Question

Let x =2sin(theta), -pi/2< theta 10 years ago

Answer 1

\[\frac{ x }{ \sqrt{4-x^2} }\]

Answer 2

\[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\]

Answer 3

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

Answer 4

hmm try factoring the 2 out of the denominator

Answer 5

er..how do I do this xP

Answer 6

is it
2(1-sintheta) ?

Answer 7

in the denominator of course.

Answer 8

\[\frac{ 2\sin(\theta) }{ 2(1-\sin(\theta)) }\] then cancel the 2

Answer 9

and we're left with
(sin(theta))/(1-sin(theta))

Answer 10

and then it turns into a mess... geez...

Answer 11

haha aii

Answer 12

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.

Answer 13

what if we ^2 the whole thing?

Answer 14

wait...

Answer 15

we can add 180 later.. right? to get it into the correct domain.

Answer 16

\[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\]

\[\frac{ 2\sin(\theta) }{ \sqrt{4-4\sin^2(\theta))} }\]
\[\[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]\]

Answer 17

\[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]

Answer 18

\[\cos^2x+\sin^2x=1 \]
\[\cos^2x = 1-\sin^2x\]
\[\frac{ 2\sin(\theta) }{ \sqrt{4(\cos^2(\theta))} }\]

Answer 19

\[\frac{2\sin(\theta)}{2\cos(\theta)}\]

Answer 20

\[\Large\rm \sqrt{4-4\sin^2x}\ne 2-2\sin x\]You silly billy Miriam -_-

Answer 21

I got tangent theta in return?!

Answer 22

I saw something weird when I saw that I mean can't we yank the 4 out and use a trig identity

Answer 23

Woah how'd you get tangent out of there?
because the sin and cos are the same here?

Answer 24

\[\sqrt{4-4\sin^2x} \rightarrow \sqrt{4(1-\sin^2x)}\]

Answer 25

\[\sqrt{4\cos^2x} \rightarrow 2cosx \]

Answer 26

\[\frac{2sinx}{2cosx} \rightarrow \frac{sinx}{cosx} \rightarrow tanx\]

Answer 27

oou..I see!

Answer 28

you've skipped a step and that trickled down later on

Answer 29

\[(2\sin(\theta))^2 \rightarrow (2\sin(\theta))(2\sin(\theta))\]

Answer 30

\[4\sin^2(\theta)\]

Answer 31

you've neglected to take the 2sin(theta) to the second power... that's why everything became nonsense

Answer 32

Sorry, my internets really bad.
yeah I was jumping around too much without checking o.0

Answer 33

Sowriee