Answer: \(\dfrac{-\pi}{6}\)
Step-by-step explanation:
To find : The exact value of \(\sin^{-1}(-0.5)\) .
We know that \(\sin^{-1}(\dfrac{1}{2})=\dfrac{\pi}{6}\).
The range of \(\sin^{-1}(x)\) is between \(\dfrac{-\pi}{2}\text{ and }\dfrac{\pi}{2}\)
Let \(t=\sin^{-1}(-0.5)\)
\(\sin t=-0.5=-\sin(\dfrac{\pi}{6})=\sin(\dfrac{-\pi}{6})\ [\text{Since }}\sin(-x)=-\sin x]\)
\(\sin(\dfrac{-\pi}{6})=0.5\)
Hence, the principal value of \(\sin^{-1}(-0.5)=\dfrac{-\pi}{6}\)
