1. **Problem:** Prove the operator property: $$F(D^2) \cos(ax + b) = F(-a^2) \cos(ax + b)$$ where $a$ and $b$ are constants. 2. **Formula and rules:** The operator $D$ represents differentiation with respect to $x$, so $D = \frac{d}{dx}$. Applying $D^2$ to $\cos(ax + b)$: $$D^2 \cos(ax + b) = \frac{d^2}{dx^2} \cos(ax + b)$$ Recall that: $$\frac{d}{dx} \cos(ax + b) = -a \sin(ax + b)$$ $$\frac{d^2}{dx^2} \cos(ax + b) = -a^2 \cos(ax + b)$$ 3. **Intermediate work:** Since $D^2 \cos(ax + b) = -a^2 \cos(ax + b)$, applying any function $F$ of the operator $D^2$ to $\cos(ax + b)$ means: $$F(D^2) \cos(ax + b) = F(-a^2) \cos(ax + b)$$ because $\cos(ax + b)$ is an eigenfunction of $D^2$ with eigenvalue $-a^2$. 4. **Explanation:** This means that when you apply the operator $D^2$ to $\cos(ax + b)$, it acts like multiplying by $-a^2$. Therefore, any function $F$ of $D^2$ acts like the same function $F$ evaluated at $-a^2$ times $\cos(ax + b)$. **Final answer:** $$F(D^2) \cos(ax + b) = F(-a^2) \cos(ax + b)$$