# How do you evaluate sin(arccos(1/2))?

Jul 3, 2015

$\sin \left(\arccos \left(\frac{1}{2}\right)\right) = \frac{\sqrt{3}}{2}$

#### Explanation:

To calculate this, you need to know two identities:
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1 \iff {\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$$\setminus \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .}$$\left(1\right)$
$\cos \left(\arccos \left(x\right)\right) = x$$\setminus \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .}$ $\left(2\right)$

From $\left(2\right)$, we know that we need to have a cosine instead of a sine. We can convert a sine into a cosine by using $\left(1\right)$. If we want to use $\left(1\right)$, we need ${\sin}^{2}$, so let's square the expression.
To allow us to square, we'll also immediately take the square root:
$\sqrt{{\left(\sin \left(\arccos \left(\frac{1}{2}\right)\right)\right)}^{2}} = \sqrt{{\sin}^{2} \left(\arccos \left(\frac{1}{2}\right)\right)}$
Now, we can use $\left(1\right)$
$\sqrt{1 - {\cos}^{2} \left(\arccos \left(\frac{1}{2}\right)\right)} = \sqrt{1 - \cos \left(\arccos \left(\frac{1}{2}\right)\right) \cdot \cos \left(\arccos \left(\frac{1}{2}\right)\right)}$
Now, we can use $\left(2\right)$
$\sqrt{1 - \frac{1}{2} \cdot \frac{1}{2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$