Domain of Inverse Trig Functions

[Jason76]

Is this right?

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

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

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

 

[srmichael]

Is this right?

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

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

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

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.

 

[lookagain]

Is this right?

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

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

The domain of \(\displaystyle \arctan\) 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

 

[Deleted member 4993]

Is this right?

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

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

The range of \(\displaystyle \arctan\) 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.

 

[Jason76]

I got it mixed up.

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

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

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

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

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

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

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

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

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

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

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

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


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

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

\(\displaystyle [0,\pi]\) if looking at \(\displaystyle \cos\)

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):

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

\(\displaystyle (-\infty,\infty)\) if \(\displaystyle \arctan\) - 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.

 

[mmm4444bot]

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: