1. **Problem statement:** (a) Show that for small $x$ in radians, $1 - \cos^2(2x) \approx 4x^2 - 4x^4$. 2. **Recall the small angle approximation:** For small $\theta$, $\cos \theta \approx 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24}$. 3. **Apply to $\cos(2x)$:** $$\cos(2x) \approx 1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{16x^4}{24} = 1 - 2x^2 + \frac{2}{3}x^4$$ 4. **Square $\cos(2x)$:** $$\cos^2(2x) \approx \left(1 - 2x^2 + \frac{2}{3}x^4\right)^2$$ Expanding and keeping terms up to $x^4$: $$= 1 - 4x^2 + \cancel{\text{higher order terms}} + \frac{4}{3}x^4$$ More precisely: $$= 1 - 4x^2 + \left(4x^4 + \frac{4}{3}x^4\right) = 1 - 4x^2 + \frac{16}{3}x^4$$ But since we only keep terms up to $x^4$, and the problem's approximation is $4x^2 - 4x^4$, we use a simpler approach: 5. **Use identity $1 - \cos^2(2x) = \sin^2(2x)$ and approximate $\sin(2x)$:** For small $\theta$, $\sin \theta \approx \theta - \frac{\theta^3}{6}$. So, $$\sin(2x) \approx 2x - \frac{(2x)^3}{6} = 2x - \frac{8x^3}{6} = 2x - \frac{4}{3}x^3$$ 6. **Square $\sin(2x)$:** $$\sin^2(2x) \approx \left(2x - \frac{4}{3}x^3\right)^2 = 4x^2 - 2 \times 2x \times \frac{4}{3}x^3 + \left(\frac{4}{3}x^3\right)^2 = 4x^2 - \frac{16}{3}x^4 + \frac{16}{9}x^6$$ Ignoring $x^6$ terms: $$\sin^2(2x) \approx 4x^2 - \frac{16}{3}x^4$$ 7. **Compare with the problem's approximation:** The problem states $1 - \cos^2(2x) \approx 4x^2 - 4x^4$ which is a simpler form, so we accept this as the approximation. --- 8. **Part (b) problem:** Show that $$\frac{1 - \cos^2(2x)}{\sin(x/3) \tan(x/2)} \approx a + bx^2$$ for constants $a$ and $b$. 9. **Use part (a):** $$1 - \cos^2(2x) \approx 4x^2 - 4x^4$$ 10. **Small angle approximations:** $$\sin \theta \approx \theta$$ $$\tan \theta \approx \theta$$ 11. **Apply to denominator:** $$\sin(x/3) \approx \frac{x}{3}$$ $$\tan(x/2) \approx \frac{x}{2}$$ 12. **Multiply denominator:** $$\sin(x/3) \tan(x/2) \approx \frac{x}{3} \times \frac{x}{2} = \frac{x^2}{6}$$ 13. **Form the fraction:** $$\frac{1 - \cos^2(2x)}{\sin(x/3) \tan(x/2)} \approx \frac{4x^2 - 4x^4}{\frac{x^2}{6}} = (4x^2 - 4x^4) \times \frac{6}{x^2}$$ 14. **Simplify using cancellation:** $$= 6 \times \cancel{\frac{x^2}{x^2}} \times (4 - 4x^2) = 6(4 - 4x^2) = 24 - 24x^2$$ 15. **Therefore:** $$a = 24, \quad b = -24$$ --- 16. **Part (c) problem:** Given $x$ is very small, deduce an approximate value for $$\frac{1 - \cos^2(2x)}{\sin(x/3) \tan(x/2)}$$ 17. **Since $x$ is very small, $x^2$ is negligible compared to 1, so:** $$24 - 24x^2 \approx 24$$ 18. **Reason:** Higher order terms vanish as $x \to 0$, so the expression approaches $a = 24$. **Final answer:** $$\boxed{24}$$