1. **Stating the problems:** - Problem 214: Simplify $$a \cos^2(-30^\circ) \left(4b^2 \cos(-45^\circ) + 1\right)$$ and verify it equals $$\frac{3}{4} a (1 + 2\sqrt{2} b^2)$$. - Problem 215: Simplify $$\cos \alpha - \frac{\cos^2(\pi + \alpha) - \sin^2(2\pi - \alpha)}{\sin(2\pi - \alpha) - \cos(2\pi - \alpha)}$$ and verify it equals $$2 \cos \alpha - \sin \alpha$$. - Problem 217: Simplify $$\sin(\pi + \alpha) \csc(2\pi - \alpha) \left(\tan(-\alpha) + \cot \alpha\right)$$ and verify it equals $$\frac{1 - 2 \sin^2 \alpha}{\sin \alpha \cos \alpha}$$. - Problem 219: Simplify $$\cos(2\pi + \alpha) \tan(\pi - \alpha) + \sin(\pi - \alpha) \cot(2\pi + \alpha)$$ and verify it equals $$\cos \alpha - \sin \alpha$$. - Problem 220: Simplify $$b^3 \cos(-\pi/4)(1 + \tan^2(\pi/4)) + \sqrt[3]{b} \cot^2(-\pi/3)$$ and verify it equals $$\sqrt{2} b^3 + \frac{1}{3} \sqrt[3]{b}$$. 2. **Important trigonometric identities and rules:** - $$\cos(-\theta) = \cos \theta$$, $$\sin(-\theta) = -\sin \theta$$. - $$\cos(\pi + \alpha) = -\cos \alpha$$, $$\sin(\pi + \alpha) = -\sin \alpha$$. - $$\cos(2\pi + \alpha) = \cos \alpha$$, $$\sin(2\pi + \alpha) = \sin \alpha$$. - $$\tan^2 \theta + 1 = \sec^2 \theta$$. - $$\cot \theta = \frac{1}{\tan \theta}$$. - $$\csc \theta = \frac{1}{\sin \theta}$$. --- ### Problem 214 3. Calculate $$\cos^2(-30^\circ) = \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$. 4. Calculate $$\cos(-45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2}$$. 5. Substitute into expression: $$a \cdot \frac{3}{4} \left(4b^2 \cdot \frac{\sqrt{2}}{2} + 1\right) = a \cdot \frac{3}{4} \left(2 \sqrt{2} b^2 + 1\right) = \frac{3}{4} a (1 + 2 \sqrt{2} b^2)$$ This matches the given answer. --- ### Problem 215 3. Use identities: $$\cos^2(\pi + \alpha) = (-\cos \alpha)^2 = \cos^2 \alpha$$ $$\sin^2(2\pi - \alpha) = \sin^2 \alpha$$ $$\sin(2\pi - \alpha) = -\sin \alpha$$ $$\cos(2\pi - \alpha) = \cos \alpha$$ 4. Substitute numerator: $$\cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha$$ 5. Substitute denominator: $$-\sin \alpha - \cos \alpha = -(\sin \alpha + \cos \alpha)$$ 6. Expression becomes: $$\cos \alpha - \frac{\cos 2\alpha}{-(\sin \alpha + \cos \alpha)} = \cos \alpha + \frac{\cos 2\alpha}{\sin \alpha + \cos \alpha}$$ 7. Use identity $$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$$ and factor numerator and denominator: Rewrite numerator as $$\cos^2 \alpha - \sin^2 \alpha$$. 8. Multiply numerator and denominator by $$\sin \alpha - \cos \alpha$$ to simplify: $$\frac{\cos 2\alpha}{\sin \alpha + \cos \alpha} \cdot \frac{\sin \alpha - \cos \alpha}{\sin \alpha - \cos \alpha} = \frac{(\cos^2 \alpha - \sin^2 \alpha)(\sin \alpha - \cos \alpha)}{\sin^2 \alpha - \cos^2 \alpha}$$ 9. Note denominator $$\sin^2 \alpha - \cos^2 \alpha = -(\cos^2 \alpha - \sin^2 \alpha) = -\cos 2\alpha$$. 10. Cancel $$\cos^2 \alpha - \sin^2 \alpha$$ in numerator and denominator: $$\frac{\cancel{(\cos^2 \alpha - \sin^2 \alpha)} (\sin \alpha - \cos \alpha)}{-\cancel{(\cos^2 \alpha - \sin^2 \alpha)}} = - (\sin \alpha - \cos \alpha) = \cos \alpha - \sin \alpha$$ 11. So expression is: $$\cos \alpha + (\cos \alpha - \sin \alpha) = 2 \cos \alpha - \sin \alpha$$ Matches the given answer. --- ### Problem 217 3. Use identities: $$\sin(\pi + \alpha) = -\sin \alpha$$ $$\csc(2\pi - \alpha) = \frac{1}{\sin(2\pi - \alpha)} = \frac{1}{-\sin \alpha} = -\csc \alpha$$ $$\tan(-\alpha) = -\tan \alpha$$ $$\cot \alpha = \frac{1}{\tan \alpha}$$ 4. Sum inside parentheses: $$-\tan \alpha + \cot \alpha = \frac{-\tan^2 \alpha + 1}{\tan \alpha}$$ 5. Use identity $$1 + \tan^2 \alpha = \sec^2 \alpha$$, so: $$1 - \tan^2 \alpha = \cos 2\alpha$$ (alternative identity) But better to keep as is for now. 6. Expression becomes: $$(-\sin \alpha)(-\csc \alpha) \cdot \frac{1 - \tan^2 \alpha}{\tan \alpha} = (\sin \alpha)(\csc \alpha) \cdot \frac{1 - \tan^2 \alpha}{\tan \alpha} = 1 \cdot \frac{1 - \tan^2 \alpha}{\tan \alpha}$$ 7. Rewrite $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$, so: $$\frac{1 - \tan^2 \alpha}{\tan \alpha} = \frac{1 - \frac{\sin^2 \alpha}{\cos^2 \alpha}}{\frac{\sin \alpha}{\cos \alpha}} = \frac{\frac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha}}{\frac{\sin \alpha}{\cos \alpha}} = \frac{\cos^2 \alpha - \sin^2 \alpha}{\cos^2 \alpha} \cdot \frac{\cos \alpha}{\sin \alpha} = \frac{\cos^2 \alpha - \sin^2 \alpha}{\cos \alpha \sin \alpha}$$ 8. Use identity $$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$$: $$= \frac{\cos 2\alpha}{\sin \alpha \cos \alpha}$$ 9. Use identity $$\cos 2\alpha = 1 - 2 \sin^2 \alpha$$: $$= \frac{1 - 2 \sin^2 \alpha}{\sin \alpha \cos \alpha}$$ Matches the given answer. --- ### Problem 219 3. Use identities: $$\cos(2\pi + \alpha) = \cos \alpha$$ $$\tan(\pi - \alpha) = -\tan \alpha$$ $$\sin(\pi - \alpha) = \sin \alpha$$ $$\cot(2\pi + \alpha) = \cot \alpha$$ 4. Substitute: $$\cos \alpha (-\tan \alpha) + \sin \alpha \cot \alpha = -\cos \alpha \tan \alpha + \sin \alpha \cot \alpha$$ 5. Rewrite $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$ and $$\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$$: $$-\cos \alpha \cdot \frac{\sin \alpha}{\cos \alpha} + \sin \alpha \cdot \frac{\cos \alpha}{\sin \alpha} = -\sin \alpha + \cos \alpha$$ 6. Rearrange: $$\cos \alpha - \sin \alpha$$ Matches the given answer. --- ### Problem 220 3. Use identities: $$\cos(-\pi/4) = \cos \pi/4 = \frac{\sqrt{2}}{2}$$ $$\tan^2(\pi/4) = 1$$ $$\cot(-\pi/3) = -\cot \pi/3 = -\frac{1}{\sqrt{3}}$$ So: $$\cot^2(-\pi/3) = \left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$$ 4. Substitute: $$b^3 \cdot \frac{\sqrt{2}}{2} (1 + 1) + \sqrt[3]{b} \cdot \frac{1}{3} = b^3 \cdot \frac{\sqrt{2}}{2} \cdot 2 + \frac{1}{3} \sqrt[3]{b} = \sqrt{2} b^3 + \frac{1}{3} \sqrt[3]{b}$$ Matches the given answer. **Final answers verified for all problems.**