Question

Solve:$\sin \left(2{\cos }^{-1}(-\frac{3}{5})\right)$

Answer 1

$\sin \left(2{\cos }^{-1}(-\frac{3}{5})\right)$

Let ${\cos }^{-1}(-\frac{3}{5})=\alpha$; $\alpha \in [0,\pi ]$

$\Rightarrow \cos \left(\alpha \right)=-\frac{3}{5}$

$\because \alpha \in [0,\pi ]$ and $\cos \alpha <0$

$\Rightarrow \alpha$ is in II quadrant.

$\therefore \sin \left(\alpha \right)>0$

$\sin \left(\alpha \right)=\sqrt{1-{\cos }^2\left(\alpha \right)}=\sqrt{1-\frac{9}{25}}$

$\sin \left(\alpha \right)={\left(\frac{16}{25}\right)}^{\frac{1}{2}}$

$\sin \left(2{\cos }^{-1}\right(3/5\left)\right)=\sin \left(2\alpha \right)$

$=2\sin \left(\alpha \right)\cos \left(\alpha \right)$

$=2\cdot \frac{4}{5}\cdot (-\frac{3}{5})$

$=-\frac{24}{25}$.