is there a methodical way to find all the possible angles of a given trig ratio? that is a regular method of working through all the known laws of that ratio (such as sinA= (pi - A) or -sinA = (2pi - A))? personally i often get tangled up in these properties and am constantly turning up with the same or negative angles.
Are you talking about reference angles?
If you have an angle in Quadrant I, we don't have to change any signs -- sine, cosine, and tangent are all positive.
If you have an angle in Quadrant II, its reference angle is . Then take the trig function of the reference angle. Remember that sine is positive, and cosine & tangent are negative. In other words,
If you have an angle in Quadrant III, its reference angle is . Then take the trig function of the reference angle. Remember that tangent is positive, and sine & cosine are negative. In other words,
If you have an angle in Quadrant IV, its reference angle is . Then take the trig function of the reference angle. Remember that cosine is positive, and sine & tangent are negative. In other words,
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see this link Math Forum: Ask Dr. Math FAQ: Trigonometry Formulas
Hello furor celticaYes, there is. This is it!
For , what you need is the general solution of the equation , and it is:
So, for example, if (which is pretty obvious)
Then if (which means that , which I'm sure you knew.)
If (so )
If , and so on.
So you can find, for example, the general solution to the equation , as follows:
We know that , so we need to solve . Using the above formula, the general solution is:
For the corresponding result is:
has the general solution
Put some values in for , as I did above, and you'll see how it works.
Finally,
I think you'll find that answers your question without reference to any diagrams.
Grandad