all possible angles of a trig ratio

is there a methodical way to find all the possible angles of a given trig ratio? (Thinking)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.(Angry)

General Solution of trig equations

Hello furor celtica Quote:

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**furor celtica** is there a methodical way to find all the possible angles of a given trig ratio? (Thinking)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.(Angry)

Yes, there is. This is it!

For $\displaystyle \sin x$, what you need is the general solution of the equation $\displaystyle \sin x = \sin A$, and it is:

$\displaystyle x = n\pi + (-1)^nA, n = 0, \pm1, \pm2, ...$

So, for example, if $\displaystyle n=0, x = A$ (which is pretty obvious)

Then if $\displaystyle n = 1, x = \pi - A$ (which means that $\displaystyle \sin A = \sin(\pi - A)$, which I'm sure you knew.)

If $\displaystyle n = -1, x = -\pi - A$ (so $\displaystyle \sin A = \sin(-\pi- A)$)

If $\displaystyle n = 2, x = 2\pi + A$, and so on.

So you can find, for example, the general solution to the equation $\displaystyle \sin x = 0.5$, as follows:

We know that $\displaystyle \sin\tfrac{\pi}{6} = 0.5$, so we need to solve $\displaystyle \sin x = \sin\tfrac{\pi}{6}$. Using the above formula, the general solution is:

$\displaystyle x = n\pi +(-1)^n\tfrac{\pi}{6}, n = 0, \pm1\pm2, ...$

For $\displaystyle \cos x$ the corresponding result is:

$\displaystyle \cos x = \cos A$ has the general solution $\displaystyle x = 2n\pi \pm A, n = 0, \pm1, \pm2, ...$

Put some values in for $\displaystyle n$, as I did above, and you'll see how it works.

Finally, $\displaystyle \tan x = \tan A \Rightarrow x = n\pi + A, n = 0, \pm1, \pm2, ...$

I think you'll find that answers your question without reference to any diagrams.

Grandad