# 1 Trig Problem with inverse

• Nov 6th 2006, 10:20 AM
qbkr21
1 Trig Problem with inverse
1. Given: f(x)=arcsin(x/3)+5, and -3<=x<=3

Unknown: f(x)^-1=?

I got y= sin^-1(x-5)*5

Any suggestions?
• Nov 6th 2006, 11:20 AM
Jameson
I'm confused on your notation. What does f(x)^1 mean? Is that supposed to be the inverse, or $\displaystyle f^{-1}(x)$?
• Nov 6th 2006, 12:19 PM
qbkr21
Re:
yes I am sorry
• Nov 6th 2006, 12:44 PM
Jameson
Mmk. So to do this problem, just switch x and y.

Original: $\displaystyle y=\sin^{-1} \left(\frac{x}{3} \right) +5$

Switch x's and y's. $\displaystyle x=\sin^{-1} \left(\frac{y}{3} \right) +5$

$\displaystyle \sin^{-1} \left(\frac{y}{3} \right) = x-5$

Take the sine of both sides. Can you finish it?
• Nov 6th 2006, 12:47 PM
qbkr21
Re:
No I got to thant point, I can finish it.
• Nov 6th 2006, 12:49 PM
qbkr21
Re:
What happens when you take a sin^-1 back across the equal's sign. Does it change back to just plain sin ?
• Nov 6th 2006, 12:56 PM
Jameson
$\displaystyle \sin^{-1} \left(\frac{y}{3} \right) = x-5$

Take the sine of both sides.

$\displaystyle \frac{y}{3}=\sin(x-5)$

$\displaystyle f^{-1}(x)=3\sin(x-5)$
• Nov 6th 2006, 12:59 PM
qbkr21
Re:
Thanks Jameson, you are the Master of Math!!! And God's Gift to us all!!!