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:
Originally Posted by
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