Quick and Easy Question

  • Thread starter Karma
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  • #1
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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)
 
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  • #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)?
 
  • #3
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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.
 
  • #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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