# inverse function question.

• May 6th 2008, 12:23 PM
augustfai
inverse function question.
For the problem:

sin(arccos(-6/7))

Would it be correct that I figure out arccos(-6/7) first and then plug that answer into sin? That's what I've been doing, but it's out of domain since the answer is around 2.6 when I plug the inverse of cos(-6/7) into the calculator.
• May 6th 2008, 12:28 PM
Moo
Hello,

Quote:

Originally Posted by augustfai
For the problem:

sin(arccos(-6/7))

Would it be correct that I figure out arccos(-6/7) first and then plug that answer into sin? That's what I've been doing, but it's out of domain since the answer is around 2.6 when I plug the inverse of cos(-6/7) into the calculator.

$\sin(\arccos(x))=\sqrt{1-x^2}$

This can be shown with the identity $\cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)=\sqrt{1-\cos^2(x)}$
• May 6th 2008, 12:50 PM
flyingsquirrel
Hi
Quote:

Originally Posted by Moo
This can be shown with the identity $\cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)={ \color{red}\pm}\sqrt{1-\cos^2(x)}$

One will choose either $\sin x =+\sqrt{1-\cos^2 x}$ if the sine is positive (that is to say $x\in[0,\pi]$) either $\sin x =-\sqrt{1-\cos^2 x }$ if $x$ lies in $[\pi,2\pi]$ where the sine is negative.

Now, you've to choose between $\sin(\arccos x )=+\sqrt{1-x^2}$ and $\sin(\arccos x )=-\sqrt{1-x^2}$ :D (What's the range of $x\mapsto\arccos x$ ? )