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### Course: Get ready for AP® Calculus>Unit 3

Lesson 5: Inverse trigonometric functions

# 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?

• At . Is there an analytical way to determine the angle? Looks like Sal just eyeballs the triangle and declares it 30,60,90.
• 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.
• So the first thing to do would be to figure out what you don't understand.

How is an inverse function defined. How do you prove f^-1(f(x)) = x and f(f^-1(x)) =x? Also important How are arcsin, arctan and arccos defined i.e. what is the domain and the range of each function.

As a side note if you want to evaluate an expression involving the arcsin, arccos or arctan then you should use a calculator. This is what you will need to do for the "Evaluate inverse trig functions" exercise.

You need to also know the unit circle definitions of the trig functions. Know the special triangles and understand SOHCAHTOA.
• Is the inverse of sin the same as the cosecant. I'm a little confused, isn't the cosecant just the reciprocal?
• 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
• How did Sal know that the arcsin domain had to be in between -1 and 1 at ?
• 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

isn't sin^-1 = 1/sin = cosecant???
• 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.
• Is -pi/3 equivalent to 5pi/3? Because I got the second result and I want to know if it's a good solution.
• 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.
• 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!
• 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.
• Got questions for you:

1) At , 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)At , 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) At , 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!
• 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.
• At , 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?
• 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?

Regards,

APD