Find Angles for sin \theta=-0.0135

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To find angles θ in the interval [0, 2π) that satisfy sin θ = -0.0135, the inverse sine gives approximately -0.01, leading to a reference angle of 0.01. The angles in the fourth and third quadrants are calculated as 6.27 radians and 3.15 radians, respectively. However, the sine of 3.15 radians does not equal -0.0135, indicating a need for more precision in calculations. The correct angles, considering the properties of sine, should be approximately 6.27 and 3.1551 radians, with adjustments for the required decimal accuracy. Proper handling of negative angles and understanding sine's behavior in different quadrants is crucial for accurate results.
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Homework Statement



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

sin \theta= -0.0135


Homework Equations



Inverse sin, reference angle formulas



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. :)
 
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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?
 
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
 
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.
 
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 \pi< x< 2\pi and sin(2\pi+ x)= sin(x). If your calculator gives you a negative x, ad 2\pito it.
 
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