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Is this solvable

  1. Jun 8, 2007 #1
    tan(30)=(sin(theta))/(1+cos(theta))

    The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I cant find it. Maybe Im just not thinking straight but its really getting to me.
     
  2. jcsd
  3. Jun 8, 2007 #2
    What about

    [tex]\cot (30) = \frac {1 + \cos (\theta)}{\sqrt {1 - \cos^{2} \theta}} [/tex]

    ?
     
  4. Jun 8, 2007 #3

    Hurkyl

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    Isn't that original equation exactly in the form of one of the trig identities?
     
  5. Jun 8, 2007 #4
    I tried to find a trigonometric identity that reduces the expression to one unknown, but couldn't.
     
  6. Jun 8, 2007 #5
    I think I found it. It came to me when I went to get the mail. :)
    basically I have this
    [tex] tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}[/tex]
    From here I just put it in a quadratic form and solved.
     
  7. Jun 8, 2007 #6

    Hurkyl

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    The original expression is essentially one of the half-angle formulas for tangent.
     
  8. Jun 8, 2007 #7
    Almost, but not quite. It looks pretty tough to put it in that particular form. The way I mentioned works so I guess Ill just stick with that, especially since remembering all those identities is a pain.

    Thanks for the suggestions though.
     
  9. Jun 8, 2007 #8
    but not knowing them is obviously more painful.
     
  10. Jun 9, 2007 #9

    Hurkyl

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    It's tough to put
    [tex]\tan 30^\circ = \frac{\sin \theta}{1 + \cos \theta}[/tex]

    into the form

    [tex]\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}[/tex]

    ?
     
    Last edited: Jun 9, 2007
  11. Jun 9, 2007 #10

    Hurkyl

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    By the way, that expression isn't right -- you don't know that the numberator is the positive root. You also have to consider

    [tex]tan(30)=-\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}[/tex]
     
  12. Jun 9, 2007 #11
    Unless Im missing something I thought the half angle formula was [tex]

    \tan \frac{\theta}{2} = \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}[/tex]

    Thats not the same as [tex]\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}[/tex] right? Or am I making a mistake?
     
    Last edited: Jun 9, 2007
  13. Jun 9, 2007 #12
    And just what multiplying [tex] \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex] by

    [tex]\frac {\sqrt{ 1 + \cos \theta}}{\sqrt{ 1 + \cos \theta}} [/tex]

    gives?
     
  14. Jun 10, 2007 #13

    VietDao29

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    Well, it gives:
    [tex]... = \frac{\sqrt{1 - \cos ^ 2 \theta}}{1 + \cos \theta} = \frac{\textcolor{red} {|} \sin \theta \textcolor{red} {|}}{1 + \cos \theta}[/tex] (Notice that it's never negative, since there's an absolute value in it)

    What you should use is Half-Angle and Power Reduction Identities:

    [tex]\sin (2 \theta) = 2 \sin ( \theta ) \cos ( \theta )[/tex]

    and: [tex]\cos ^ 2 \theta = \frac{1 + \cos (2 \theta)}{2}[/tex]
     
    Last edited: Jun 10, 2007
  15. Jun 10, 2007 #14
    ???

    One solution is found by taking the positive root and the other by taking the negative, as such

    [tex] \tan \frac{\theta}{2} = \frac{-\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}[/tex]
     
  16. Jun 10, 2007 #15

    VietDao29

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    What Hurkyl, and I are (?, is it is, am, or are should be used here?) trying to say is that, the expression:
    [tex]\tan \left( \frac{\theta}{2} \right) = \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}}[/tex] is, indeed, incorrect.

    It'll be correct if you take into account its negative part as well, i.e using the (+/-) sign as you did.
     
    Last edited: Jun 10, 2007
  17. Jun 12, 2007 #16
    Technicalities, you say tomato, I say tomato.
     
  18. Oct 7, 2007 #17
    tan30=sin(@)/(1+cos@)
    =2sin(@\2)cos(@\2)/2cossquare(@/2)
    =tan(@/2)
    implies 30=@/2
    @=60

    here @=theta
     
  19. Oct 7, 2007 #18

    Hurkyl

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    Of course, that's merely one solution.

    (p.s. use [ code ] ... [ /code ] tags for preformatted text)
     
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