English mein likhe

Differentiating with respect to $x$:

The interval is $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.
We know that $\cos x$ is an even function, and its value at the endpoints is $\cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.
At $x=0$, $\cos(0) = 1$.
For $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, the value of $\cos x$ ranges from just above $\frac{1}{2}$ up to 1 (inclusive of 1).
Thus, we have $\frac{1}{2} < \cos x \le 1$.

Using the inequality for $\cos x$ from Step 2, we add $-\frac{1}{2}$ to all parts of the inequality:

Since $0 < f'(x) \le \frac{1}{2}$, it implies that $f'(x)$ is always greater than 0 for all $x$ in the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

Therefore, the function $f(x) = -\frac{x}{2} + \sin x$ is strictly increasing on the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.