# Trig. Question

1. Sep 11, 2008

### kg90

sin-1(-1/2)

Can someone explain how to do this problem by hand. Thanks.

2. Sep 11, 2008

### tiny-tim

Welcome to PF!

Hi kg90! Welcome to PF!

May I check … do you know how to do sin-1(1/2)?

3. Sep 11, 2008

### HallsofIvy

Staff Emeritus
What is your definition of sine? (Since a right triangle cannot have an angle whose sine is -1/2, I assume you are NOT using "opposite side divided by hypotenuse" as your definition.

4. Sep 11, 2008

### kg90

Hey thanks for replying. I do know how to do most problems like this. I just don't know how you get the negative answer for the problem I posted above. I always end up with a positive.

HallsofIvy, I'm not sure..

5. Sep 11, 2008

### NoMoreExams

Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?

6. Sep 11, 2008

### kg90

Yeah, but aren't there two quadrants where that would be true?

7. Sep 11, 2008

### NoMoreExams

Yes, and therefore on the interval [0, 2pi), there will be 2 answers.

8. Sep 11, 2008

### NoMoreExams

Think about it this way, you probably know or can convince yourself (and if you can't, try to) that f(x) = sin(x) crosses the x - axis twice on the interval [0, 2pi) (which points would that be?). Now how would you do that? Well you would say f(x) = sin(x) = 0, so I need to find x such that x = sin^{-1)(0). Now in your case you have g(x) = sin(x) - 1/2 and you are trying to find those x-intercepts, so you solve g(x) = sin(x) - 1/2 = 0 or in other words x = sin^{-1}(1/2).

9. Sep 11, 2008

### tiny-tim

Hi kg90!

(have a pi: π )

"sin-1" normally means the principal value of sin-1

that's the one in (-π/2,π/2] …

in other words, nearest to 0.

it may help to think of the sin as the y coordinate of the angle …
so sin-1(-1/2) is the angle where the y coordinate is -1/2