# I have a question thatasks me to find an algebraic

I have a question thatasks me to find an algebraic expression for $\mathrm{sin}\left(\mathrm{arccos}\left(x\right)\right)$. 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 $\sqrt{1-{x}^{2}}$
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Edward Patten
What is $\mathrm{arccos}\left(x\right)$? it is the angle given by the ratio of sides of length x and 1. That is, on a triangle with adjacent side length x and hypotenuse length 1 we will find the angle $\mathrm{arccos}\left(x\right)$.
Now what is $\mathrm{sin}\left(y\right)$? It is the ratio of the opposite side to the hypotenuse of a triangle with angle y in the respective part. So if we use $y=\mathrm{arccos}\left(x\right)$, we have that $\mathrm{sin}\left(y\right)$ is the ratio of the sides with degree given by a triangle whose adjacent and hypotenuse sides are length x and 1 respectively.
Since we have the adjacent and hypotenuse lengths, we can calculate the opposite sides length by the Pythagorean theorem. This means, if we use z for the opposite side, that ${z}^{2}+{x}^{2}={1}^{2}=1$. Solving for z gives $z=\sqrt{1-{x}^{2}}$
Then $\mathrm{sin}\left(y\right)$ is the ratio of the opposite size, $z=\sqrt{1-{x}^{2}}$ and the hypotenuse, 1. We may now say $\mathrm{sin}\left(\mathrm{arccos}\left(x\right)\right)=\sqrt{1-{x}^{2}}$.
###### Not exactly what you’re looking for?
esfloravaou
Set $\alpha =\mathrm{arccos}x$. Since, by definition, $\alpha \in \left[0,\pi \right]$, you know that $\mathrm{sin}\alpha \ge 0$, so
$\mathrm{sin}\mathrm{arccos}x=\mathrm{sin}\alpha =\sqrt{1-{\mathrm{cos}}^{2}\alpha }=\sqrt{1-{\left(\mathrm{cos}\mathrm{arccos}x\right)}^{2}}=\sqrt{1-{x}^{2}}$