Question

True or False? Can anyone explain these? Thank you
The statement sin(sin^-1)=x for all real numbers in the interval -∞≤ X≤ ∞?

The statement sin^-1(sinx)=x for all real numbers in the interval -∞≤ X≤ ∞?

Answer 1

First question doesn't make sense sin(sin^-1)=x? is that really what it says?

For the second one ask yourself what is the domain of sin(x), then ask yourself what is the range of sin(x), then ask yourself is the range of sin(x) either a subset or equal to that of the domain of the sine inverse function.

Answer 2

Sorry, I left out an X. It says sin(sin^-1X)=x

Answer 3

what is the domain of the sine inverse function?

Answer 4

sin domain all real numbers
sin^-1 I'm not sure -1,1?

Answer 5

Right!
\[y=\sin(x) \text{ has domain all real numbers } \text{ or you can say } \\ y=\sin(x) \text{ where} -\infty <x< \infty \\ \text{ and yes } \\ y=\arcsin(x) \text{ where } -1 \le x \le 1 \]

Answer 6

now look at the first one the inside function is arcsin(x) which means we can only input what numbers?

Answer 7

0

Answer 8

you just said the domain was for arcsin(x) was [-1,1]
why do you change it to just the number 0?

Answer 9

arcsin(x) has domain [-1,1]
this means you can input any number in the interval [-1,1]
and since the domain for sin(x) is all real numbers it doesn't matter what the output of arcsin(x) is...
the domain of sin(arcsin(x)) will just be the domain of arcsin(x)

Answer 10

-1 ≤ 0 ≤ 1 Because 0 is greater than -1, but less than 1

Answer 11

you do know there are infinitely many numbers between -1 and 1?

Answer 12

0 is not the only number

Answer 13

Oh okay. so problem 1 is true and problem 2 will be false.

Answer 14

examples you have
-1,-.99999,-.95,-.55, -.5, 0,.4, so on...
you can not just write on the numbers in a listing type roster

Answer 15

No...
the domain of sin(arcsin(x)) is the domain of arcsin(x)

Answer 16

you can see reason above

Answer 17

and so what is the domain of sin(arcsin(x))?

Answer 18

The domain of sin(arcsin(x)) is all real numbers between -1 and 1

Answer 19

right so sin(arcsin(x))=x when x is in [-1,1]
choosing an x outside that interval for example x=2 would give us arcsin(2) which is undefined...
so for the first one we only needed to consider the domain of arcsin(x)

Answer 20

so while arcsin(sin(x)) has domain all real number we will not have arcsin(sin(x))=x
for all real numbers because arcsin doesn't have range all real numbers


arcsin(sin(x))=x.....
now this one sin(x) does have all real numbers
but here you have to consider the domain restriction you put on y=sin(x) so that y=arcsin(x) could exist. what is that domain restriction?

Answer 21

I'm asking you what was the domain restriction put on y=sin(x) so y=arcsin(x) could exist?
This question is also equivalent to what is the range of y=arcsin(x)?

Answer 22

-1 to 1

Answer 23

no that is the domain of y=arcsin(x) and the range of y=sin(x)

Answer 24

the restricted domain of y=sin(x) and the range of y=arcsin(x) is [-pi/2,pi/2]

Answer 25

Thank you

Answer 26

some of the words got cut off