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Approximating Angles

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Approximate, to the nearest 0.01 radian, all angles [tex]\theta[/tex] in the interval [0,2pi) that satisfy the equation.

    sin [tex]\theta[/tex]= -0.0135


    2. Relevant equations

    Inverse sin, reference angle formulas



    3. The attempt at a solution

    The inverse sin is -.01, so the reference angle for this is .01. If I subtract this from 2pi, I'll get 6.27 radians which checks out. I know the other angle should be in the third quadrant because sin is negative. I'm stuck here. I thought to get an angle in the third quadrant, you take the reference angle and subtract pi from it. This give me a negative number. I don't know what I've done wrong.

    Thanks for the help in advance. :)
     
  2. jcsd
  3. Sep 16, 2009 #2
    I think I might have figured it out on my own, but want to check. I'm getting angles (in radians) 6.27 and 3.15. When I try and find the sin of 3.15, I get -.0099998333... which is not -0.0135, but is close.

    Is this right?
     
  4. Sep 16, 2009 #3

    lurflurf

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    Homework Helper

    Notice sin(x) is small so the approximation x=sin(x) holds
    you need a few more digits
    Arcsin(-0.0135)=-0.0135
    3.14159+0.0135=3.1551
    6.28319-0.0135=6.2697
     
  5. Sep 16, 2009 #4
    I'm guessing it's off because the answers asked for is to the hundredth decimal place. I think the idea that a negative inverse should be treated as positive regardless.
     
  6. Sep 17, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm not sure what you mean by "a negative inverse should be treated as positive regardless". If you mean, "use a calculator and, if the answer is negative, just drop it the negative sign", that's wrong: sin(-x)= -sin(x) not sin(x). What is true is that sin x is negative for [itex]\pi< x< 2\pi[/itex] and [itex]sin(2\pi+ x)= sin(x)[/itex]. If your calculator gives you a negative x, ad [itex]2\pi[itex] to it.
     
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