Inverse Trig functions

[Violagirl]

I was wondering how you would say this property? I know what it means but am not sure how to say it aloud?

f^-1(f(x))=sin^-1(sin x)=x where -pi/2<x<pi/2

f(f^-1(x))=sin(sin^-1 x)=x where -1<x<1

I also wasn't sure if that was for the domain or range it falls between and if it's the same with other functions or not?

Also how would you do these two problems:

Find the inverse function f^-1 of each function f. State the domain and range of f and f^-1.

1. f(x)=3 sin(2x+1)

Find the exact solution of the equation:

2. 4 cos^-1x-2pi=2 cos^-1 x

 

[fasteddie65]

2. 4 cos^-1 x - 2? = 2 cos^-1 x
2 cos^-1 x - 2? = 0
2 cos^-1 x = 2?
cos^-1 x = ?
x = cos ? = -1

 

[Violagirl]

Thanks for the help on the second one!

I think I was able to figure out the first one, I was wondering if someone could check it?

f(x)=3 sin (2x-1)

y=3 sin (2x-1)

x=3 sin (2y-1)

x/3+1=sin (2y)

sin^-1 x/3 + 1=2y

y=1/2 sin^1 x/3+1

f(x): D: (-oo, oo)

I'm not sure how to find the domain of f(-1). Can anyone show me how to do this? Thanks!

 

[mmm4444bot]

x=3 sin (2y-1) <<<< Take the inverse sine of both sides of this equation, instead.

x/3+1=sin (2y) <<<< We are not allowed to do this.

The expression 2y - 1 is the input to the sine function; we may not change it. After you take the inverse sine, the expression will no longer be the input to any function, so you can add 1 to both sides at that point.


I'm not sure how to find the domain of f(-1). Can anyone show me how to do this?

The domain of f[sup:5mhqj54n]-1[/sup:5mhqj54n](x) is the range of f(x).