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

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