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Babynini

  • one year ago

Let x =2sin(theta), -pi/2< theta<pi/2 Simplify the expression x/(sqroot(4-x^2)

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  1. Babynini
    • one year ago
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    \[\frac{ x }{ \sqrt{4-x^2} }\]

  2. Babynini
    • one year ago
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    \[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\]

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

  4. UsukiDoll
    • one year ago
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    hmm try factoring the 2 out of the denominator

  5. Babynini
    • one year ago
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    er..how do I do this xP

  6. Babynini
    • one year ago
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    is it 2(1-sintheta) ?

  7. Babynini
    • one year ago
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    in the denominator of course.

  8. UsukiDoll
    • one year ago
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    \[\frac{ 2\sin(\theta) }{ 2(1-\sin(\theta)) }\] then cancel the 2

  9. Babynini
    • one year ago
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    and we're left with (sin(theta))/(1-sin(theta))

  10. UsukiDoll
    • one year ago
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    and then it turns into a mess... geez...

  11. Babynini
    • one year ago
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    haha aii

  12. UsukiDoll
    • one year ago
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    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.

  13. Babynini
    • one year ago
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    what if we ^2 the whole thing?

  14. UsukiDoll
    • one year ago
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    wait...

  15. Babynini
    • one year ago
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    we can add 180 later.. right? to get it into the correct domain.

  16. UsukiDoll
    • one year ago
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    \[\frac{ 2\sin(\theta) }{ \sqrt{4-(2\sin(\theta))^2} }\] \[\frac{ 2\sin(\theta) }{ \sqrt{4-4\sin^2(\theta))} }\] \[\[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]\]

  17. UsukiDoll
    • one year ago
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    \[\frac{ 2\sin(\theta) }{ \sqrt{4(1-\sin^2(\theta))} }\]

  18. UsukiDoll
    • one year ago
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    \[\cos^2x+\sin^2x=1 \] \[\cos^2x = 1-\sin^2x\] \[\frac{ 2\sin(\theta) }{ \sqrt{4(\cos^2(\theta))} }\]

  19. UsukiDoll
    • one year ago
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    \[\frac{2\sin(\theta)}{2\cos(\theta)}\]

  20. zepdrix
    • one year ago
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    \[\Large\rm \sqrt{4-4\sin^2x}\ne 2-2\sin x\]You silly billy Miriam -_-

  21. UsukiDoll
    • one year ago
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    I got tangent theta in return?!

  22. UsukiDoll
    • one year ago
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    I saw something weird when I saw that I mean can't we yank the 4 out and use a trig identity

  23. Babynini
    • one year ago
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    Woah how'd you get tangent out of there? because the sin and cos are the same here?

  24. UsukiDoll
    • one year ago
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    \[\sqrt{4-4\sin^2x} \rightarrow \sqrt{4(1-\sin^2x)}\]

  25. UsukiDoll
    • one year ago
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    \[\sqrt{4\cos^2x} \rightarrow 2cosx \]

  26. UsukiDoll
    • one year ago
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    \[\frac{2sinx}{2cosx} \rightarrow \frac{sinx}{cosx} \rightarrow tanx\]

  27. Babynini
    • one year ago
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    oou..I see!

  28. UsukiDoll
    • one year ago
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    you've skipped a step and that trickled down later on

  29. UsukiDoll
    • one year ago
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    \[(2\sin(\theta))^2 \rightarrow (2\sin(\theta))(2\sin(\theta))\]

  30. UsukiDoll
    • one year ago
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    \[4\sin^2(\theta)\]

  31. UsukiDoll
    • one year ago
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    you've neglected to take the 2sin(theta) to the second power... that's why everything became nonsense

  32. Babynini
    • one year ago
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    Sorry, my internets really bad. yeah I was jumping around too much without checking o.0

  33. Babynini
    • one year ago
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    Sowriee

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