 2\sin(\theta) }{ \sqrt{4(\cos^2(\theta))} }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2\sin(\theta)}{2\cos(\theta)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large\rm \sqrt{44\sin^2x}\ne 22\sin x\]You silly billy Miriam _

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I got tangent theta in return?!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I saw something weird when I saw that I mean can't we yank the 4 out and use a trig identity

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Woah how'd you get tangent out of there? because the sin and cos are the same here?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{44\sin^2x} \rightarrow \sqrt{4(1\sin^2x)}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{4\cos^2x} \rightarrow 2cosx \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{2sinx}{2cosx} \rightarrow \frac{sinx}{cosx} \rightarrow tanx\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2you've skipped a step and that trickled down later on

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[(2\sin(\theta))^2 \rightarrow (2\sin(\theta))(2\sin(\theta))\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2you've neglected to take the 2sin(theta) to the second power... that's why everything became nonsense

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, my internets really bad. yeah I was jumping around too much without checking o.0
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