1. **State the problem:** (a) Prove the double angle formulas: (i) $\sin 2A = 2 \sin A \cos A$ (ii) $\cos 2A = 2 \cos^2 A - 1$ Given function: $$f(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$$ (b) Show that $f(\theta) = \sin 2\theta$ (c) Solve for $x$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$: $$5 \tan(x + \frac{\pi}{6}) = [1 + \tan^2(x + \frac{\pi}{6})] [1 - 2 \cos^2 (x + \frac{\pi}{6})]$$ (d) Find the exact value of the integral: $$\int_0^{\pi/2} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} - \cos 5\theta + 2 \right) d\theta$$ --- 2. **Part (a)(i): Prove $\sin 2A = 2 \sin A \cos A$** By the sine addition formula, $$\sin(2A) = \sin(A + A) = \sin A \cos A + \cos A \sin A = 2 \sin A \cos A$$ 3. **Part (a)(ii): Prove $\cos 2A = 2 \cos^2 A - 1$** By the cosine addition formula, $$\cos(2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A$$ Using Pythagorean identity $\sin^2 A = 1 - \cos^2 A$, $$\cos 2A = \cos^2 A - (1 - \cos^2 A) = 2 \cos^2 A - 1$$ 4. **Part (b): Show that $f(\theta) = \sin 2\theta$** Recall $\tan \theta = \frac{\sin \theta}{\cos \theta}$, so: $$f(\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{2 \sin \theta / \cos \theta}{\frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}} = \frac{2 \sin \theta / \cos \theta}{\frac{1}{\cos^2 \theta}} = 2 \sin \theta \cos \theta$$ From part (a)(i), we know $\sin 2\theta = 2 \sin \theta \cos \theta$, so: $$f(\theta) = \sin 2\theta$$ 5. **Part (c): Solve** $$5 \tan\left(x + \frac{\pi}{6}\right) = \left[1 + \tan^2\left(x + \frac{\pi}{6}\right)\right] \left[1 - 2 \cos^2 \left(x + \frac{\pi}{6}\right)\right]$$ Rewrite the right bracket using the double angle identity from part (a)(ii): $$1 - 2 \cos^2 A = -\cos 2A$$ Let $A = x + \frac{\pi}{6}$, $$5 \tan A = (1 + \tan^2 A)(-\cos 2A)$$ Recall $1 + \tan^2 A = \sec^2 A = \frac{1}{\cos^2 A}$, so: $$5 \tan A = -\frac{\cos 2A}{\cos^2 A}$$ Multiply both sides by $\cos^2 A$: $$5 \tan A \cos^2 A = - \cos 2A$$ Using $\tan A = \frac{\sin A}{\cos A}$, this becomes: $$5 \frac{\sin A}{\cos A} \cos^2 A = 5 \sin A \cos A = - \cos 2A$$ From part (a)(i), $2 \sin A \cos A = \sin 2A$, so: $$5 \sin A \cos A = \frac{5}{2} \sin 2A = - \cos 2A$$ Rearranged: $$\frac{5}{2} \sin 2A + \cos 2A = 0$$ Divide both sides by $\cos 2A$ (assuming $\cos 2A \neq 0$): $$\frac{5}{2} \tan 2A + 1 = 0 \implies \tan 2A = - \frac{2}{5}$$ Recall $A = x + \frac{\pi}{6}$: $$2A = 2x + \frac{\pi}{3}$$ So the equation is: $$\tan(2x + \frac{\pi}{3}) = - \frac{2}{5}$$ The general solution for $\tan y = m$ is: $$y = \arctan(m) + n\pi, \quad n \in \mathbb{Z}$$ Calculate $2x + \frac{\pi}{3} = \arctan(-\frac{2}{5}) + n\pi$ Since $-\frac{2}{5}$ is approximately $-0.4$, $$\arctan(-0.4) \approx -0.380506 \, \text{radians}$$ So: $$2x + \frac{\pi}{3} = -0.380506 + n\pi$$ Solve for $x$: $$x = \frac{-0.380506 + n\pi - \frac{\pi}{3}}{2} = \frac{n\pi - 1.4661}{2}$$ (approximate $\frac{\pi}{3} \approx 1.0472$) Find all $x$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$: - For $n=0$: $$x = \frac{0 - 1.4661}{2} = -0.73305$$ (approx $-0.733$) valid since $>-\frac{\pi}{2} \approx -1.5708$ - For $n=1$: $$x = \frac{3.1416 - 1.4661}{2} = \frac{1.6755}{2} = 0.83775$$ (approx 0.838) valid since $<1.5708$ - For $n=-1$: $$x = \frac{-3.1416 - 1.4661}{2} = -2.30385$$ not in domain Thus solutions to 3 s.f.: $$x \approx -0.733, \quad 0.838$$ 6. **Part (d): Evaluate the integral** $$I = \int_0^{\pi/2} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} - \cos 5\theta + 2 \right) d\theta$$ Rewrite $\frac{4 \tan \theta}{1 + \tan^2 \theta}$ using the identity from part (b): $$\frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta \implies \frac{4 \tan \theta}{1 + \tan^2 \theta} = 2 \sin 2\theta$$ So the integral simplifies: $$I = \int_0^{\pi/2} \left( 2 \sin 2\theta - \cos 5\theta + 2 \right) d\theta$$ Split integral: $$I = \int_0^{\pi/2} 2 \sin 2\theta \, d\theta - \int_0^{\pi/2} \cos 5\theta \, d\theta + \int_0^{\pi/2} 2 \, d\theta$$ Evaluate each: - $\int 2 \sin 2\theta d\theta = - \cos 2\theta + C$ - $\int \cos 5\theta d\theta = \frac{\sin 5\theta}{5} + C$ - $\int 2 d\theta = 2\theta + C$ Calculate definite integrals: - $$\int_0^{\pi/2} 2 \sin 2\theta d\theta = \left[-\cos 2\theta \right]_0^{\pi/2} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$$ - $$\int_0^{\pi/2} \cos 5\theta d\theta = \left[ \frac{\sin 5\theta}{5} \right]_0^{\pi/2} = \frac{\sin(\frac{5\pi}{2})}{5} - \frac{\sin 0}{5} = \frac{1}{5} - 0 = \frac{1}{5}$$ - $$\int_0^{\pi/2} 2 d\theta = 2 \times \frac{\pi}{2} = \pi$$ Therefore, $$I = 2 - \frac{1}{5} + \pi = \frac{10}{5} - \frac{1}{5} + \pi = \frac{9}{5} + \pi$$ **Final answer:** $$\boxed{ \int_0^{\pi/2} \left( \frac{4 \tan \theta}{1 + \tan^2 \theta} - \cos 5\theta + 2 \right) d\theta = \pi + \frac{9}{5} }$$