# How can you find two different values for sin^-1(0.750)?

1. Oct 13, 2007

### ND3G

I am working on a statics question and one of the angles is sin^-1(0.750) which I calculate as 48.6 degrees though the book gives 131.41 degrees

If I calculate sin(48.6) it equals 0.750
If I calculate sin(131.41) it equals 0.750

Clearly there is some relation or rule that I am unaware of. How do I ensure that I find the proper angle?

2. Oct 13, 2007

### malawi_glenn

Just draw the unit circle and joggle a bit then all becomes clear.

sin^1(0.75) means "what angle gives y = 0.75" if you do the unit circle. You'll see that there is two angles in the range 0 - 360 degree that gives you that value.

Which is correct depends on what the problem is, what physical situation it is and so on.

3. Oct 13, 2007

### Hurkyl

Staff Emeritus
sin x = a has infinitely many solutions for x.

If you're working on a problem and expect a unique answer, then you'll have to find some other constraint on x.

4. Oct 13, 2007

### ND3G

Ok, I figured it out. Because of the direction of the vector it is not 0 degrees + 48.6 degrees but 180 degrees - 48.6 degrees which equals 131.4 degrees

5. Oct 13, 2007

### HallsofIvy

Staff Emeritus
You realize that you said nothing at all about vectors in your original post. And even now, we do not know WHY the direction is "180 degrees- 48.6 degrees"!

6. Oct 13, 2007

### dynamicsolo

Your calculator cannot help you with interpreting the output in relation with the problem you are working on. The inverse trig functions on a calculator will only give output in the ranges

arcsin or arctan: -90º to +90º (not quite 90º, of course, for arctan)

arccos: 0º to 180º .

In order to define these as functions, it is necessary to have restricted ranges on
y=f(x) and the calculator is programmed to follow these definitions (this is why the inverse trig functions are sometimes identified as Sin^{-1}, Cos^{-1}, etc., rather than using lower-case).

As you note that the positive sine value is associated with two angles (in the first and second quadrants), it is important to have a picture in mind for the physical situation in order to decide which angle is appropriate to the solution. (I particularly warn engineering students about this all the time: you have to know what the result should look like because the calculator can't.)

7. Oct 13, 2007

Thanks