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

Jun 27, 2015

$\frac{24}{25} = \frac{24 \times 4}{25 \times 4} = \frac{96}{100} = 0.96$

#### Explanation:

$\arcsin \left(\frac{3}{5}\right)$ is some $\theta$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\sin \theta = \frac{3}{5}$.
Furthermore, with $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ and $\sin \theta$ a positive number, we conclude that $\theta$ is between $0$ and $\frac{\pi}{2}$.

We want to find $\sin 2 \theta$ and we already know $\sin \theta$, so if we find $\cos \theta$, then we can ue the double angle formula for sine.

You've probably done this kind of problem many times by now. $\theta$ is in the first quadrant and $\sin \theta = \frac{3}{5}$, find $\cos \theta$.

Use your favorite method -- draw a triangle, or a unit circle, or an angle in standard position, or skip the picture and use $\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$ (recall that our $\theta$ is in Quadrant 1, so its cosine is positive.)

All of the above is really explanation of our thought process.

All we really need to write is something like:

Let $\theta = \arcsin \left(\frac{3}{5}\right)$, then $\sin \theta = \frac{3}{5}$ and
$\cos \theta = \frac{4}{5}$

And $\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$.

So, putting it all together we get:

$\sin \left(2 \arcsin \left(\frac{3}{5}\right)\right) = 2 \left(\frac{3}{5}\right) \left(\frac{4}{5}\right) = \frac{24}{25}$