# How do you evaluate sin[2cos^(-1)(3/5)] ?

$\pm \frac{24}{25} = \pm 0.96$.
Let $a = {\cos}^{- 1} \left(\frac{3}{5}\right)$. Then $\cos a = \frac{3}{5} > 0$.. So,a is in the 1st or un
the 4th quadrant. Accordingly, $\sin a = \pm \frac{4}{5}$.
Now, the given sine is $\sin 2 a = 2 \sin a \cos a = 2 \left(\pm \frac{4}{5}\right) \left(\frac{3}{5}\right) = \pm \frac{24}{25}$..