# How do you evaluate sin(2arccos(3/5))?

Oct 17, 2015

It's $\frac{24}{25}$

#### Explanation:

We need to make use of the property that $\cos \left(\arccos \left(x\right)\right) = x$. At the moment we have a sine instead of a cosine. We also know the formula: ${\cos}^{2} x + {\sin}^{2} x = 1$.
To make use of both of these, we need to square the expression and immediately take the square root (so we don't do anything illegal):

sqrt(sin^2(2arccos(3/5)))=sqrt(1-cos^2(2arccos(3/5))

The only problem that we have now is the $2$ in front of the $\arccos$. We can solve this by ussing the double angle formula:

$\cos \left(2 a\right) = {\cos}^{2} a - {\sin}^{2} a = 2 {\cos}^{2} a - 1 = 1 - 2 {\sin}^{2} a$

We need it in therms of the cosine, so let's take the second one:

sqrt(1-(2cos^2(arccos(3/5))-1)^2
Simplifying this:
$\sqrt{1 - 4 \cdot {\cos}^{4} \left(\arccos \left(\frac{3}{5}\right)\right) + 4 {\cos}^{2} \left(\arccos \left(\frac{3}{5}\right)\right) - 1}$
Now we can replace $\cos \left(\arccos \left(x\right)\right)$ by $x$:
sqrt(4*(3/5)^2-4*(3/5)^4)=sqrt(4*(3/5)^2*(1-(3/5)^2)
$= 2 \cdot \frac{3}{5} \cdot \sqrt{1 - \frac{9}{25}} = \frac{6}{5} \cdot \sqrt{\frac{16}{25}} = \frac{6}{5} \cdot \frac{4}{5} = \frac{24}{25}$