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

Its just the feeling man, I felt that too

@@mind-h4i Haha!

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.

same idk what the heck going on

why you homeschooled?

samee

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.

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.

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.

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)

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°

when dealing with trig functions, always use radians cuz the graphs of these functions are in radians

yep!

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.

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.

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 .

Yes

And in the second one it can be 270° as well

​@@saumitchandhok5730 😂 indians are intelligent 🧠 indeed

@@shubhamsingh9785 thanks 😇

@@saumitchandhok5730 welcome buddy

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

I did that! Check my entire mathematics playlist.

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.

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.

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

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.

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 KZfaq 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.

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 💀

I do not choose the ads that run.

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.

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).

L

@@somerandomguyplayingwithtoys I finished my exam and ate HAHAHA

@@-3lnmv7.99 finished*

​@@somerandomguyplayingwithtoys man stfu please this is not english lesson

I like your username

sine of 45 is root 2 over 2.

@@ProfessorDaveExplains root 2 over 2 and 1 over root 2 are equal, professor ...

correct.

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.