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Sine, cosine and tangency of angle 11pi/12

  1. Aug 18, 2005 #1
    When I worked this problem, I came up with the following:

    sin 11pi/12= -sq root 2/4 (sq root3-1)
    cos 11pi/12= -sq root 2/4 (sq root3+1)
    tan 11pi/12= 2-sq root3

    am I far off?
     
  2. jcsd
  3. Aug 18, 2005 #2

    TD

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    How did you get that? Does't seem right to me.
     
  4. Aug 18, 2005 #3

    Tide

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    How did you arrive at those?

    For the first one i get

    [tex]sin \frac {11 \pi}{12} = \frac {\sqrt 2}{4} (\sqrt 3 - 1)[/tex]
     
  5. Aug 18, 2005 #4
    I started by working in radians.
     
  6. Aug 18, 2005 #5

    Tide

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    And then what?
     
  7. Aug 18, 2005 #6
    11pi/12=165 degrees

    11pi/12=pi/4+2pi/3
    sq root 3/3+sq root 2/2

    If sin 11pi/12=sq root 2/4(sq root 3-1)

    I came up with:
    cos 11pi/12=sq root 2/4(sq root 3+1)
    tan 11pi/12=2-sq root3
     
  8. Aug 18, 2005 #7

    TD

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    This is correct: [tex]\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}[/tex]

    But [itex]\sin \left( {\alpha + \beta } \right) \ne \sin \left( \alpha \right) + \sin \left( \beta \right)[/tex] so you can't just take the sine of both angles!

    Use [itex]\sin \left( {\alpha + \beta } \right) = \sin \left( \alpha \right) \cdot \cos \left( \beta \right) + \cos \left( \alpha \right) \cdot \sin \left( \beta \right)[/itex]
     
  9. Aug 18, 2005 #8
    I am still confused.....
     
  10. Aug 18, 2005 #9
    P.S. what program are you using so I don't have to keep writing out the equations long hand?
     
  11. Aug 18, 2005 #10

    TD

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    Well, we have that [tex]\frac{{11\pi }}{{12}} = \frac{\pi }{4} + \frac{{2\pi }}{3}[/tex]

    And we know that:
    [tex]\sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}[/tex]
    [tex]\sin \left( {\frac{2\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}[/tex]

    But you cannot say now that:
    [tex]\sin \left( {\frac{\pi }{4} + \frac{{2\pi }}{3}} \right) = \sin \left( {\frac{\pi }{4}} \right) + \sin \left( {\frac{{2\pi }}{3}} \right) = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2}[/tex]

    That's wrong, you have to use [tex]\sin \left( {\alpha + \beta } \right) = \sin \left( \alpha \right) \cdot \cos \left( \beta \right) + \cos \left( \alpha \right) \cdot \sin \left( \beta \right)[/tex]

    Just fill in the formula :smile:
     
  12. Aug 18, 2005 #11
    so by plugging in cos....

    I come up with:
    cos11pi/12=-sq rt2/4(sq rt3+1)
     
  13. Aug 18, 2005 #12

    TD

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    That seems correct, [tex]\cos \left( {\frac{{11\pi }}{{12}}} \right) = - \frac{{\sqrt 2 }}{4}\left( {\sqrt 3 + 1} \right)[/tex]
     
  14. Aug 18, 2005 #13
    Again, thank you for your assistance and patience.

    :smile:
     
  15. Aug 18, 2005 #14

    TD

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    No problem, you were on the right track for a longer time but I didn't understand your notation at first hehe.
     
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