Trig Resource: Transferring Angles (in Radians)

• Oct 17th 2009, 08:39 AM
ginarific
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
• Oct 17th 2009, 10:22 AM
skeeter
Quote:

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?
• Oct 19th 2009, 10:07 AM
ginarific
I.e Solve the following question:

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

So if you solve for x, you get:

$\displaystyle x = \pi/6$

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

$\displaystyle x = 5\pi/6$

I don't know what else to call this than "transferring" angles (uh, co-related angles?)
• Oct 19th 2009, 05:10 PM
skeeter
learn this and you will never need to "solve" or transfer angles again ...

http://galileo.math.siu.edu/%7Emsull...unitcircle.gif
• Oct 20th 2009, 05:19 PM
ginarific
Thank you!

Yes, that's fine for the "nice" trig ratios. However ... what if you have decimal answers? :s
• Oct 20th 2009, 05:48 PM
skeeter
Quote:

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 $\displaystyle \sin{x} = 0.4567$ and x is in quadrant II ...

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

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

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

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

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

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

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

you should be able to generalize the above examples.