Disambiguating arccosine/arcsine functions

1. Jul 19, 2007

techninja

Hi all,

I'm working on a program, and it seems that I can't get the trigonometry in my head right. I have two equations, sin(omega) = something and cos(omega) = something, and I need to find what omega is.

Given that there are two trig functions, I should be able to disambiguate what quadrant omega is in.

On the other hand, I don't quite understand the process of how this would be done.

Any help or input would be greatly appreciated!

2. Jul 19, 2007

Staff: Mentor

Thread moved to Homework Help forums (where homework and coursework should be posted).

What are the +/- signs of the sin and cos functions in the 4 quadrants?

3. Jul 19, 2007

techninja

Well, if we're going to go thataway, sine is positive in I and II, cosine is positive in I and III. Arcsine is defined in I and IV, arccosine is defined in I and II.

Thanks!

4. Jul 19, 2007

Staff: Mentor

5. Jul 19, 2007

techninja

Nope; not at all. :rofl:

6. Jul 19, 2007

VietDao29

Well, firstly, you have to check for the existence of omega. You know the relation between sine, and cosine function, right?

$$\sin ^ 2 \omega + \cos ^ 2 \omega = 1$$

If the above equation holds, then omega exists, if not, it doesn't. Do you know why?

We don't need the arcsin, and arccos part here.

Ok, so, say, if sin(omega) is positive, and cos(omega) is negative, what quadrant is omega in?

:)

7. Jul 19, 2007

techninja

That's a trig identity, I believe.

And, if sine is positive, it would be arccos(cos(omega)), and if not, it would be... 2*pi-arccos(cos(omega))?

Would that be right?

Thanks. (:

8. Jul 20, 2007

VietDao29

Yup, correct. :)

However, does your omega has any restriction? i.e, say, must it be on the interval [0; 2pi[? Or anything along those line?

If omega must be on [0; 2pi[, then your solution would be:

$$\left[ \begin{array}{ll} \omega = \arccos (\cos (\omega)) , & \quad \mbox{for non-negative } \sin \omega \\ \omega = 2 \pi - \arccos \cos ( \omega ), & \quad \mbox{for negative } \sin \omega \end{array} \right.$$

If, omega can be anything, then the general solution for omega would be:

$$\left[ \begin{array}{ll} \omega = \arccos (\cos (\omega)) + \2 k \pi , & \quad \mbox{for non-negative } \sin \omega \\ \omega = - \arccos ( \cos \omega ) + 2 k' \pi , & \quad \mbox{for negative } \sin \omega \end{array} \right.$$, where k, and k' are both integers.

You got it correctly. Congratulations. ^.^

Can you complete the programme? :)

Last edited: Jul 20, 2007