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### Course: Trigonometry > Unit 4

Lesson 1: Inverse trigonometric functions- Intro to arcsine
- Intro to arctangent
- Intro to arccosine
- Evaluate inverse trig functions
- Restricting domains of functions to make them invertible
- Domain & range of inverse tangent function
- Using inverse trig functions with a calculator
- Inverse trigonometric functions review
- Trigonometric equations and identities: FAQ

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# Domain & range of inverse tangent function

Sal finds the formula for the inverse function of g(x)=tan(x-3π/2)+6, and then determines the domain of that inverse function. Created by Sal Khan.

## Want to join the conversation?

- Inspite of more than 5 to 6 answers to the question, I still dont understand the first step Sal did around0:50-0:59. Please help. Is there any video regarding how to find the inverse of a function ?(56 votes)
- I'm writing this answer even though the question was asked five years ago in the hopes that another KA user may learn from it. I eventually realized that Sal is
**inputting**g⁻¹(x) into g(x). This is so that g⁻¹(x) is the x in g(x), just like how we can input 2 into g(x) like g(2).

So we have:

g(x) = tan(x - (3*Pi / 2)) + 6

Input g⁻¹(x):

g(g⁻¹(x)) = tan( (g⁻¹(x)) - (3*Pi / 2) ) + 6

g(g⁻¹(x)) is simply x. So we get:

x = tan( (g⁻¹(x)) - (3*Pi / 2) ) + 6(62 votes)

- Why is the domain of arcsin and arccos "restricted" -- [-1,1] -- while the domain for arctan is "all reals"?(6 votes)
- That is because sine and cosine range between [-1,1] whereas tangent ranges from (−∞,+∞). Thus their inverse functions have to have their domains restricted in that way.

If you extend cosine and sine into the complex plane, then arcsin and arccos can similarly be extended. But such an extension will give you complex numbers. Obviously, this extension would be no use for ordinary geometric shapes and their angles. It is mostly useful in advanced mathematical topics and almost certainly won't be covered in a basic trigonometry class.(14 votes)

- So raising a number to the negative one power "flips" the fraction over (creates the reciprocal), but raising a function to the negative one power denotes the inverse of that function, right?(7 votes)
- Hello Emma,

Yes, this is a bit confusing. You have correctly summed up the situation...

If it helps for trigonometry you can use the older ACRSIN, ARCCOS and ACRTAN terms.

See also https://en.wikipedia.org/wiki/Inverse_function

Regards,

APD(6 votes)

- why can we replace x with the g inverse of x and g(x) with x?(7 votes)
- So that's from the definition of inverse function.

Definition: If (x, y) is a point of f then (y, x) is a point of f^-1.(3 votes)

- for anyone wondering "why can we replace x with the g inverse of x and g(x) with x?"

you can also do it like:(6 votes)- y=tan(x-3pi/2)+6

x+6=tan(y-3pi/2)

tan^-1(x+6)=y-3pi/2

y=tan^-1(x+6)+3pi/2(6 votes)

- Can anyone point me in the direction of an explanation on how to solve the following practice question for this section.

"The following are all angle measures (in radians, rounded to the nearest hundredth) whose tangent is 3.7.

Which is the principal value of arctan(3.7)"

Answers are

-4.98

-1.83

1.31

4.45(5 votes)- The principal value of arctan is between -π/2 and π/2;i.e.-between -1.57 and 1.57 radians. So the solution to your question is 1.31 radians.(4 votes)

- At around9:15Sal says that you add 3pi/2 to The low end of our range restriction. Why are we using the Low end and not the high end (-pi/2 instead of pi/2) ?(3 votes)
- tan⁻¹ has a range of (-π/2, π/2), but the equation is
`g⁻¹(x) = tan⁻¹(x-6) + 3π/2`

. So, in order to find the range of the whole thing, we need to add 3π/2 to both ends of the range of tan⁻¹, giving us (-π/2+3π/2, π/2+3π/2) or (π, 2π).(5 votes)

- At8:36, he said the domain is negative infinity to infinity, but earlier in the video (5:28)) he says the domain is all reals except for pi/2 + pi(k). If the domain isn't ALL values, how can it go from negative infinity to infinity? Also, the range is all real numbers (6:00), but at8:51, he writes down the range is -pi/2 to pi/2

Could you explain to me why? I'm not understanding how this works(3 votes)- From3:50-6:12, Sal was talking about the normal tan(theta) function. Later, he talks about the inverse of tan and the restrictions.(4 votes)

- Why can't arctan's range be (0-pi)? Bad notation, but if you're restricted to only the first and second quadrant, then isn't that the same thing as restricting from -pi/2-pi/2?(3 votes)
- because that will include pi/2 as a value for a function and tangent is undefined for pi/2. and if a function is undefined, it will cease to exist for that value.(2 votes)

- at the starting of the video, how are we swapping x with g^-1(x) ?(4 votes)
- We do it so that we understand at first which value we are solving for, which is g^-1x, and then once we have found the value, we switch back to x based on the domain.(1 vote)

## Video transcript

Voiceover:We're told given g of x is equal to ten of x minus
three pi over two plus six, find the g inverse of x. They want us to type that in here and then they also want us to figure out what is the domain of g inverse, the domain of g inverse of x. I've got my little scratch pad here to try to work that through. Let's figure out what g inverse of x is. This is g of x, so g inverse of x. Essentially, let me just read this is g of x right over here, g of x is equal to tangent of x minus three pi over two plus six. G inverse of x, I
essentially can swap the … I can replace the x
with the g inverse of x and replace to g of x with an x and then solve for g inverse of x. I could write that x is equal
to tangent of g inverse of x minus three pi over two plus six. Let's just solve for g inverse of x. I actually encourage
you to pause this video and try to work through this
out or work it out on your own. Let's subtract six from both sides to at least get rid of this six here so I'm left with x minus
six is equal to the tangent of g inverse of x minus three pi over two. Now let's take the inverse tangent of both sides of this equation so the inverse tangent
on the left hand side is the inverse tangent of x minus six and on the right hand side the
inverse tangent of tangent. If we restrict the
domain in the proper way and we'll talk about that in a little bit is just going to be what the input into the tangent function is. If you restrict the
domain in the right way, inverse tangent of the
tangent of something, let's say theta is just
going to be equal to theta. Once again if we restrict the domain, if we restrict what the possible values of theta are in the right way. Let's just assume that we're doing that and so the inverse
tangent of the tan, of this is going to be just this
stuff right over here. It's just going to be that, it's going to be g inverse of x minus three pi over two. Now we're in the [home] stretch to solve for g inverse of x we could just add three
pi over two to both sides so we get and actually let
me just swap both sides. We get g inverse of x is equal to the inverse tangent of x minus six and then we're adding three
pi over two to both sides so this side is now on this side so plus three pi over two. Let me actually type that and I'm going to see if I can remember it because I'm about to
lose this on my screen so inverse tangent of x minus
six plus three pi over two so let me write that down. Let me type this. G inverse of x is going
to be the inverse tangent so I can write it like this, the inverse tangent of x minus six and yes it interpreted it correctly. Inverse tangent you can do that as arctangent of x minus six
plus three pi over two and it did interpret it correctly but then we have to think about what is the domain of g inverse? What is the domain of g inverse of x? Let's think about this a little bit more. The domain of g inverse of x, so let's just think about
what tangent is doing. The tangent function if
we imagine a unit circle, so that's a unit circle right over there. Guess we can imagine to be a unit circle. My pen tool is acting up a little bit it's putting this little gaps and things but I think we can power through that. Let's just say for the sake of argument that that's a unit circle, that's the x axis and that's the y axis. If you form an angle theta. If you form some angle
theta right over here, the tangent of theta is
essentially the slope of this terminal ray
of the angle or the ... Or I guess we can call it the
terminal ray of the angle. The angles form by that ray and this ray along the positive x axis. The tangent of theta is
the slope right over there and you can get a tangent of
any theta except for a few. You can find the tangent of that, you could find the slope there, you could find the slope there, you could also find the slope there, you could find the slope there but the place where you
can't find the slope is when this ray goes straight up, or this ray goes straight down. Those were the cases where
you can't find the slope. They are the slope you
could say is approaching positive or negative infinity. The domain of tangent, so tangent domain so the domain is essentially
all real numbers, all reals except multiples of pi over … I guess you can say pi over
two plus multiples of pi, except pi over two plus multiples of pi where k could be any integer so you could also be subtracting pi because if you have pi over two, if you add pi, you go straight down here. You add another pi you go up there, if you subtract pi you go down here, add, subtract another
pi you go over there. This is the domain but given this domain you
can get any real number. The range here is all reals because you can get any slope here, you can increase theta if
you want a really high slope, decrease theta if you want
a really negative slope right over there. You can really get to anything. Now when you're taking
about the inverse tangent, by convention you're going to … Well to make tangent invertible so that you don't have multiple elements of your domain all mapping
to the same element of the range because for example, this angle right over here
has the exact same slope as this angle right over here. If you have two theta's
mapping to the same tangent then that's not, if you
don't restrict your domain so that you only have one of them, it's not going to be invertible so the convention is is that
to make tangent invertible you restrict its domain to the interval from negative pi over two to pi over two in order to
construct the inverse tangent. The inverse tangent, you can input any real number into it so the inverse tangent's domain, this is just the convention. They could have restricted
tangents domains as long as for any theta,
there's only one theta in that domain that maps to a
specific element of the range but the convention is,
well inverse tangent can … The convention is to
restrict tangents domain between negative pi over
two and pi over two. Inverse tangents domain is all reals but its range is restricted. Its range and this is by convention it's going to be between
negative pi over two and pi over two and not including them. Le'’s go back to our original
question right over here, what is the domain of g inverse? Let's look at our domain of g inverse, well g inverse the domain of this I could put any real number in here. Now what this is going to pop out is going to be something between negative pi
over two and pi over two but they're not asking us
the range of g inverse. Actually would have been a
more interesting question. They're asking us what's
the domain of g inverse and I could put in any real
number right here for x, so let's put that in here. Domain of g inverse of x, it's negative infinity to infinity but actually just for fun and let's just verify that
we got the question right and we did but just for fun. Actually I am curious, let's think about what
the range of g inverse is. The range of this thing right over here is going to be between
negative pi over two to pi over two, that's for
this part right over here and then we're going to add
three pi over two's to it. The range for the entire function, so the range for this
thing is going to be, what the low end if we add
three pi over two to this, this is going to give us two pi over two which is just going to be, so three pi over two minus pi over two is going to be two pi
over two which is just pi, just pi all the way to three pi over two plus another pi over two is
going to be four pi’s over two or two pi. The range of g inverse
of x is pi to two pi and it's an open interval. Doesn't include the boundaries but its domain you could
put any value for x here and it will be defined.