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### Course: Trigonometry>Unit 4

Lesson 1: Inverse trigonometric functions

# 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 ?
• 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
• 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.
• 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 we replace x with the g inverse of x and g(x) with x?
• 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.
• 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:
• 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
• 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 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).
• 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.