all possible angles of a trig ratio

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Jul 1st 2009, 04:02 AM
furor celtica

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)
Jul 1st 2009, 04:33 AM
running-gag

Hi

You can use the trig circle
You can see for instance that sin(pi-A)=sinA

http://nsa07.casimages.com/img/2009/...3428459608.jpg
Jul 1st 2009, 04:51 AM
furor celtica

anything non-graphic? i just mean suggestions, like 'this law should be used first and then you can jump to this one bcos it gives an egative angle which can then be passed through this law which gives a new positive one..'
Jul 1st 2009, 05:03 AM
yeongil

Are you talking about reference angles?

If you have an angle $\theta$ in Quadrant I, we don't have to change any signs -- sine, cosine, and tangent are all positive.

If you have an angle $\theta$ in Quadrant II, its reference angle is $\pi - \theta$ . Then take the trig function of the reference angle. Remember that sine is positive, and cosine & tangent are negative. In other words,
$\sin \theta = \sin(\pi - \theta)$
$\cos \theta = -\cos(\pi - \theta)$
$\tan \theta = -\tan(\pi - \theta)$

If you have an angle $\theta$ in Quadrant III, its reference angle is $\theta - \pi$ . Then take the trig function of the reference angle. Remember that tangent is positive, and sine & cosine are negative. In other words,
$\sin \theta = -\sin(\theta - \pi)$
$\cos \theta = -\cos(\theta - \pi)$
$\tan \theta = \tan(\theta - \pi)$

If you have an angle $\theta$ in Quadrant IV, its reference angle is $2\pi - \theta$ . Then take the trig function of the reference angle. Remember that cosine is positive, and sine & tangent are negative. In other words,
$\sin \theta = -\sin(2\pi - \theta)$
$\cos \theta = \cos(2\pi - \theta)$
$\tan \theta = -\tan(2\pi - \theta)$

01
Jul 1st 2009, 06:35 AM
Amer

see this link Math Forum: Ask Dr. Math FAQ: Trigonometry Formulas
Jul 1st 2009, 10:03 AM
Grandad

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

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

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

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

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

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

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

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

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

For $\cos x$ the corresponding result is:

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

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

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

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

Grandad