# inverse function for arcsin

• Nov 9th 2006, 04:25 PM
kcsteven
inverse function for arcsin
I have a webworks problem that I am not doing correctly. here it is
f(x)=arcsin(x/7)+10
This is what I did. Y=arcsin(x/7)+10
X=arcsin(y/7)+10
x-10=arcsin(y/7)
y/7=arcsin^-1(x-10)
the answer is supposed to be in this form, webworks said this is incorrect. I did the rest of the problems this way but they were tan and cos and I am sure this one must be done differently or I must have skipped or omitted something. Please let me know if you can help me understand how this one is solved.
Thank you!
keith stevens
keithstevens@ctc.net
• Nov 9th 2006, 05:44 PM
ThePerfectHacker
Quote:

Originally Posted by kcsteven
I have a webworks problem that I am not doing correctly. here it is
f(x)=arcsin(x/7)+10
This is what I did. Y=arcsin(x/7)+10
X=arcsin(y/7)+10
x-10=arcsin(y/7)
y/7=arcsin^-1(x-10)
the answer is supposed to be in this form, webworks said this is incorrect. I did the rest of the problems this way but they were tan and cos and I am sure this one must be done differently or I must have skipped or omitted something. Please let me know if you can help me understand how this one is solved.
Thank you!
keith stevens
keithstevens@ctc.net

It is okay but!
$\displaystyle \arcsin^{-1}(x)=\sin x$
Because the inverse of the inverse is the original function.
So you have,
$\displaystyle y=7\sin (x-10)$

WARNING. If you want to be stirctly correct it is not really the regular sine curve (along the x-axis) this one must be restricted on some domain because the original inverse sine you had was itself restricted.
• Nov 9th 2006, 05:47 PM
Soroban
Hello, Keith!

Quote:

Find the inverse of: $\displaystyle f(x) \:=\:\arcsin\left(\frac{x}{7}\right)+10$

This is what I did.
$\displaystyle y\:=\:\arcsin\left(\frac{x}{7}\right)+10$
$\displaystyle x \:=\:\arcsin\left(\frac{y}{7}\right)+10$
$\displaystyle x-10\:=\:\arcsin\left(\frac{y}{7}\right)$
$\displaystyle \frac{y}{7}\:=\:\arcsin^{-1}(x-10)$ . . . ??

Answer: .$\displaystyle y\:=\:7\arcsin^{-1}(x-10)$

You just invented some new notation: "the inverse of the inverse sine" . . .

You had: .$\displaystyle x - 10 \:=\:\arcsin\left(\frac{y}{7}\right)$

Take the sine of both sides:
. . $\displaystyle \sin(x - 10) \:=\:\sin\left[\arcsin\left(\frac{y}{7}\right)\right]$

. . $\displaystyle \sin(x-10) \:=\:\frac{y}{7}\quad\Rightarrow\quad 7\sin(x-10)\:=\:y$

Therefore: .$\displaystyle f^{-1}(x)\:=\:7\sin(x-10)$