If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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 around -. Please help. Is there any video regarding how to find the inverse of a function ?
• Essentially you set g(x) equal to y. Now you have y = tan(x - (3*Pi / 2) ) + 6
Then you solve for x:
y - 6 = tan(x - (3*Pi / 2) )
tan^-1(y - 6) = x - (3*Pi / 2)
x = tan^-1(y - 6) + (3*Pi / 2)

Since the domain is usually referred to as 'x' and the function as 'y' we now swap the x and y:
y = tan^-1(x - 6) + (3*Pi / 2) = g'(x)
g'(x) = tan^-1(x - 6) + (3*Pi / 2)

As you can see the former x is now g'(x) and the g(x) is x .. I don't like how he quickly brushes over these swaps but it makes sense to save some time in inverting functions !
• Why is the domain of arcsin and arccos "restricted" -- [-1,1] -- while the domain for arctan is "all reals"?
• 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.
• At around Sal 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) ?
• 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).
• 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?
• 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.

Regards,

APD
• 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?
• 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.
• 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)"

-4.98
-1.83
1.31
4.45
• 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.
• at the starting of the video, how are we swapping x with g^-1(x) ?
• 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)
• At , he said the domain is negative infinity to infinity, but earlier in the video ()) 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 (), but at , he writes down the range is -pi/2 to pi/2

Could you explain to me why? I'm not understanding how this works
• From -, Sal was talking about the normal tan(theta) function. Later, he talks about the inverse of tan and the restrictions.
• At , why can't we find the slope of a vertical line?
• Hello Hannah,

As Sal said in the video, that would be approaching positive or negative infinity. But I guess a bit of an intuitive approach, we know that slope is rise over run. When it is vertical, the run is 0, and we know that you can't divide by 0. So, if the slope, by its own definition, for a vertical line, would be a fraction with a 0 in the denominator, you would not be able to define it.

Hope this helps! :)