What is the domain of ascsin(x+y)

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The discussion focuses on determining the domain of the function arcsin(x+y) without using a formula. It emphasizes that the sine function outputs values between -1 and 1, which means its inverse, arcsine, must accept inputs within this range. The conversation highlights the importance of understanding functions and their inverses, illustrating this with an analogy of a reverse phone book. The domain of arcsin(x+y) is established as the set of valid inputs, specifically values that keep x+y within the interval of -1 to 1. The participants conclude that recognizing these relationships clarifies how to approach the problem.
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Hey

Try to figure out the way to find domain of the function ascsin (x+y) without using the formula can someone give the clue how I should start ? Always confuse about inverse of trigonometry function. Thanks in advance.
 
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The sine function takes an angle as an input and gives out a number between -1 to 1 as an output. Therefore, what will its inverse take as inputs and outputs?
 
Thanks for your reply. :kI think that the donain should be in the interval. π/2 and -π/2. The thing is in order to be able to plot graph which shows the relation between x and y , I have absolutely no idea how I can do without formula. Can you give me some idea ?
 
I'll elaborate more. In this situation, we are calling x + y the "input" of the function. This means that, no matter what, x + y will be in the correct domain--i.e. it will be a valid input for the function.

It's necessary to look at what an inverse function means. But to do that, we have to be sure you know what a function is. A function is something that you give an input to, and it gives you an output. Sine is a function. We give it an angle, and it gives us a number between -1 and 1. The inverse of sine is arcsine, but what does that mean? In a sense, it means that arcsine "undoes" whatever sine does. Let's look at some concrete example that don't involve these functions first.

Consider the function "Reverse phone book". Yes, this is a function. Why? Because we give it an input, and it assigns the input exactly one output. I give it the number 555-5555, and it outputs the name "John Smith". What would the inverse function look like? It would just be "phone book". I give it a name (John Smith), and it outputs a number (555-5555). (there is a complication with using this analogy--just assume for now that there are no repeated names in the book, and it will not be important for now).

So let's apply the same logic to sine. Sine takes in an angle, and it gives us a number between -1 and 1. So applying our logic, we see that the inverse function (arcsine) should take in a number between -1 and 1, and it should output an angle. The domain is just another name for the "list of valid inputs", so by extension right here, what's the domain?

Note: We limit the output to numbers between -pi/2 and pi/2, because otherwise, each input would have an infinite number of different outputs, and that's a no-no for functions.
 
Ohhh thank you so much ..I know now how it should be. . You are awesome. ! :k
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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