Inverse Trigonometric Functions
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- Опубликовано: 5 сен 2024
- We know about inverse functions, and we know about trigonometric functions, so it's time to learn about inverse trigonometric functions! These are functions where you plug in valid values that trig functions can possess, and they spit out the angles that produce them. There's a little more to it than that, so let's get into it!
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Extremely helpful. My teacher encounters some inability to teach this stuff for some reason.
Lol same
My university professor has the same problem apparently
i think its just a public school math teacher thing... i'm on my third bum one in the past 2 years.
My teacher does a great job at teaching stuff like this 😸
stg bruh
I'm beginning to feel like a Genius.
Its just the feeling man, I felt that too
@@mind-h4i Haha!
This was so much more useful than your teacher writing notes directly off a textbook
I understand all what you said and most of the concepts than my current teacher's lessons. The funny fact about this, is that im not even a native english speaker (spanish btw) . Thanks a lot for helping me!
I honestly think writing inverse trig functions as arcsin, arccos, arctan, arccsc, arcsec and arccot is way better than using function^-1(x), as the exponent may lead to a lot of misconception and confusion.
*(Obviously, this is my personal preference, and i wont try applying it over other people like a tyrant).*
This was much easier to understand thanks prof ♥️
Thank you for the video. One quibble: the domain and range of some of these inverse functions should be corrected to have closed brackets [ , ], not open parentheses ( , ).
True that the domain and range of inverse Sine/Cosine should be closed interval, but inverse Tangent's should be open, since infinity is unreachable and there are asymptotes at -pi/2 and pi/2.
I was just thinking the same thing.
luckily, we can always plug inverse trig functions into our calculators!
can't relate :(
You have to understand the concept of restricting the domain of these functions, otherwise you can get incorrect results, when the context of the answer only makes sense in a different quadrant. The inverse trig functions on your calculator, are set up to always return an answer in Quadrant 1, when the input is positive. When the input is negative, it will return an answer in Quadrant 2 for inverse cosine, and Quadrant 4 for inverse sine or inverse tangent.
Unfortunately, I watched the whole video, understood it, and still got the comprehension questions wrong. Maybe there's something I'm not getting?
Yep grade 10 got me mcf###ed up just threw me into this with nothing and I’m homeschooled so I ain’t got no teach
same idk what the heck going on
why you homeschooled?
samee
Gonna take my horse to the ol town road, I'm gonna ride... till I can't no mo.
Fr
Hey man out of all the videos this one explains why inverse trigonometric functions work the way they do.
amazing explanation of concept
Easy to understand. Thanks.
Also, when he explained that the inverse of sine is different from cosine because it moves from -1 to 1 through quadrants 1 and 4 didn't make sense to me because it doesn't move through quadrant 4 but quadrant 3. At least, that's what the graph is showing. Can someone maybe clarify in laymen's terms?
Cosine is x-axis, therefore only the angles in Q1-Q2 are relevant because Q4-Q1 have positive x values.
Similarly sine is y-axis, therefore only the angles on Q4-Q1 are relevant because Q1-Q2 have positive y values. We don't care about Q3.
For example, y = sin^-1 (-√3/2) can be 4pi/3 (240°) or 5pi/3 (300°), but we dgaf about Q3 so it's 300°. However we say it's -60° because we can move only 90° ccw or cw in this case.
Why -45 the last result can someone explain me ?
i am confused
So basically inv functions in a nutshell,
Let y=4, adding 4 to it results in 8. The inverse function here is -4
It undoes the action. Similarly, in trigno, y=sin^-1 x is basically finding the angle y where x= sin y.
@@pog16384 I think she has already graduated from university
@@Nick-qs6chLOL
@@Nick-qs6chstill helps new people looking
Remember the unit circle 💀
3:10 3pi/4 also gives sin root 2 over 2...?
4:10 4pi/3 also gives cos -1/2 ...?
3:19 - 3:36 why?
i was having trouble on the first two questions too. But if u plug these into the calculator u get specific points that correspond to each other.
In degree mode, plug in arcsine(root 2 over 2). You will get 45 degrees, which is (root 2 over 2, root 2 over 2), or pi/4.
Think about how the functions are defined on the unit circle. Given an angle in standard position, i.e. starting at the x-axis and measuring counterclockwise to the y-axis, what are the coordinates of the point where it intersects the unit circle?
Sine answers this question for the y-coordinate.
Cosine answers this question for the x-coordinate
Tangent gives the slope of the line joining the point and the origin, which is sine over cosine.
The functions are periodic, and will repeat the same outputs multiple times. There are multiple points on the unit circle with the same y-coordinate, and multiple points on the unit circle with the same x-coordinate. On top of that, once you make a complete cycle and exceed 2*pi, you repeat the same pattern of answers.
pi/4 is equivalent to 45 degrees, and 3*pi/4 is equivalent to 135 degrees. These two angles are mirror images of each other about the y-axis. Because of the symmetry of the circle, they should have the same sine.
one thing boyz, i thought -1/2 was the sine of both 7π /6 and 11π /6, then why on the check comprehension exercise the solution of the first is only 11π /6?
Yes
And in the second one it can be 270° as well
@@saumitchandhok5730 😂 indians are intelligent 🧠 indeed
@@shubhamsingh9785 thanks 😇
@@saumitchandhok5730 welcome buddy
Thank you so much for this video!
holy moly! Such a great video! I do love your presentations!
Why cos^-1(0) is 90 degrees? cos (0) = 1. Shouldn't it be 0 as inverse?
And did I got right - if we asked that cos^-1 (+/-X) - then we have to search in I and II quaters for inverses and if sin^-1 (+/-X) then in I and IV? (I'm talking about unit circle)
so the inverse cosine of zero is asking what angle gives a cosine of zero. 90 degrees has a cosine of zero, so the inverse cosine of zero is 90 degrees.
Here is the proof
Let cos^-1(0)=Q
0=cosQ
Q=90°
I need to ask you that, is it nessasary that range must be equal if we are taking an inverse, like
If Sin x = y; y belongs to [-π/2,π/2]
Then
Sin^-1 y=x ; x belongs to [-π/2,π/2]
when analyzing an inverse function, the domain and range are always switched (between the function and the inverse) and the x values correspond to the domain and y is range
in this case the range for y=sin x is [-π/2,π/2], and so the domain for y=sin^-1 x is [-π/2,π/2].
(also sin^-1 y=x is just equal to y=sin x, not the inverse)
I guess that's where those trigg identities come in handy!
The video is amazing well explained but there’s just something I don’t get why do we have to inverse X and Y in order to find the inverse function I do get it somehow I can apply it without even being concerned but why ?
oh never mind i’ve just noticed the video in the description box
Why do you need to spread butter over bread. It is useful to differentiate the different functions
Mainly useful in solutions to equations .
Horizontal line test:- If a horizontal line touches a function more than once , then this function does not have a inverse function (refer to org. Chem tutor channel about this topic)
YOU HELPED US SURVIVE PHYSICS - SAIS CLASS OF 2022
THANK YOU VERY MUCH!!!
Perfect! All the necessary info included
Thanks Professor ❤
But my prof. I can't understand what is vertical or horizontal line test..
Basically when you take a line (or a pencil) and pass it across the graph. If it the graphed line touches the pencil in multiple places it is not a function. From left to right is the Vertical line test, and that tells you if it is a function or not. Up to down is a horizontal line test and that tells you if it is a one to one function.
Sir can you make videos on vectors..
I did that! Check my entire mathematics playlist.
best youtube tutor.
You are the best teacher ever❤❤❤
Are there inverse secant, cosecant, and cotangents
yep!
thank you math jesus
I'm surprised by the advertisement he chose to show.
I do not choose the ads that run.
How do I calculate sin^-1(1/√5) =theta? How do I find theta without a calculator?
Thank you professor Dave
You just saved my grade, thanks!
Thx
what is horizontal and vertical line test?
check out my tutorials on functions and inverse functions!
Professor Dave Explains thnk u
@@kasamuthukarthik5228 The vertical line test is to determine whether a relationship between two variables, can be considered a function. That is, a relationship that gives exclusively one output for any given input.
As an example, the equation of a circle is a non-function. x^2 + y^2 = R^2, where R is a constant indicating the radius. There is a positive half of the circle, and a negative half of the circle. You can re-write it as y=sqrt(r^2 - x^2), but that will not make it a function. Only one half of the circle can pass the vertical line test, when sqrt assumes the positive root, aka the principle root. The full circle fails the vertical line test, and is not considered a function.
The horizontal line test, is to determine whether a function's inverse also qualifies as a function. A function that passes the horizontal line test is called a one-to-one function. Meaning no individual output of the function is ever duplicated.
Consider y=x^2, and y=2^x as two examples. For y=x^2, there are two possible values of x that can make y equal 4, which are -2 and +2. This function passes the vertical line test, but fails the horizontal line test. It is a function, but not a one-to-one function. Moving on to y=2^x. For all real number inputs, there is a unique output of this function that never occurs more than once. All vertical lines only intersect it once, and all horizontal lines only intersect it once. You can find an inverse function for it, which is y = log(x)/log(2). The inverse function has a restricted domain of x>0, because the original function never returns a negative value for a real number input.
why do you remind me of a young sir issac newton?
Omg chill. What an idiot you are. He’s not even that good at teaching
The organic chemistry lab tutor is the best math teacher on RUclips by light years
@@ronaldruizdeluzuriaga2649 People have different preferences, and besides, they each have their strengths. Consider the differences in how people learn, their preferred teaching styles, and the level of their prior knowledge. Those and many factors affect what people choose to watch.
Regardless, these YT educators make life easier for many students. That's the point. There is simply no need to pit them against each other.
How do we know that "y= sin^-1(√2/2)" is "y=pi/4" and not "y=3pi/4"?
good question. I was think gin the same thing. did u find an answer?
I suppose its because the x value with sqrt2 bit is positive. so the answer where x is positive. Which happens to be pi/4
We know, because the range of sin-1 must be on quadrant I and IV (-pi/2 and pi/2), then the 3pi/4 is on the quadrant II which is outside the range... meaning the result must be pi/4.
Professor! you saved me again
Well explained
Thank you!
how to find the value of sine^-1(sine 60),where 60 is in radian?plz give an easiest method by solving it.
well inverse sine of sine just gives you the angle. so it's 60 radians, but then convert that to degrees and find a simpler way to express it.
Sir ,this question is very hard for me so i request u to make a video solution for my question
@@AmitKumar-ho3mr sin 60 rad will be in the quad 1V. Take the reference angle and that's going to be the answer. I think.
Thanks dave
Great video!!
Great , thank you
Nice approach
#ITF
Done.
Very good.
THANK YOU!
5:15 but what about 4pi/3
At minute 5:49, video shows set to radian mode, shouldn't it be in degree mode?
when dealing with trig functions, always use radians cuz the graphs of these functions are in radians
Big leap with the "this will cancel the sin on y"
Do we have to use our calculator for comprehension
All of those examples are ones that you can determine without a calculator, since the results will be angles that are part of your special right triangles
It's the rest of the quadrant?
1:14 important
when you get to evaluate them like for the inverse sine of x graph where did the root 2 over 2 come from and same with the -1/2 for cos x and root 3 for tan x.
please help
check out the tutorial earlier in this trigonometry playlist about the unit circle
@@vancehayes9319 30 degrees, 45 degrees, and 60 degrees are special cases, where the sine and cosine are either rational numbers or algebraic numbers, that can be easily memorized. The numerator is either sqrt(1), sqrt(2), or sqrt(3), and the denominator is 2. Obviously sqrt(1) = 1, so you don't need to keep writing the sqrt. You can express the exact values of all trig functions as algebraic expressions for every integer degree from 0 to 90 degrees, but most of them are complicated and don't simplify easily. 30, 45, and 60 are the ones that simplify the easiest, and you are expected to memorize. .
It comes from your special right triangles, where the 45-45-90 triangle introduces only the square root of 2 as the only irrational number among the sides. And the 30-60-90 triangle, that is exactly half of an equilateral triangle.
We have to pick two quadrants where every value on that part of the unit circle has a unique y-coordinate, in order for inverse sine to be considered a function. Otherwise, it would have multiple outputs, which would make it a non-function. We could either pick Q-1 and Q-4, or we could pick Q-2 and Q-3. But because we are most likely interested in Q-1 most of the time, we pick Q-1 and Q-4.
Likewise for cosine, we have to pick two quadrants where every value on that part of the circle has a unique x-value. We could either pick Q-1 and Q-2, or we could pick Q-3 and Q-4. Because we are most likely interested in Q-1, we pick Q-1 and Q-2.
2:13~2:18 WHAT. Is that why when dealing with inverse trig function, you limit the domain??? So that it can pass the horizontal line test???? WOW IM DEAD.
My Calculus teacher NEVER taught me this.
That's insane.
"Domain is restricted so as to sustain the bijectivity of the inverse function."
Didn't he tell even something like this?
Sine and cosine are functions that repeat their output periodically. This is why their inverse functions are not functions, unless you restrict their domain. Given a sine of 1/2, there are an infinite number of angles that have the same sine. You have to restrict the domain of sine from -pi/2 to +pi/2 radians, to get an inverse that actually is a function.
This is why you really need an angle resolver function that accepts two inputs, to determine an angle on a circle given the x and y coordinates. In programming languages, this is what arctan2 or atan2 is for. It accepts two inputs, x and y, and takes the signs of both inputs in to consideration, so that it can give you any angle on the full circle.
Your Calculus teacher probably never taught this, because this is usually a topic you learn before you get to Calculus.
Great vid
Thank you Math Jesus😊😊😊
my god, thank you so much
what if given is Arcsin2? is it undefined?
The arcsin of 2 is undefined, when you restrict your output to real numbers. Only inputs from -1 to +1 have real number outputs in inverse trig functions.
There is a way to evaluate it when you are allowed to have complex numbers as your output, which involves rearranging the trig function with Euler's formula to express it as an exponential of a complex number. It turns out that in radians, the principal result in quadrant Q1 is:
arcsin(2) = pi/2 + ln(2 + sqrt(3))*i
And it's just that easy.
but the explanation is good
The derivative of sin x is 1/√1-x^2
don't worry buddy i'll be getting to calculus very soon. but by the way the derivative of sin x is cos x.
No, that’s the derivative of arcsin(x).
Subscribed! 👍
Sir i wanna see you in real life once..
Trigonometry
Wow
Finally I understand TAN ❤ thank you
Math Jesus
Like how are people not ashamed of not including the cotangent?
Haha lol. Calculaters lol!
- some indian.
I was expecting how to solve without a calculator. :3
I hate it when coordinates are labeled.
thanks math Jesus !
Idk why I have a problem on focusing… because you look like MoistCrit lmao 😭😭😭 (my opinion)
I don’t really get it
L
@@somerandomguyplayingwithtoys I finished my exam and ate HAHAHA
@@-3lnmv7.99 finished*
@@somerandomguyplayingwithtoys man stfu please this is not english lesson
As a 14 yo it's hard to get it but I'll manage
I AM FUCKED PROFESSOR DAVE
American's you don't realise how lucky you are to be allowed to use a calculator
5:27 lucky bastards, you guys get to use calculators...
I like your username
What the fuck
Haha at 45° or pi÷4 sine is 1÷√2 not 2√2
sine of 45 is root 2 over 2.
@@ProfessorDaveExplains root 2 over 2 and 1 over root 2 are equal, professor ...
correct.
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bro that intro has to change makes me want to click off the video immediately
Way to over complicate things just like every other video out here. Why can't there just be a 1 min vid saying when to use it and why. Ended up googling it and had to search around however the inverse function is used to find the angle when both sides are known. Literally just that.
There is no one minute video because this concept can't be explained in one minute, and incidentally it can't be explained any clearer than in this video.
@@ProfessorDaveExplains no you could have started the video by saying what I said (when to use it and how). This information would be useful to people who want to know it but you could have said the basics briefly then went on about the ins and outs. Imo
The video literally begins with a derivation of the functions. You have to know what something is in order to understand how and why to use it.
@@ProfessorDaveExplains going on a out the root of x and y etc isn't explaining where to use inverse trig rule. Not Ur fault I just don't know why there are no videos that sum everything up into simple, brief words
I was reviewing the definition of an inverse function, so that the viewer can understand the inverse trig functions. That's the whole point. Again, this video is absolutely as simple as it gets. Sorry.