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Find the exact value of this inverse function

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the exact value of cos(2arcsin(-1/8))



    2. Relevant equations
    make use of the double angle formula


    3. The attempt at a solution
    let arcsin(-1/8)=theta
    then sin theta= -1/8

    a=sqrt63=3sqrt7
    and that is as far as i could go...please help? Thank you.
     
  2. jcsd
  3. Apr 21, 2010 #2

    Mark44

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    So now evaluate cos(2theta).
    Where does a come in?
     
  4. Apr 22, 2010 #3
    'a' was a side i named
     
  5. Apr 22, 2010 #4

    Mark44

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    OK. Still, you need to evaluate cos(2theta). One identity is cos(2theta) = cos^2(theta) - sin^2(theta). This identity appears in two other forms, one of which will be useful for this problem.
     
  6. Apr 23, 2010 #5
    i'm still lost...
     
  7. Apr 24, 2010 #6

    LCKurtz

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    Draw a picture of the angle θ and label the sides of a little triangle appropriately to have a sine of -1/8. You can read all six trig functions off that triangle and should be able to calculate cos(2θ) from there.
     
  8. Apr 24, 2010 #7

    HallsofIvy

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    Since you were told "make use of the double angle formula", [itex]cos(2\theta)= cos^2(\theta)- sin^2(\theta)= 1- 2sin^2(\theta)[/itex].

    You know that [itex]sin(\theta)= \frac{1}{8}[/itex].
     
  9. Apr 26, 2010 #8
    what does cos^2 equal?
     
  10. Apr 26, 2010 #9

    LCKurtz

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    cos2(θ) is just an abbreviation for (cos(θ))2.
     
  11. Apr 27, 2010 #10

    Mentallic

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    Once you draw your right-triangle and label an angle [itex]\theta[/itex] then since sin[itex]\theta[/itex] = -1/8, you know what your opposite and hypotenuse sides are, then you can use pythagoras' theorem to find the 3rd side. This will give you cos[itex]\theta[/itex].
     
  12. Apr 27, 2010 #11

    HallsofIvy

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    cos(2x)[itex]= cos^2(x)- sin^2(x)[/itex][itex]= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)[/itex]

    If x= arcsin(-1/8) what is sin(x)?

    I just notice that I had said this same thing several days ago!
     
    Last edited: Apr 28, 2010
  13. Apr 27, 2010 #12
    sin x is -1/8
     
  14. Apr 27, 2010 #13

    Mentallic

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    Right, so what is cos(2x) given that cos(2x)=cos^2(x)-sin^2(x) ?
     
  15. Apr 28, 2010 #14

    HallsofIvy

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    Which, as I pointed out in the post that Irishman's post was in response to (and, I just noticed, several days ago!) , is equal to [itex]1- 2sin^2(x)[/itex].

    Irishman, if sin(x)= -1/8, what is [itex]sin^2(x)[/itex]? What is [itex]1- 2sin^2(x)[/itex]?
     
  16. Apr 28, 2010 #15

    Mentallic

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    Yes, I know, this identity has been mentioned at least 3 times in this thread now. It just seems like 1irishman is only considering and responding to the most recent reply in this thread each time he logs back on, so I felt the need to hold his hand as we guide him through the process...

     
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