Domain of Inverse Trig Functions

Jason76

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Is this right? :confused:

The domain of arcsin\displaystyle \arcsin is restricted to [π2,π2]\displaystyle [-\dfrac{\pi}{2},-\dfrac{\pi}{2}]

The domain of arccos\displaystyle \arccos is restricted to [0,π]\displaystyle [0,\pi]

The domain of arctan\displaystyle \arctan is restricted to [π2,π2]\displaystyle [-\dfrac{\pi}{2},-\dfrac{\pi}{2}]
 
Is this right? :confused:

The domain of arcsin\displaystyle \arcsin is restricted to [π2,π2]\displaystyle [-\dfrac{\pi}{2},-\dfrac{\pi}{2}]

The domain of arccos\displaystyle \arccos is restricted to [0,π]\displaystyle [0,\pi]

The domain of arctan\displaystyle \arctan is restricted to [π2,π2]\displaystyle [-\dfrac{\pi}{2},-\dfrac{\pi}{2}]

It's actualy the range, not the domain. That being said, arcsin and arccos are correct, but for arctan the range is (-π/2, π/2). Open parentheses, not closed parentheses since tan of -π/2 and π/2 are undefined.
 
Is this right? :confused:

The domain of arcsin\displaystyle \arcsin is restricted to [1,1]\displaystyle [-1, 1]

The domain of arccos\displaystyle \arccos is restricted to [1,1]\displaystyle [-1, 1]

The domain of arctan\displaystyle \arctan is restricted to (- oo, oo).
Jason76, specifically for the questions about the domains of those, see the amendments above in the quote box.     \displaystyle \ \ \ \ \ Source: http://mathworld.wolfram.com/InverseTrigonometricFunctions.html
 
Is this right? :confused:

The range of arcsin\displaystyle \arcsin is restricted to [π2,π2]\displaystyle [-\dfrac{\pi}{2},\dfrac{\pi}{2}]

The range of arccos\displaystyle \arccos is restricted to [0,π]\displaystyle [0,\pi]

The range of arctan\displaystyle \arctan is restricted to (,\displaystyle -\infty,\infty)

Those are the "understood" ranges when it is not specified. You can specify any range you wish. Thus when you calculate 'arc' function through the calculator or software like 'excel' - the answer is restricted to those ranges.

The "arc" expressions are multivalued expressions - not real functions. So while working with those - you must specify ranges (and/or domains) to treat those as functions.
 
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I got it mixed up.

Domain of sin=(,)\displaystyle \sin = (-\infty,\infty)

Domain of cos=(,)\displaystyle \cos = (-\infty,\infty)

Domain tan=(,)\displaystyle \tan = (-\infty, \infty)Except at multiples of π2\displaystyle \dfrac{\pi}{2}

Range of sin=[1,1]\displaystyle \sin = [-1,1]

Range of cos=[1,1]\displaystyle \cos = [-1,1]

Range tan=(,)\displaystyle \tan = (-\infty, \infty)

Domain of arcsin=[1,1]\displaystyle \arcsin = [-1,1]

Domain of arccos=[1,1]\displaystyle \arccos = [-1,1]

Domain arctan=(,)\displaystyle \arctan = (-\infty, \infty)

Range of arcsin=(,)\displaystyle \arcsin = (-\infty,\infty)

Range of arccos=(,)\displaystyle \arccos = (-\infty,\infty)

Range arctan=(,)\displaystyle \arctan = (-\infty, \infty)Except at multiples of π2\displaystyle \dfrac{\pi}{2}


But if evaluating some composite function with trig function in the middle (not an inverse one), see if your evaluation (answer) is within (on the unit circle

[π2,π2]\displaystyle [-\dfrac{\pi}{2}, -\dfrac{\pi}{2}] if looking at sin\displaystyle \sin or tan\displaystyle \tan

[0,π]\displaystyle [0,\pi] if looking at cos\displaystyle \cos

If evaluating a composite function and an inverse trig function is in the middle, see if your evaluation (answer) is within (on the unit circle):

[1,1]\displaystyle [-1,1] if arcsin\displaystyle \arcsin or arccos\displaystyle \arccos

(,)\displaystyle (-\infty,\infty) if arctan\displaystyle \arctan - which is basically anything

If it isn't in it's respective domain, then evaluate the inside part again to be undefined or something else. If it is something else, then see if it now is within the domain's range.
 
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If evaluating a composite function and an inverse trig function is in the middle, see if your evaluation (answer) is within (on the unit circle):

If it isn't in it's respective domain, then evaluate the inside part again to be undefined or something else. If it is something else, then see if it now is within the domain's range.

Will you please post an example composite function, to illustrate what you're trying to say above?

I'm not sure what you're thinking, when you say that the output of a composite function is "on" or "within" a circle.

The condition "if it isn't in its respective domain" is also unclear.

You ought to consider spelling out the nouns in a re-statement. Using the same pronoun ('it') to reference different nouns simultaneously often leads to comprehension issues.

Thank you. :cool:
 
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