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## Precalculus

### Course: Precalculus > Unit 2

Lesson 3: 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

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# Intro to arcsine

Sal introduces arcsine, which is the inverse function of sine, and discusses its principal range. Created by Sal Khan.

## Want to join the conversation?

- At08:14. Is there an analytical way to determine the angle? Looks like Sal just eyeballs the triangle and declares it 30,60,90.(237 votes)
- I was wondering the same. The videos are here http://www.khanacademy.org/video/intro-to-30-60-90-triangles?playlist=Geometry and here http://www.khanacademy.org/video/30-60-90-triangles-ii?playlist=Geometry.(155 votes)

- I have no idea how to actually figure out arcsin, arccos or arctan after watching these three videos. All of Sal's videos have been very helpful to me but it seems as though he's began rushing in these videos and uses patterns he already knows rather than teaching how to solve for any of this.

For example: I feel like he is teaching 5x=10 by saying you know x=2 because 5 times 2 equals 10. Great, but aren't we skipping something about division? If you know the pattern, great, but I don't know the patterns yet so I need the by-the-numbers way to solving.

Anyone have any ideas or any thoughts on this? Maybe another place I could look for this particular portion of trig.(75 votes) - Is the inverse of sin the same as the cosecant. I'm a little confused, isn't the cosecant just the reciprocal?(21 votes)
- Cosecant is the
**multiplicative**inverse of sin. That is, if you multiplied sin and csc, the product would be 1.

A**function's**inverse is much different. The inverse f^-1(x) of a function f(x) flips the x and y values of f(x). So if f(.25)=√π, then f^-1(√π)=.25(72 votes)

- How did Sal know that the arcsin domain had to be in between -1 and 1 at5:31?(31 votes)
- If you take the sine function of any angle, you can only get values between -1 and 1 (including-1 and 1). This means that all the possible outputs of the sine function are between -1 and 1 (in other words, the range is between -1 and 1).

Now if you take the inverse function (arcsin), the original possible outputs become the possible inputs of this inverse function. Hence, the domain of arcsin is between -1 and 1(20 votes)

- PLEASE ANSWER

isn't sin^-1 = 1/sin = cosecant???(13 votes)- Not necessarily; it depends on where your parentheses are, since sin^-1 (x) is different from (sin x)^-1. Sin^-1 (x) -- read "inverse sine of x," and note that the parentheses here are not necessary if you can write the exponent as a superscript -- is the same as arcsin x. Here the input would be a sine ratio and the output would be an angle measure. But that is NOT the same as (sin x)^-1, parentheses absolutely necessary, which would be the reciprocal of sin x, or 1/(sin x), or csc x, which has an angle input and a ratio output.(28 votes)

- Is -pi/3 equivalent to 5pi/3? Because I got the second result and I want to know if it's a good solution.(15 votes)
- I am having the same trouble with these problems, and as far as I'm told, yes they are equivalent, but only the negative answer is CORRECT because of the domain restriction. your answers should only be between -pi/2 and pi/2. the reason why is like he said- a function can not have multiple outputs (such as -pi/3 and 5pi/3) so they restricted the domain to only a piece of the graph. The restricton however, is arbitrary.(24 votes)

- For the radian thing, there seems to be times when the word 'radian' follows pi. When does 'radian' follow pi? Usually Sal doesn't mention 'radian' but just writes pi/3 but in certain cases he does... I'm confused!(12 votes)
- Yeah, a radian is a length around the circle that is equal to the length of the circle's radius. So pi radians, which equals 180 degrees, is pi times the length of the radius.(3 votes)

- Got questions for you:

1) At1:20, how does "rational form" work? I Googled it and it is says it is basically a "proper" fractional form, is that correct? How would you know to multiply 1/sqrt(2) by sqrt(2)/sqrt(2) to get the rational form?

2)At5:40, why is arcsin restricted only the 1st and 4th quadrant? Why not 1st and 2nd? What about for arc-tan and arc-cos? How does this all relate?

3) At6:10, does the restriction of the range from -pi/2 to pi/2 mean that the restriction is set at 180 degrees or half the circle, making it valid this way?

4) Could this all be easily solved without any calculation if one memorized the unit circle intuitively?

5) So sine is asking for the y-coordinate so then the arc-sine is asking for the unknown angle (theta) that would give you the y-coordinate if plugged into sin(theta)?

Thank you!(7 votes)- 1) A lot of teachers do not like seeing square roots in the denominator. It is sometimes more practical and cleaner to find a way to get the square root out of the dominator. That's why he calls it rational form and multiples by sqrt(2)/sqrt(2). There's nothing wrong with the original answer of 1/sqrt(2), but this is just more 'proper', if you will.

2) Arcsin is restricted to the 1st and 4th quadrant because the value of sine goes from all possible values that way. Think about the unit circle. In quadrants 1 and 2 sin will have the same value. An example being: sin(0) = sin(pi) = 0. But, if you take quadrants 1 and 4, then the sin function hits all possible values. That's why there is that restriction. The same logic follows for arctan and arc cos.

3) Well, it's set at -90 degrees to 90 degrees.

4) Somewhat. If it's all simple degree or radian measurements that you are working with, then yes, it can be memorized. It takes some time working with it, but it can be done. A lot of questions will ask you the arcsin(4/9) or something for example and that would be quite difficult to memorize (near impossible). So it just depends on the question.

5) Yes, absolutely correct.

arcsin(1/2) = pi/6 for example. Pi/6 is the radian measure that has a sine value of 1/2.(11 votes)

- At5:54, why does the range of arcsin have to be within the first and fourth quadrants? Will arcsin never be in the 2nd or 3rd quadrant?(6 votes)
- Hi Anna,

A simple answer is to try with your calculator. Take the 45 degree angle as an example. Make a table and calculate SIN of 45, 135, 225, 315, 405 degrees. Now that you have these use the calculator to take ASIN of the results.

You have just arrived at a fundamental concept in trig. The calculator thinks about the principal answer (1st and 4th quadrants for SIN). Later you will be introduced to the concept of a general answer...

Before I forget, try the same experiment for COS and TAN. Do they also follow the 1st a4th quadrant pattern?

Leave a comment if you would like clarification of any of my ramblings...

Regards,

APD(3 votes)

- I don't quite understand why we must restrict the range to the 1st and 4th quadrants. How does the restricted range become (-pi/2)<= theta <= (pi/2) at6:09?(4 votes)
- We restrict the range so we don’t repeat y-values. The range you mentioned sweeps over all the possible values from -1 to 1 without any repeats.

Why don’t we want repeats? Say you want to find arcsin(0.5). There are many angles that have a sine of 0.5, so there is no definite answer. So we restrict the range so we get one answer.

Hope this helps.(5 votes)

## Video transcript

If I were to walk up to you on
the street and say you, please tell me what-- so I didn't want
to write that thick --please tell me what sine
of pi over 4 is. And, obviously, we're assuming
we're dealing in radians. You either have that memorized
or you would draw the unit circle right there. That's not the best
looking unit circle, but you get the idea. You'd go to pi over 4
radians, which is the same thing as 45 degrees. You would draw that
unit radius out. And the sine is defined
as a y-coordinate on the unit circle. So you would just want to
know this value right here. And you would
immediately say OK. This is a 45 degrees. Let me draw the triangle
a little bit larger. The triangle looks like this. This is 45. That's 45. This is 90. And you can solve a
45 45 90 triangle. The hypotenuse is 1. This is x. This is x. They're going to be
the same values. This is an isosceles
triangle, right? Their base angles are the same. So you say, look. x squared
plus x squared is equal to 1 squared, which is just 1. 2x squared is equal to 1. x squared is equal to 1/2. x is equal to the square root
of 1/2, which is one over the square root of 2. I can put that in rational form
by multiplying that by the square root of 2 over 2. And I get x is equal to the
square root of 2 over 2. So the height here is
square root of 2 over 2. And if you wanted to know
this distance too, it would also be the same thing. But we just cared
about the height. Because the sine value,
the sine of this, is just this height right here. The y-coordinate. And we got that as the
square root of 2 over 2. This is all review. We learned this in the
unit circle video. But what if someone else--
Let's say on another day, I come up to you and I say you,
please tell me what the arcsine of the square
root of 2 over 2 is. What is the arcsine? And you're stumped. You're like I know what the
sine of an angle is, but this is some new trigonometric
function that Sal has devised. And all you have to realize,
when they have this word arc in front of it-- This is also
sometimes referred to as the inverse sine. This could have just as easily
been written as: what is the inverse sine of the
square root of 2 over 2? All this is asking is what
angle would I have to take the sine of in order to get the
value square root of 2 over 2. This is also asking what angle
would I have to take the sine of in order to get square
root of 2 over 2. I could rewrite either of
these statements as saying square-- Let me do it. I could rewrite either of these
statements as saying sine of what is equal to the
square root of 2 over 2. And this, I think, is a
much easier question for you to answer. Sine of what is square
root of 2 over 2? Well I just figured out that
the sine of pi over 4 is square root of 2 over 2. So, in this case, I know that
the sine of pi over 4 is equal to square root of 2 over 2. So my question mark is
equal to pi over 4. Or, I could have rewritten
this as, the arcsine-- sorry --arcsine of the square root of
2 over 2 is equal to pi over 4. Now you might say so, just as
review, I'm giving you a value and I'm saying give me an angle
that gives me, when I take the sine of that angle that
gives me that value. But you're like hey Sal. Look. Let me go over here. You're like, look
pi over 2 worked. 45 degrees worked. But I could just keep adding
360 degrees or I could keep just adding 2 pi. And all of those would work
because those would all get me to that same point of
the unit circle, right? And you'd be correct. And so all of those values, you
would think, would be valid answers for this, right? Because if you take the sine of
any of those angles-- You could just keep adding 360 degrees. If you take the sine of any
of them, you would get square root of 2 over 2. And that's a problem. You can't have a function where
if I take the function-- I can't have a function, f
of x, where it maps to multiple values, right? Where it maps to pi over 4, or
it maps to pi over 4 plus 2 pi or pi over 4 plus 4 pi. So in order for this to be a
valid function-- In order for the inverse sine function to
be valid, I have to restrict its range. And the way that-- We'll
just restrict its range to the most natural place. So let's restrict its range. Actually, just as a
side note, what's its domain restricted to? So if I'm taking the
arcsine of something. So if I'm taking the arcsine of
x, and I'm saying that that is equal to theta, what's the
domain restricted to? What are the valid values of x? x could be equal to what? Well if I take the sine of
any angle, I can only get values between 1 and
negative 1, right? So x is going to be greater
than or equal to negative 1 and then less than or equal to 1. That's the domain. Now, in order to make this
a valid function, I have to restrict the range. The possible values. I have to restrict the range. Now for arcsine, the convention
is to restrict it to the first and fourth quadrants. To restrict the possible angles
to this area right here along the unit circle. So theta is restricted to being
less than or equal to pi over 2 and then greater than or
equal to minus pi over 2. So given that, we now
understand what arcsine is. Let's do another problem. Clear out some space here. Let me do another arcsine. So let's say I were to ask you
what the arcsine of minus the square root of 3 over 2 is. Now you might have
that memorized. And say, I immediately know
that sine of x, or sine of theta is square
root of 3 over 2. And you'd be done. But I don't have
that memorized. So let me just draw
my unit circle. And when I'm dealing with
arcsine, I just have to draw the first and fourth
quadrants of my unit circle. That's the y-axis. That's my x-axis. x and y. And where am I? If the sine of something is
minus square root of 3 over 2, that means the y-coordinate on
the unit circle is minus square root of 3 over 2. So it means we're
right about there. So this is minus the
square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about
it a little bit. My y-coordinate is minus
square root of 3 over 2. This is the angle. It's going to be a negative
angle because we're going below the x-axis in the
clockwise direction. And to figure out-- Let me just
draw a little triangle here. Let me pick a better
color than that. That's a triangle. Let me do it in
this blue color. So let me zoom up
that triangle. Like that. This is theta. That's theta. And what's this
length right here? Well that's the same as
the y-height, I guess we could call it. Which is square
root of 3 over 2. It's a minus because
we're going down. But let's just figure
out this angle. And we know it's a
negative angle. So when you see a square root
of 3 over 2, hopefully you recognize this is a
30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course,
this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the
side opposite to the square root of 3 over 2 is 60 degrees. This side over here
is 30 degrees. So we know that our theta
is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to
minus 60 degrees. But if we're dealing
in radians, that's not good enough. So we can multiply that times
100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with
theta is equal to minus pi over 3 radians. And so we can say-- We can now
make the statements that the arcsine of minus square root
of 3 over 2 is equal to minus pi over 3 radians. Or we could say the inverse
sign of minus square root of 3 over 2 is equal to
minus pi over 3 radians. And to confirm this, let's
just-- Let me get a little calculator out. I put this in radian
mode already. You can just check that. Per second mode. I'm in radian mode. So I know I'm going to get,
hopefully, the right answer. And I want to figure
out the inverse sign. So the inverse sine-- the
second and the sine button --of the minus square
root of 3 over 2. It equals minus 1.04. So it's telling me that this is
equal to minus 1.04 radians. So pi over 3 must
be equal to 1.04. Let's see if I can
confirm that. So if I were to write minus pi
divided by 3, what do I get? I get the exact same value. So my calculator gave me the
exact same value, but it might have not been that helpful
because my calculator doesn't tell me that this is
minus pi over 3.