Disambiguating arccosine/arcsine functions

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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!
 
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What are the +/- signs of the sin and cos functions in the 4 quadrants?
 
Oh, sorry about that.

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!
 
techninja said:
Oh, sorry about that.

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!

You're welcome. So does that mean your question is answered?
 
techninja said:
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.

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

[tex]\sin ^ 2 \omega + \cos ^ 2 \omega = 1[/tex]

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

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!

techninja said:
Well, if we're going to go thataway, sine is positive in I and II, cosine is positive in I and III...

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?

:)
 
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. (:
 
techninja said:
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. (:

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:

[tex]\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.[/tex]

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

[tex]\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.[/tex], where k, and k' are both integers.

You got it correctly. Congratulations. ^.^

Can you complete the programme? :)
 
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