1. ## 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.

2. Hello,

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

This can be shown with the identity $\displaystyle \cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)=\sqrt{1-\cos^2(x)}$

3. Hi
Originally Posted by Moo
This can be shown with the identity $\displaystyle \cos^2(x)+\sin^2(x)=1 \Longleftrightarrow \sin(x)={ \color{red}\pm}\sqrt{1-\cos^2(x)}$
One will choose either $\displaystyle \sin x =+\sqrt{1-\cos^2 x}$ if the sine is positive (that is to say $\displaystyle x\in[0,\pi]$) either $\displaystyle \sin x =-\sqrt{1-\cos^2 x }$ if $\displaystyle x$ lies in $\displaystyle [\pi,2\pi]$ where the sine is negative.

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