1. **Stating the problem:** (a) Show that $\cos 2x = 1 - 2 \sin^2 x$ using a known formula. (b) Find the coordinates of points A, B, and C where the curves $y=\sin x + 2$ and $y=\cos 2x + 2$ intersect. (c) Use calculus to find the ratio of the areas of regions R1 to R2 formed between these curves. 2. **Part (a): Prove the identity** Recall the double-angle formula for cosine: $$\cos 2x = \cos^2 x - \sin^2 x$$ Using the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$, substitute: $$\cos 2x = (1 - \sin^2 x) - \sin^2 x = 1 - 2 \sin^2 x$$ This completes the proof for part (a). 3. **Part (b): Find points of intersection (A, B, C)** Points common to both curves satisfy: $$\sin x + 2 = \cos 2x + 2$$ Simplify by subtracting 2: $$\sin x = \cos 2x$$ Using the identity from part (a): $$\sin x = 1 - 2 \sin^2 x$$ Rearranged: $$2 \sin^2 x + \sin x - 1 = 0$$ Let $s = \sin x$, then: $$2 s^2 + s - 1 = 0$$ Use the quadratic formula: $$s = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ Solutions: - $s = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5$ - $s = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$ For $\sin x = 0.5$, solutions in $[0, 2\pi]$ are: $$x = \frac{\pi}{6}, \frac{5\pi}{6}$$ For $\sin x = -1$, solution is: $$x = \frac{3\pi}{2}$$ Find $y$ for each $x$ point using $y=\sin x + 2$: - At $x=\frac{\pi}{6}$, $y = 0.5 + 2 = 2.5$ - At $x=\frac{5\pi}{6}$, $y = 0.5 + 2 = 2.5$ - At $x=\frac{3\pi}{2}$, $y = -1 + 2 = 1$ Thus, points are: $$A\left(\frac{\pi}{6}, 2.5\right), B\left(\frac{5\pi}{6}, 2.5\right), C\left(\frac{3\pi}{2}, 1\right)$$ 4. **Part (c): Find ratio area R1 : area R2** - R1 is between $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ - R2 is between $x=\frac{5\pi}{6}$ and $x=\frac{3\pi}{2}$ The curves intersect at these points. For R1, the top curve is $y=\sin x + 2$ and bottom is $y=\cos 2x + 2$ because $\sin x > \cos 2x$ there. For R2, the top curve is $y=\cos 2x + 2$ and bottom is $y=\sin x + 2$ because $\cos 2x > \sin x$ in that interval. Calculate areas: $$\text{Area R1} = \int_{\pi/6}^{5\pi/6} \left[(\sin x + 2) - (\cos 2x + 2)\right] dx = \int_{\pi/6}^{5\pi/6} (\sin x - \cos 2x) dx$$ $$\text{Area R2} = \int_{5\pi/6}^{3\pi/2} \left[(\cos 2x + 2) - (\sin x + 2)\right] dx = \int_{5\pi/6}^{3\pi/2} (\cos 2x - \sin x) dx$$ Evaluate Area R1: $$\int \sin x dx = -\cos x$$ $$\int \cos 2x dx = \frac{\sin 2x}{2}$$ So, $$\text{Area R1} = \left[-\cos x - \frac{\sin 2x}{2}\right]_{x=\pi/6}^{5\pi/6}$$ Calculate at bounds: - At $5\pi/6$: $$-\cos\left(\frac{5\pi}{6}\right) - \frac{\sin\left(2 \times \frac{5\pi}{6}\right)}{2} = -\left(-\frac{\sqrt{3}}{2}\right) - \frac{\sin\frac{5\pi}{3}}{2} = \frac{\sqrt{3}}{2} - \frac{-\frac{\sqrt{3}}{2}}{2} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$$ - At $\pi/6$: $$-\cos\left(\frac{\pi}{6}\right) - \frac{\sin\left(2 \times \frac{\pi}{6}\right)}{2} = -\frac{\sqrt{3}}{2} - \frac{\sin\frac{\pi}{3}}{2} = -\frac{\sqrt{3}}{2} - \frac{\frac{\sqrt{3}}{2}}{2} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{4} = -\frac{3\sqrt{3}}{4}$$ Therefore, $$\text{Area R1} = \frac{3\sqrt{3}}{4} - \left(-\frac{3\sqrt{3}}{4}\right) = \frac{3\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} = \frac{6\sqrt{3}}{4} = \frac{3\sqrt{3}}{2}$$ Evaluate Area R2: Similarly, $$\text{Area R2} = \int_{5\pi/6}^{3\pi/2} (\cos 2x - \sin x) dx = \left[\frac{\sin 2x}{2} + \cos x\right]_{5\pi/6}^{3\pi/2}$$ Calculate at bounds: - At $3\pi/2$: $$\frac{\sin(3\pi)}{2} + \cos\left(\frac{3\pi}{2}\right) = 0 + 0 = 0$$ - At $5\pi/6$: $$\frac{\sin\left(2 \times \frac{5\pi}{6}\right)}{2} + \cos\left(\frac{5\pi}{6}\right) = \frac{\sin\frac{5\pi}{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right) = \frac{-\frac{\sqrt{3}}{2}}{2} - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{4}$$ Therefore, $$\text{Area R2} = 0 - \left( -\frac{3\sqrt{3}}{4} \right) = \frac{3\sqrt{3}}{4}$$ 5. **Ratio of areas R1 : R2** $$\frac{\text{Area R1}}{\text{Area R2}} = \frac{\frac{3\sqrt{3}}{2}}{\frac{3\sqrt{3}}{4}} = \frac{3\sqrt{3}}{2} \times \frac{4}{3\sqrt{3}} = 2$$ \textbf{Final answer:} \boxed{\cos 2x = 1 - 2 \sin^2 x;\quad A\left(\frac{\pi}{6}, 2.5\right), B\left(\frac{5\pi}{6}, 2.5\right), C\left(\frac{3\pi}{2}, 1\right);\quad \text{Ratio } R1:R2=2:1}