1. **Problem:** (a) Show that $$\cos 2A = 1 - 2 \sin^2 A$$ using a formula from page 2. (b) Find the volume of the solid formed by rotating the region bounded by $$y = 3 + 2 \sin x$$, the x-axis, y-axis, and the line $$x = \frac{\pi}{4}$$ about the x-axis. 2. **Part (a): Proof of identity** Using the double-angle formula for cosine: $$\cos 2A = \cos^2 A - \sin^2 A$$ Recall that $$\cos^2 A = 1 - \sin^2 A$$, substitute this: $$\cos 2A = (1 - \sin^2 A) - \sin^2 A = 1 - 2 \sin^2 A$$ Thus, the identity $$\cos 2A = 1 - 2 \sin^2 A$$ is proven. 3. **Part (b): Calculate volume of solid of revolution** - The region R is bounded by the curve $$y = 3 + 2 \sin x$$ - Limits of integration are $$x=0$$ to $$x = \frac{\pi}{4}$$ - The solid is formed by rotating R about the x-axis. The volume $$V$$ is given by the formula for rotation about x-axis: $$V = \pi \int_0^{\pi/4} (y)^2 dx = \pi \int_0^{\pi/4} (3 + 2 \sin x)^2 dx$$ Expand the integrand: $$(3 + 2 \sin x)^2 = 9 + 12 \sin x + 4 \sin^2 x$$ So, $$V = \pi \int_0^{\pi/4} (9 + 12 \sin x + 4 \sin^2 x) dx$$ Split the integral: $$V = \pi \left[ \int_0^{\pi/4} 9 dx + \int_0^{\pi/4} 12 \sin x dx + \int_0^{\pi/4} 4 \sin^2 x dx \right]$$ Calculate each integral: - $$\int_0^{\pi/4} 9 dx = 9x \Big|_0^{\pi/4} = 9 \times \frac{\pi}{4} = \frac{9\pi}{4}$$ - $$\int_0^{\pi/4} 12 \sin x dx = 12(-\cos x) \Big|_0^{\pi/4} = 12(-\cos \frac{\pi}{4} + \cos 0) = 12\left(- \frac{\sqrt{2}}{2} + 1\right) = 12\left(1 - \frac{\sqrt{2}}{2}\right)$$ - For $$\int_0^{\pi/4} 4 \sin^2 x dx$$, use the identity: $$\sin^2 x = \frac{1 - \cos 2x}{2}$$ So, $$4 \int_0^{\pi/4} \sin^2 x dx = 4 \int_0^{\pi/4} \frac{1 - \cos 2x}{2} dx = 2 \int_0^{\pi/4} (1 - \cos 2x) dx$$ Calculate inside: $$2 \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi/4} = 2 \left( \frac{\pi}{4} - \frac{\sin \frac{\pi}{2}}{2} - 0 \right) = 2 \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{2} - 1$$ Combine results: $$V = \pi \left( \frac{9\pi}{4} + 12\left(1 - \frac{\sqrt{2}}{2}\right) + \frac{\pi}{2} - 1 \right)$$ Simplify terms with $$\pi$$: $$\frac{9\pi}{4} + \frac{\pi}{2} = \frac{9\pi}{4} + \frac{2\pi}{4} = \frac{11\pi}{4}$$ So, $$V = \pi \left( \frac{11\pi}{4} + 12 - 6 \sqrt{2} - 1 \right) = \pi \left( \frac{11\pi}{4} + 11 - 6\sqrt{2} \right)$$ Approximate numeric values: - $$\pi \approx 3.1416$$ - $$\frac{11\pi}{4} = 11 \times \frac{3.1416}{4} = 11 \times 0.7854 \approx 8.6394$$ - $$11 - 6\sqrt{2} = 11 - 6 \times 1.4142 = 11 - 8.4853 = 2.5147$$ So the inside parentheses: $$8.6394 + 2.5147 = 11.1541$$ Then volume: $$V \approx \pi \times 11.1541 = 3.1416 \times 11.1541 \approx 35.04$$ Rounded to nearest integer: $$\boxed{35}$$ Final answers: (a) $$\cos 2A = 1 - 2 \sin^2 A$$ is proven. (b) The volume of the solid generated is approximately $$35$$ cubic units.