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Math Help - Trig Resource: Transferring Angles (in Radians)

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by ginarific View Post
    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?
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    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?)
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  4. #4
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    learn this and you will never need to "solve" or transfer angles again ...

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    Thank you!

    Yes, that's fine for the "nice" trig ratios. However ... what if you have decimal answers? :s
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  6. #6
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    Quote Originally Posted by ginarific View Post
    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.
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