inverse of sin

Vali

Junior Member
Joined
Feb 27, 2018
Messages
87
The inverse of the function \(\displaystyle f:[\frac{\pi}{2},\frac{3\pi}{2}]\rightarrow [-1,1], f(x)=\sin(x)\), is the function \(\displaystyle f^{-1}:[-1,1]\rightarrow [\frac{\pi}{2},\frac{3\pi}{2}]\) which is defined: ( right answer: pi-arcsinx)
pi+arcsinx
pi-arcsinx
arcsinx
2pi-arcsinx
-pi+arcsinx
I know the inverse of sin is arcsin but why is pi-arcsin ?
 
The inverse of the function \(\displaystyle f:[\frac{\pi}{2},\frac{3\pi}{2}]\rightarrow [-1,1], f(x)=\sin(x)\), is the function \(\displaystyle f^{-1}:[-1,1]\rightarrow [\frac{\pi}{2},\frac{3\pi}{2}]\) which is defined: ( right answer: pi-arcsinx)
pi+arcsinx
pi-arcsinx
arcsinx
2pi-arcsinx
-pi+arcsinx
I know the inverse of sin is arcsin but why is pi-arcsin ?
I simply do not understand any of the above.
The function \(\displaystyle \arcsin\) has domain of \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) and range \(\displaystyle [-1,1]\).
Where do you find \(\displaystyle \pi-\arcsin(x)~?\)
 
The inverse of the function \(\displaystyle f:[\frac{\pi}{2},\frac{3\pi}{2}]\rightarrow [-1,1], f(x)=\sin(x)\), is the function \(\displaystyle f^{-1}:[-1,1]\rightarrow [\frac{\pi}{2},\frac{3\pi}{2}]\) which is defined: ( right answer: pi-arcsinx)
pi+arcsinx
pi-arcsinx
arcsinx
2pi-arcsinx
-pi+arcsinx
I know the inverse of sin is arcsin but why is pi-arcsin ?
Sketch a graph of f (that is, the sine function restricted to the domain [MATH][\frac{\pi}{2},\frac{3\pi}{2}][/MATH]). Its inverse must take any number in [-1, 1] to a number in [MATH][\frac{\pi}{2},\frac{3\pi}{2}][/MATH]. The function arcsin has the wrong range to accomplish this, namely [MATH][-\frac{\pi}{2},\frac{\pi}{2}][/MATH]. But if you subtract any number in this interval from [MATH]\pi[/MATH], you get a number in the required interval; that is, the range of [MATH]\pi - \arcsin(x)[/MATH] is [MATH][\frac{\pi}{2},\frac{3\pi}{2}][/MATH] as required. Moreover, [MATH]\sin(\pi - x) = \sin(x)[/MATH], so [MATH]\sin(\pi - \arcsin(x)) = \sin(\arcsin(x) = x[/MATH]. Therefore, this is the inverse of f.
 
Top