Show that the function given by $f(x) = \sin x$ is strictly increasing in the interval $\left(0, \frac{\pi}{2}\right)$.

(N/A) The given function is $f(x) = \sin x$.
To determine the intervals of increase or decrease,we find the derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$.
We analyze the sign of $f'(x)$ in the interval $\left(0, \frac{\pi}{2}\right)$.
For any $x \in \left(0, \frac{\pi}{2}\right)$,the value of $\cos x$ is always positive (since the angle lies in the first quadrant).
Therefore,$f'(x) = \cos x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$.
Since the derivative $f'(x) > 0$ for all $x$ in the given interval,the function $f(x) = \sin x$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.