# What is cos (2 arcsin (3/5))?

Jul 21, 2015

$\frac{7}{25}$

#### Explanation:

First consider that : $\epsilon = \arcsin \left(\frac{3}{5}\right)$

$\epsilon$ simply represents an angle.

This means that we are looking for color(red)cos(2epsilon)!

If $\epsilon = \arcsin \left(\frac{3}{5}\right)$ then,

$\implies \sin \left(\epsilon\right) = \frac{3}{5}$

To find $\cos \left(2 \epsilon\right)$ We use the identity : $\cos \left(2 \epsilon\right) = 1 - 2 {\sin}^{2} \left(\epsilon\right)$

$\implies \cos \left(2 \epsilon\right) = 1 - 2 \cdot {\left(\frac{3}{5}\right)}^{2} = \frac{25 - 18}{25} = \textcolor{b l u e}{\frac{7}{25}}$

Jul 22, 2015

We have:

$y = \cos \left(2 \arcsin \left(\frac{3}{5}\right)\right)$

I will do something similar to Antoine's method, but expand on it.
Let $\arcsin \left(\frac{3}{5}\right) = \theta$

$y = \cos \left(2 \theta\right)$

$\theta = \arcsin \left(\frac{3}{5}\right)$
$\sin \theta = \frac{3}{5}$

Using the identity $\cos \left(\theta + \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$, we then have:

$\cos \left(2 \theta\right) = \left(1 - {\sin}^{2} \theta\right) - {\sin}^{2} \theta = 1 - 2 {\sin}^{2} \theta$
(I didn't remember the result, so I just derived it)

$= 1 - 2 {\left\{\sin \left[\arcsin \left(\frac{3}{5}\right)\right]\right\}}^{2}$

$= 1 - 2 {\left(\frac{3}{5}\right)}^{2}$

$= \frac{25}{25} - 2 \left(\frac{9}{25}\right)$

$= \frac{25}{25} - \frac{18}{25} = \textcolor{b l u e}{\frac{7}{25}}$