Find the exact value of the expression. \[ \sin \left(2 \cos ^{-1}\left(-\frac{4}{5}\right)\right) \] \( \sin \left(2 \cos ^{-1}\left(-\frac{4}{5}\right)\right)=\square \)

To find the exact value of the expression \( \sin \left(2 \cos ^{-1}\left(-\frac{4}{5}\right)\right) \), we can use the double angle identity for sine: \[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \] Let \( \theta = \cos^{-1}\left(-\frac{4}{5}\right) \). Therefore, we have \( \cos(\theta) = -\frac{4}{5} \). Next, we need to find \( \sin(\theta) \). Using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Substituting \( \cos(\theta) \): \[ \sin^2(\theta) + \left(-\frac{4}{5}\right)^2 = 1 \] \[ \sin^2(\theta) + \frac{16}{25} = 1 \] \[ \sin^2(\theta) = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \] \[ \sin(\theta) = \pm \frac{3}{5} \] Since \( \theta = \cos^{-1}\left(-\frac{4}{5}\right) \) corresponds to the angle in the second quadrant where sine is positive, we take: \[ \sin(\theta) = \frac{3}{5} \] Now, we can substitute back into the double angle formula: \[ \sin\left(2 \theta\right) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) \] Calculating this gives: \[ \sin\left(2 \theta\right) = 2 \cdot \frac{3}{5} \cdot -\frac{4}{5} = -\frac{24}{25} \] Thus, the exact value of the expression is: \[ \sin \left(2 \cos ^{-1}\left(-\frac{4}{5}\right)\right) = -\frac{24}{25} \] Final answer: \[ -\frac{24}{25} \]