# What is \sin(2\arcsin(\frac{3}{5})) ?

What is $\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{3}{5}\right)\right)$ ?
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Frauental91
$\mathrm{arcsin}\left(\frac{3}{5}\right)$ is some $\theta$ between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ with $\mathrm{sin}\theta =\frac{3}{5}$
Furthermore, with $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ and $\mathrm{sin}\theta$ a positive number, we conclude that $\theta$ is between 0 and $\frac{\pi }{2}$.
We want to find $\mathrm{sin}2\theta$ and we already know $\mathrm{sin}\theta$. So if we find $\mathrm{cos}\theta$, 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 $\mathrm{sin}\theta =\frac{3}{5}$, find $\mathrm{cos}\theta$.
Use your favorite method -- draw a triangle, or a unit circle, or an angle in standard position, or skip the picture and use $\mathrm{cos}\theta =±\sqrt{1-{\mathrm{sin}}^{2}\theta }$ (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 $\theta =\mathrm{arcsin}\left(\frac{3}{5}\right)$, then $\mathrm{sin}\theta =\frac{3}{5}$ and
$\mathrm{cos}\theta =\frac{4}{5}$
And $\mathrm{sin}\left(2\theta \right)=2\mathrm{sin}\theta \mathrm{cos}\theta$
So, putting it all together we get:
$\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{3}{5}\right)\right)=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}$
###### Not exactly what you’re looking for?
Jacob Trujillo
Evaluate $\mathrm{arcsin}\left(\frac{3}{5}\right)$.
$\mathrm{sin}\left(2\cdot 0.6435011\right)$
Multiply 2 by 0.6435011
$\mathrm{sin}\left(1.28700221\right)$
The result can be shown in multiple forms.
Exact Form:
$\mathrm{sin}\left(2\left(\mathrm{arcsin}\left(\frac{3}{5}\right)\right)\right)$
Decimal Form: 0.96