Question

what is the range of 2*sin^-1(x)

Answer 1

hmmm

Answer 2

range of arcsine is not minus infinity to infinity, it is \([-\frac{\pi}{2},\frac{\pi}{2}]\)

Answer 3

@Sheng i dont think so.
if inverse circular function is restricted

Answer 4

isnt that domain?

Answer 5

multiply by two to get your answer

Answer 6

no that is the range

Answer 7

@satellite73 i dont care about the 2 infront of arcsine?

Answer 8

domain is \([-1,1]\) because that is the range of sine
@ksw19372 yes you do care about the two

Answer 9

that is why i said multiply by 2

Answer 10

then how do i get [-pi/2,pi/2]??
i dont know why but i am keep getting [-pi/4,pi/4]

Answer 11

if arcsine is \(-\frac{\pi}{2}\) then 2arcsine is \(-\pi\)

Answer 12

i thought range was bound of y values?

Answer 13

i think you are confusing domain and range

Answer 14

@cheng its inverse

Answer 15

domain of arcsine is the range of sine

Answer 16

ooh! thx for pointing that out, inverse functions domain/range are flipped :)

Answer 17

range of sine is \([-1,1]\) so domain of arcsine is \([-1,1]\)

Answer 18

@satellite73 what about the range of sin^-1(x)?

Answer 19

now to make sine one to one you restrict the domain to \([-\frac{\pi}{2},\frac{\pi}{2}]\) so that now becomes the range of arsine

Answer 20

so range of arcsine is \([-\frac{\pi}{2},\frac{\pi}{2}]\) and range of \(2\arcsin(x)\) is
\[[-\pi,\pi]\]

Answer 21

since you multiply your answer by two

Answer 22

so the answer of my question is [-pi,pi]?

Answer 23

yes

Answer 24

i get it !

Answer 25

\[\arcsin(-1)=-\frac{\pi}{2}\] and so
\[2\arcsin(-1)=-\pi\]

Answer 26

ok good!

Answer 27

@satellite73 thank you so much. please help in the further questions later on! :)