TITLE: Find an algebraic expression for sin(arccos(x))
QUESTION [6 upvotes]: I have a question that asks me to find an algebraic expression for sin(arccos(x)). From the lone example in the book I seen they're doing some multistep thing with the identities, but I'm just not even sure where to start here. It's supposed to be $\sqrt{1-x^2}$

REPLY [13 votes]: Set $\alpha=\arccos x$. Since, by definition, $\alpha\in[0,\pi]$, you know that $\sin\alpha\ge0$, so
$$
\sin\arccos x=\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-(\cos\arccos x)^2}=\sqrt{1-x^2}
$$