1. **Problem:** Show that $f(x) = \cos^2 x$ is decreasing on $(0, \frac{\pi}{2})$. 2. **Formula and rules:** To determine if a function is increasing or decreasing, we use the first derivative test. If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing there. 3. **Find the derivative:** $$f(x) = \cos^2 x = (\cos x)^2$$ Using the chain rule: $$f'(x) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x$$ 4. **Analyze the sign of $f'(x)$ on $(0, \frac{\pi}{2})$:** - On $(0, \frac{\pi}{2})$, $\cos x > 0$ and $\sin x > 0$. - Therefore, $f'(x) = -2 \cos x \sin x < 0$ on this interval. 5. **Conclusion:** Since $f'(x) < 0$ on $(0, \frac{\pi}{2})$, the function $f(x) = \cos^2 x$ is decreasing on this interval. Final answer: $f(x) = \cos^2 x$ is decreasing on $(0, \frac{\pi}{2})$ because its derivative is negative there.