1. The problem statement, all variables and given/known data I have a problem, I don't know to substantiate, why arcsin(sin(x)) = sin(arcsin(x)) = x ? Thank you very much for each advice. 2. Relevant equations 3. The attempt at a solution
Re: arcsin(sin(x)) i didnt exactly get your question. Are you trying to find the range of values of x for which the above eq is valid? what are u asking? can u be clearer.
Re: arcsin(sin(x)) My question is: why arcsin(sin(x)) is possible to regulate for x. Why arcsin(sin(x)) = x? Why graph for y = arcsin(sin(x)) is y = x? Thank you.
Re: arcsin(sin(x)) Do you know about the composition of inverse functions? Also, arcsin(sin(x)) = sin(arcsin(x)) = x only for -1≤x≤1 for all three.
Re: arcsin(sin(x)) [itex]\sin(\arcsin(x)) = x[/itex] for all real [itex]x[/itex] but [itex]\arcsin(\sin(x))[/itex] only for [itex]-\pi/2 \le x \le \pi/2[/itex] . That's with one standard way of defining [itex]\arcsin[/itex]
Re: arcsin(sin(x)) well look at it this way : sin( asin(x ) ) let x = 1; the asin(1) = P; then sin(P) = 1;
Re: arcsin(sin(x)) Did you mean all x in [-1,1]? (Or are you making an assertion about the complex Arcsin function?)
Re: arcsin(sin(x)) You are right. For my equation you have to use complex arcsin. Wherever arcsin is defined, we have sin(arcsin(x)) = x , that is what we mean by arcsin.
Re: arcsin(sin(x)) y=arcsin[sin[x]] implies sin[y] = sin[x] and y=x. Forgetting the intervals, I don't the question had much to do with the interval which this valid for. It was just how does the graph look like y=x.