The minimum value

The given function is $f(x)=\sin x, x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$f(x)=\sin x$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=\cos x$

For maxima or minima,

$f^{\prime}(x)=0$

$\Rightarrow \cos x=0$

$\Rightarrow x=-\frac{\pi}{2}$ or $x=\frac{\pi}{2}$, for all $x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Now,

$f^{\prime \prime}(x)=-\sin x$

At $x=\frac{\pi}{2}$, we have

$f^{\prime \prime}\left(\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}=-1<0$

So, $x=\frac{\pi}{2}$ is the point of local maximum of $f(x)$.

At $x=-\frac{\pi}{2}$, we have

$f^{\prime \prime}\left(-\frac{\pi}{2}\right)=-\sin \left(-\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1>0$         $[\sin (-\theta)=-\sin \theta]$

So, $x=-\frac{\pi}{2}$ is the point of local minimum of $f(x)$.

$\therefore$ Minimum value of $f(x)=f\left(-\frac{\pi}{2}\right)=\sin \left(-\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}=-1$

Thus, the minimum value of $f(x)=\sin x$ in $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is $-1$.

The minimum value of $f(x)=\sin x$ in $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is  ___−1___.