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Trigonometry help wanted

  1. Jun 20, 2009 #1
    Okay, I must admit, my trigonometry is rather awful...

    anyway, I would like to write out what 1 / 2^2 is equal to in terms of sine, cosine, tangent, and the like

    is the following correct, or how do I write it correctly (or what would be the correct figures for 0.5^0.5):

    0.5^0.5 = a 45-degree angle = cos (45) = sin (45) = a tangent of 1
  2. jcsd
  3. Jun 21, 2009 #2


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    Yes this is correct. What you are looking for is a value [itex]\theta[/itex] where
    And yes, you correctly noted that the isosceles right-angled triangle has adjacent and opposite sides (to the angle [itex]\theta[/itex]) of value 1 and hypotenuse of value [itex]\sqrt{2}[/itex].

    Did you mean 1/2^(1/2)? If you actually meant 1/4 then you wont have a 'nice' simple value for [itex]\theta[/itex]. Don't worry, this isn't uncommon.

    The best answer you can give for [itex]\theta[/itex] to

    Is: [tex]\theta=arcsin(\frac{1}{4})\approx 14.48^o[/tex]

    this answer is just an acute angle, and I'm sure you're aware that there are more (actually, infinite) values of [itex]\theta[/itex] that satisfy this result? :smile:
  4. Jun 21, 2009 #3
    thanks... but I did mean exactly what I wrote 0.5^0.5, which equates to a value around .71something or other
  5. Jun 21, 2009 #4


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    Yeah I thought so. It just put me off when you wrote:
    and the value .71 something IS [tex]\frac{1}{\sqrt{2}}[/tex] and isn't 1/2^2 :wink:
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