1. ## Trig Resource: Transferring Angles (in Radians)

Hello,

I still get a little bit muddled whenever transferring angles from one quadrant to another. I was wondering if anybody knew any resource/tutorial that teaches you how to do this, preferably in radians? I've been trying to look this up, but I always come across tutorials that are in degrees.

Any help would be much appreciated!

Thank you,
Gina

2. Originally Posted by ginarific
Hello,

I still get a little bit muddled whenever transferring angles from one quadrant to another. I was wondering if anybody knew any resource/tutorial that teaches you how to do this, preferably in radians? I've been trying to look this up, but I always come across tutorials that are in degrees.

Any help would be much appreciated!

Thank you,
Gina
what do you mean by "transferring" angles from one quadrant to another?

3. I.e Solve the following question:

$sin(x) = 1/2, (\pi)/2 < x < 3\pi/2$

So if you solve for x, you get:

$x = \pi/6$

However, you'd have to "transfer" this into the domain, so it would be

$x = 5\pi/6$

I don't know what else to call this than "transferring" angles (uh, co-related angles?)

4. learn this and you will never need to "solve" or transfer angles again ...

5. Thank you!

Yes, that's fine for the "nice" trig ratios. However ... what if you have decimal answers? :s

6. Originally Posted by ginarific
Thank you!

Yes, that's fine for the "nice" trig ratios. However ... what if you have decimal answers? :s
it's all about understanding how a calculator determines the value of inverse trig functions.

say you have $\sin{x} = 0.4567$ and x is in quadrant II ...

$x = \pi - \arcsin(0.4567)$ because the calculator calculates the value of $\arcsin(0.4567)$ as a positive quad I angle.

say you have $\sin{x} = -0.8765$ and x is in quad III ...

$x = \pi - \arcsin(-0.8765)$ because the calculator calculates $\arcsin(-0.8765)$ as a negative angle in quad IV.

say you have $\cos{x} = 0.9876$ and x is in quad IV ...

$x = 2\pi - \arccos(0.9876)$ because the calculator calculates $\arccos(0.9876)$ as a positive angle in quad I

say you have $\cos{x} = -0.3456$ and x is in quad III

$x = 2\pi - \arccos(-0.3456)$ because the calculator calculates the value of $\arccos(-0.3456)$ as a positive quad II angle.

you should be able to generalize the above examples.