anonymous
  • anonymous
what is arcsin?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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petiteme
  • petiteme
is it the inverse function of sin?
anonymous
  • anonymous
Can you explain that?
anonymous
  • anonymous
yes,I think it's about function! :)

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anonymous
  • anonymous
but I don't know what exactly is it!
dape
  • dape
It's hard to say that it's anything more than the inverse of \(\sin(x)\), since that's what it's defined to be. So it is the function that makes this true \[\arcsin(\sin(x))=x.\]
dape
  • dape
But we have to restrict which \(x\)'s it works for, since if we put in \(x=\pi\) say, we have \[\arcsin(\sin(\pi))=\arcsin(0)=\pi.\] But if we put in \(2\pi\) we also have that \(\arcsin(0)=2\pi\).
dape
  • dape
So if it should be a function we must restrict it's domain, so \(\arcsin(x)\) is only defined for \(-1≤x≤1\). It's graph is just \(\sin(x)\) tipped on it's side and cut off, so that it becomes a function.
anonymous
  • anonymous
I think that you may know that arcsin is abbreviation of arcsine which is an invers function arcsin = (opp/ hyp)= q
anonymous
  • anonymous
I havn't study about function so can I understand arcsin?
anonymous
  • anonymous
arcsin is part of the inverse function in calculus
anonymous
  • anonymous
I have been seen some of it, but not all so that is what I can tell you about it
anonymous
  • anonymous
if arcsin is part of the inverse function in calculus,and I don't know anything about function so I can't understand this?
UnkleRhaukus
  • UnkleRhaukus
The Sine of an angle is equal to, the ratio of the opposite side to the hypotenuse side, of a right angled triangle. \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenues}}\) As theta goes from 0 to 2π ,(0° to 360°),the function returns to its starting value and repeats, \(\sin(\theta)=\sin(\theta+2\pi n)\) The Arcsine of the ratio of the opposite side to the hypotenues side of the right angled triangle, is equal to the angle. \(\arcsin\left(\frac{\text{opposite}}{\text{hypotenues}}\right)=\theta\) This time there is a restriction on the domain, the hypotenuse will always be larger than the opposite side and hence the ratio will never be larger in magnitude than 1. This means that \(\arcsin \theta\) will only make sense / be defined for x between -1 and 1
anonymous
  • anonymous
@UnkleRhaukus , Thank you :)
anonymous
  • anonymous
I am pretty sure that you will understand it , if you put on your part ok don't worry too much. take care ok I got to go
dape
  • dape
But that definition of \(\sin(\theta)\) breaks down when \(\pi/2≤\theta≤0\). It's more practical to define it in terms of the unit circle. In this context \(\arcsin(y)\) is just the angle \(\theta\) for which \(\sin(\theta)=y\), and can be phrased as "for which angle \(\theta\) is the y-coordinate of a point on the unit circle equal to \(y\)?"
dape
  • dape
And \(\sin(\theta)\) could be phrased "what is the y-coordinate at angle \(\theta\) on the unit circle?"
anonymous
  • anonymous
@dape , Thank you very much :)

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