Get ready for AP® Calculus
- 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
Intro to arcsine
Sal introduces arcsine, which is the inverse function of sine, and discusses its principal range. Created by Sal Khan.
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- At08:14. Is there an analytical way to determine the angle? Looks like Sal just eyeballs the triangle and declares it 30,60,90.(234 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.(154 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.(65 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(69 votes)
- How did Sal know that the arcsin domain had to be in between -1 and 1 at5:31?(30 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(19 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.(27 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.(23 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.(10 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...
Can someone explain why do we restrict theta to quadrant 1 and quadrant 4? Why not other quadrants?(4 votes)
- other quadrants could totally work, but 1 and 4 are the easiest to see why visually.
Hopefully you are familiar witht he graph of sine. arcsin is the inverse. just like x^2 is the inverse of sqrt(x). This isn't the case for all inverses, but sine and arcsine can be thought of as rotated 90 degrees from each other.
So now imagine a vertical sine wave rather than the normal horizontal one. There is a problem now, a function cannot have multiple y values be the answer to one x value.
In sine, sin(0) = sin(-180) = sin(180) = sin(-360) = sin(360) = 0 This is fine since functions can have multiple x values go to the same y value, just not the other way around. a common way to test this is the vertical line test, where if a vertical line touches a graph twice anywhere, it is not a function. what an inverse actually does is flip x and y coordinates. So without limiting the range in arcsine we would have arcsin(0) = -360, -180, 0, 180, 360 and so on. It was then chosen to pick a portion of the sine graph to focus on, and the points immediately around the (0,0) point were chosen.
I wan tyou to specifically imagine the part of the sine graph that goes from -90 degrees to 90 degrees. Kind of an s shape. If you notice it reaches every possible y value for sin(x) This symmetry and full range coverage meant it was a good choice for the domain of arcsin(x) or more specifically -90 to 90 was a good range for it.
since that is the case -90 degrees is in Q4 and 90 degrees is in Q1, which meant the range had to be restricted to those two quadrants
Now keep in mind, if youhad the range any smaller or bigger, you would start to get repeat y values for certain x inputs. This would make it no longer a function.
Let me know if this doesn't make sense. I will add, you couldn't restrict to Q1 and Q2 or Q3 and Q4 since you wouldn't get alltheta values, so it would have to be 1 and 4 or 2 and 3. 1 and 4 has the added benefit of having a portion centered on (0,0), so it is easier to work with in some regards.(3 votes)
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.