# Approximating Angles

1. Sep 16, 2009

### TrueStar

1. The problem statement, all variables and given/known data

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

sin $$\theta$$= -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. Sep 16, 2009

### TrueStar

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?

3. Sep 16, 2009

### lurflurf

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

4. Sep 16, 2009

### TrueStar

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.

5. Sep 17, 2009

### HallsofIvy

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 [itex]2\pi[itex] to it.