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I need some help with these two calculator trig problems

  1. Oct 10, 2004 #1
    I need some help with these two questions:
    [tex]tan2A=cot40[/tex](40degrees)
    and
    [tex]cosec^2x=1[/tex]
     
  2. jcsd
  3. Oct 10, 2004 #2
    Exactly what help do you need?
     
  4. Oct 10, 2004 #3

    HallsofIvy

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    Can you use your calculator (which looks to me like the only way to do the first one)?
    If you can, use to find cot (40) and then to find 2A.


    As for the second problem, do you know that cosec x is defined as 1/sin x?

    If cosec^2 x= 1, what is sin^2 x?
     
  5. Oct 10, 2004 #4
    Doing the question. I just don't know.

    How do u find cot(40)? Don't u have to change it to [tex]\frac{1}{tan\theta}[/tex] or something like that?

    So are u saying that [tex]cosec^2x=1[/tex] can be also written as [tex]\frac{1}{sin^2x} = 1[/tex]
     
  6. Oct 10, 2004 #5

    selfAdjoint

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    Yup. And then what can you conclude from that, concerning x?
     
  7. Oct 11, 2004 #6
    It is in the 1st and 2nd quadrants?
     
  8. Oct 11, 2004 #7
    tan 2A = cot40 =tan 50
    => 2A = [50 +n(180)]degrees
    =>A=25+90n deg

    cosec^2 x = 1 =>sinx=+-1=>x=180n+(-1)^n *(+-90) deg

    :bugeye:
     
  9. Oct 11, 2004 #8
    For the 2nd question.....

    [tex] cosec^2x=1 [/tex] can be expressed as,

    [tex] 1/sin^2x=1 [/tex] while multiplying [tex] sin^2x [/tex] both sides, we have,

    [tex] 1=sin^2x [/tex] and by square rooting both sides, we now have,

    [tex] sinx= \pm1 [/tex]

    and so, since sin x is both positive and negative, it must lie in all quadrants with alpha 90 degrees.
     
    Last edited: Oct 11, 2004
  10. Oct 12, 2004 #9
    Oh so now i know. I didn't know u could [tex]\sqrt{sin^2x}[/tex]. Thanks to everyone that helped. :smile:
     
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