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Exact value of Trigonometric equation.

  1. Aug 9, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]1/\sqrt{}2 cos 15 - 1/\sqrt{}2 sin 15[/tex]


    3. The attempt at a solution
    I got [tex]\sqrt{}3/2[/tex].
    If anyone could confirm this, I would appreciate it. Thanks.
     
  2. jcsd
  3. Aug 9, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi ifomoe! Welcome to PF! :smile:

    (have a square-root: √ :smile:)

    (I assume you mean (1/√2)cos15 - (1/√2)sin15?)

    Nooo … you've used + instead of - :cry:

    Which formula did you use? :smile:
     
  4. Aug 9, 2008 #3
    Thank you. :smile:
    Yes. I meant it like that: (1/√2)cos15 - (1/√2)sin15.
    I used the reduction formula but I just re-did it and got 1/2.
    Please, don't tell me I'm wrong again. My brain can't handle much more of this. :frown:
     
  5. Aug 9, 2008 #4

    tiny-tim

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    ifomoe brain good!

    :biggrin: Woohoo! :biggrin:

    (btw, what's the "reduction formula"?

    I used sin(45 - 15) = sin45 cos15 - cos45 sin15. :wink:)
     
  6. Aug 9, 2008 #5
    Yay! :smile:

    I think my lecturer said that another name for reduction formula is auxiliary angle formula.
    If I'm not mistaken.
     
  7. Aug 9, 2008 #6

    HallsofIvy

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    [tex]sin(x/2)= \sqrt{\frac{1}{2}(1- cos(x))}[/tex]
    so
    [tex]sin(15)= \sqrt{\frac{1}{2}(1- cos(30))}= \sqrt{\frac{1}{2}\left(1- \frac{\sqrt{3}}{2}\right)}[/tex]
     
  8. Aug 9, 2008 #7

    tiny-tim

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    :biggrin: Or (cosθ - sinθ)/√2 = √2 sin45 sin(45 - θ) = sin(45 - θ). :biggrin:
     
    Last edited: Aug 9, 2008
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