What is $\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{3}{5}\right)\right)$ ?

Answer & Explanation

Frauental91

Beginner2022-01-22Added 15 answers

$\mathrm{arcsin}\left(\frac{3}{5}\right)$ 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 $\mathrm{cos}\theta =\pm \sqrt{1-{\mathrm{sin}}^{2}\theta}$ (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: $\mathrm{sin}\left(2\mathrm{arcsin}\left(\frac{3}{5}\right)\right)=2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)=\frac{24}{25}$

Jacob Trujillo

Beginner2022-01-23Added 13 answers

Evaluate $\mathrm{arcsin}\left(\frac{3}{5}\right)$.
$\mathrm{sin}(2\cdot 0.6435011)$
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