1. We start by proving the double-angle formulae: (a)(i) Prove that $\cos 2A = 2 \cos^2 A - 1$. Recall the double-angle identity: $\cos 2A = \cos^2 A - \sin^2 A$. Using the Pythagorean identity $\sin^2 A = 1 - \cos^2 A$, substitute: $$\cos 2A = \cos^2 A - (1 - \cos^2 A) = \cos^2 A - 1 + \cos^2 A = 2 \cos^2 A - 1$$ Thus, (i) is shown. (a)(ii) Prove that $\sin 2A = 2 \sin A \cos A$. Using the double-angle identity for sine: $$\sin 2A = 2 \sin A \cos A$$ which is a standard formula. 2. (b) Show that $$\cos^3 A = \frac{\cos 3A + 3 \cos A}{4}$$ Start with the triple-angle formula: $$\cos 3A = 4 \cos^3 A - 3 \cos A$$ Rearranged: $$4 \cos^3 A = \cos 3A + 3 \cos A$$ Divide both sides by 4: $$\cos^3 A = \frac{\cos 3A + 3 \cos A}{4}$$ Thus, the identity is proved. 3. (c) Solve for exact values of $\theta$ in $$8 \cos^3 \frac{\theta}{2} - 6 \cos \frac{\theta}{2} - 1 = 0$$ Let $x = \cos \frac{\theta}{2}$. Rewrite the equation: $$8 x^3 - 6 x - 1 = 0$$ Divide by 2: $$4 x^3 - 3 x - \frac{1}{2} = 0$$ Recall from (b): $$4 x^3 - 3 x = \cos 3A$$ Therefore, $$\cos 3A = \frac{1}{2}$$ So, $$3 \times \frac{\theta}{2} = 3A = \pm \frac{\pi}{3} + 2 k \pi$$ $$\Rightarrow \frac{3 \theta}{2} = \pm \frac{\pi}{3} + 2 k \pi$$ Multiply both sides by $\frac{2}{3}$: $$\theta = \pm \frac{2 \pi}{9} + \frac{4 k \pi}{3}$$ Now consider $0 \leq \theta \leq 2\pi$, find values of $\theta$: - For $+$ sign: $k=0 \Rightarrow \theta = \frac{2 \pi}{9}$ $k=1 \Rightarrow \theta = \frac{2 \pi}{9} + \frac{4 \pi}{3} = \frac{2 \pi}{9} + \frac{12 \pi}{9} = \frac{14 \pi}{9}$ $k=2 \Rightarrow \theta = \frac{2 \pi}{9} + \frac{8 \pi}{3} > 2 \pi$ (ignore) - For $-$ sign: $k=0 \Rightarrow \theta = - \frac{2 \pi}{9} < 0$ (ignore) $k=1 \Rightarrow \theta = - \frac{2 \pi}{9} + \frac{4 \pi}{3} = \frac{12 \pi}{9} - \frac{2 \pi}{9} = \frac{10 \pi}{9}$ $k=2 \Rightarrow \theta = - \frac{2 \pi}{9} + \frac{8 \pi}{3} > 2 \pi$ (ignore) So the exact solutions in the interval are: $$\theta = \frac{2 \pi}{9}, \frac{10 \pi}{9}, \frac{14 \pi}{9}$$ 4. (d) Find the exact value of $$\int_0^\pi (4 \cos^3 \theta - \sin 2 \theta) \, d\theta$$ Use the identity from (b): $$4 \cos^3 \theta = \cos 3\theta + 3 \cos \theta$$ Rewrite the integral: $$\int_0^\pi (\cos 3\theta + 3 \cos \theta - \sin 2 \theta) \, d\theta = \int_0^\pi \cos 3\theta \, d\theta + 3 \int_0^\pi \cos \theta \, d\theta - \int_0^\pi \sin 2 \theta \, d\theta$$ Calculate each integral: - $\int_0^\pi \cos 3\theta \, d\theta = \left[ \frac{\sin 3\theta}{3} \right]_0^\pi = \frac{\sin 3\pi}{3} - \frac{\sin 0}{3} = 0$ - $3 \int_0^\pi \cos \theta \, d\theta = 3 \left[ \sin \theta \right]_0^\pi = 3 (\sin \pi - \sin 0) = 0$ - $- \int_0^\pi \sin 2 \theta \, d\theta = - \left[ - \frac{\cos 2\theta}{2} \right]_0^\pi = - \left( - \frac{\cos 2\pi}{2} + \frac{\cos 0}{2} \right) = - \left( - \frac{1}{2} + \frac{1}{2} \right) = 0$ Sum the results: $$0 + 0 + 0 = 0$$ Thus, the exact value of the integral is $0$.