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Quick and Easy Question

  1. Sep 29, 2007 #1
    For the question tan (cos-(1/3))... is it possible to represent the value of (1/3) in terms of radians? if not would the answer be tan(70.52877939)


    (- being the inverse of cos)
     
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2

    EnumaElish

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    a = cos-1(x) or a = arccos(x) is the solution to x = cos(a). The argument of cos (that is a) can be in radians, but x is a real number.

    What is the definition of cosine?

    How did you get tan(70.52877939)?
     
  4. Sep 29, 2007 #3
    Ahhh...well after looking at it carefully ..i have derived this as my solution...please tell me if i did it right or wrong..

    tan (cos-(1/3))

    pheta= @



    cos- means the return of an angle. so...cos(x)=@ than cos-(cos@)=0
    @=cos- (1/3) --> cos@=1/3


    Cos@= adjacent/hyponeus
    cos@=1/3

    cos is the ratio of the lengths of adjacent and hypotneuse of the triangle

    to find the other side which is the opposite side.. i used the pythagorean theorm which is
    the Hypotneuse^2 - Adjacent^2 and this becomes...

    3^2 -1^2 ( all over square root)
    so the square root of 8 is the opposite

    and to find the tangent angle now is easy since i have the opposite and adjacent side

    this becomes tan@= sqrt(8)/1

    therefore the answer is tan (cos-(1/3))=tan@=sqrt(8)/1


    is my solution correct? and thanks for the help.
     
  5. Sep 30, 2007 #4

    HallsofIvy

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    If you want to do it more formally, you can use trig identities:
    [tex]tan(\theta)= \frac{sin(\theta)}{cos(\theta)}= \frac{\sqrt{1- cos^2(\theta)}}{cos(\theta)}[/tex]
    Since know that [itex]\theta= cos^{-1}(1/3)[/itex], [itex]cos(\theta)= 1/3[/itex] so
    [tex]tan(cos^{-1}(1/3))= \frac{\sqrt{1- \frac{1}{9}}}{\frac{1}{3}}[/tex]
    [tex]= \frac{\frac{\sqrt{8}}{3}}{\frac{1}{3}}= \sqrt{8}= 2\sqrt{2}[/tex]

    But using setting it up in terms of a triangle is simpler and better.
     
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