sin$\left[\frac{\mathit{\pi }}{3}-\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(-\frac{1}{2}\right)\right]$
Let $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(-\frac{1}{2}\right)$ = x
⇒ $\left(-\frac{1}{2}\right)$ = sin x
⇒ sin x = - sin $\frac{\mathit{\pi }}{6}$ = sin $\left(-\frac{\mathit{\pi }}{6}\right)$ = sin $\left(2\mathit{\pi }-\frac{\mathit{\pi }}{6}\right)$
⇒ x = 2π - $\frac{\mathit{\pi }}{6}$
∴ sin $\left[\frac{\mathit{\pi }}{3}-\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(-\frac{1}{2}\right)\right]$ = sin $\left[\frac{\mathit{\pi }}{3}-\left(2\mathit{\pi }-\frac{\mathit{\pi }}{6}\right)\right]$
= sin $\left[-\frac{9\mathit{\pi }}{6}\right]$
= - sin $\left[\frac{3\mathit{\pi }}{2}\right]$
= - sin $\left[\mathit{\pi }+\frac{\mathit{\pi }}{2}\right]$
= - $\left[-\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{\pi }}{2}\right]$
= + sin $\frac{\mathit{\pi }}{2}$
= 1
Thus,sin$\left[\frac{\mathit{\pi }}{3}-\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(-\frac{1}{2}\right)\right]$ = 1