Trig Evaluation 5456B3

1. **Problem 1.1:** Evaluate $\frac{\cos 15^\circ \times \tan 30^\circ \times \sin 315^\circ}{\sin 30^\circ \times \cos 210^\circ \times \sin 105^\circ}$ without a calculator. 2. **Step 1:** Recall exact values: - $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$ - $\tan 30^\circ = \frac{1}{\sqrt{3}}$ - $\sin 315^\circ = \sin(360^\circ - 45^\circ) = -\sin 45^\circ = -\frac{\sqrt{2}}{2}$ - $\sin 30^\circ = \frac{1}{2}$ - $\cos 210^\circ = \cos(180^\circ + 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$ - $\sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} + \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$ 3. **Step 2:** Substitute values: $$\frac{\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \times \frac{1}{\sqrt{3}} \times \left(-\frac{\sqrt{2}}{2}\right)}{\frac{1}{2} \times \left(-\frac{\sqrt{3}}{2}\right) \times \frac{\sqrt{6} + \sqrt{2}}{4}}$$ 4. **Step 3:** Simplify numerator: $$\frac{\sqrt{6} + \sqrt{2}}{4} \times \frac{1}{\sqrt{3}} \times \left(-\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}(\sqrt{6} + \sqrt{2})}{8 \sqrt{3}}$$ 5. **Step 4:** Simplify denominator: $$\frac{1}{2} \times \left(-\frac{\sqrt{3}}{2}\right) \times \frac{\sqrt{6} + \sqrt{2}}{4} = -\frac{\sqrt{3}(\sqrt{6} + \sqrt{2})}{16}$$ 6. **Step 5:** Write full fraction: $$\frac{-\frac{\sqrt{2}(\sqrt{6} + \sqrt{2})}{8 \sqrt{3}}}{-\frac{\sqrt{3}(\sqrt{6} + \sqrt{2})}{16}} = \frac{-\frac{\sqrt{2}(\sqrt{6} + \sqrt{2})}{8 \sqrt{3}}}{-\frac{\sqrt{3}(\sqrt{6} + \sqrt{2})}{16}}$$ 7. **Step 6:** Cancel $-(\sqrt{6} + \sqrt{2})$ and negatives: $$= \frac{\frac{\sqrt{2}}{8 \sqrt{3}}}{\frac{\sqrt{3}}{16}} = \frac{\sqrt{2}}{8 \sqrt{3}} \times \frac{16}{\sqrt{3}}$$ 8. **Step 7:** Simplify: $$= \frac{\sqrt{2} \times 16}{8 \sqrt{3} \times \sqrt{3}} = \frac{16 \sqrt{2}}{8 \times 3} = \frac{16 \sqrt{2}}{24} = \frac{2 \sqrt{2}}{3}$$ --- 9. **Answer 1.1:** $\boxed{\frac{2 \sqrt{2}}{3}}$ --- 10. **Problem 1.2:** Simplify $$\frac{\tan(360^\circ - x) \cdot \sin(90^\circ + x) \cdot \sin(360^\circ - x) \cdot \tan(180^\circ - x)}{\cos(360^\circ - x) \cdot \sin(180^\circ + x)}$$ 11. **Step 1:** Use identities: - $\tan(360^\circ - x) = -\tan x$ - $\sin(90^\circ + x) = \cos x$ - $\sin(360^\circ - x) = -\sin x$ - $\tan(180^\circ - x) = -\tan x$ - $\cos(360^\circ - x) = \cos x$ - $\sin(180^\circ + x) = -\sin x$ 12. **Step 2:** Substitute: $$\frac{(-\tan x) \cdot \cos x \cdot (-\sin x) \cdot (-\tan x)}{\cos x \cdot (-\sin x)}$$ 13. **Step 3:** Simplify numerator: $$(-\tan x) \times \cos x \times (-\sin x) \times (-\tan x) = (-\tan x)(-\tan x) \times \cos x \times (-\sin x) = (\tan^2 x) \times \cos x \times (-\sin x)$$ 14. **Step 4:** So numerator is: $$- \tan^2 x \cos x \sin x$$ 15. **Step 5:** Denominator is: $$\cos x \times (-\sin x) = - \cos x \sin x$$ 16. **Step 6:** Full fraction: $$\frac{- \tan^2 x \cos x \sin x}{- \cos x \sin x}$$ 17. **Step 7:** Cancel $- \cos x \sin x$: $$= \tan^2 x$$ --- 18. **Answer 1.2:** $\boxed{\tan^2 x}$ --- 19. **Problem 1.3a:** Prove $$\frac{1 - (\cos \theta - \sin \theta)^2}{\cos \theta} = 2 \sin \theta$$ 20. **Step 1:** Expand numerator: $$(\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2 \sin \theta \cos \theta + \sin^2 \theta$$ 21. **Step 2:** Substitute: $$1 - (\cos^2 \theta - 2 \sin \theta \cos \theta + \sin^2 \theta) = 1 - \cos^2 \theta + 2 \sin \theta \cos \theta - \sin^2 \theta$$ 22. **Step 3:** Use identity $\sin^2 \theta + \cos^2 \theta = 1$: $$1 - \cos^2 \theta - \sin^2 \theta = 1 - 1 = 0$$ 23. **Step 4:** So numerator simplifies to: $$0 + 2 \sin \theta \cos \theta = 2 \sin \theta \cos \theta$$ 24. **Step 5:** Divide numerator by $\cos \theta$: $$\frac{2 \sin \theta \cos \theta}{\cos \theta} = 2 \sin \theta$$ 25. **Step 6:** Hence proved: $$\frac{1 - (\cos \theta - \sin \theta)^2}{\cos \theta} = 2 \sin \theta$$ --- 26. **Problem 1.3b:** Solve $$\frac{1 - (\cos \theta - \sin \theta)^2}{\cos \theta} = -1.532$$ 27. **Step 1:** From 1.3a, left side equals $2 \sin \theta$, so $$2 \sin \theta = -1.532$$ 28. **Step 2:** Solve for $\sin \theta$: $$\sin \theta = \frac{-1.532}{2} = -0.766$$ 29. **Step 3:** Find $\theta$ where $\sin \theta = -0.766$ and $0^\circ \leq \theta \leq 360^\circ$. 30. **Step 4:** Using inverse sine (approximate): $$\theta_1 = 180^\circ + \arcsin(0.766) \approx 180^\circ + 50^\circ = 230^\circ$$ $$\theta_2 = 360^\circ - 50^\circ = 310^\circ$$ 31. **Answer 1.3b:** $\boxed{\theta \approx 230^\circ, 310^\circ}$ --- 32. **Problem 1.4:** Given quadrilateral with $CD = 3.5$ km, $BC = 2.4$ km, $BA = 2$ km, $\angle CBA = 120^\circ$, and $\angle C = 80^\circ$, find $\angle DBC$. 33. **Step 1:** Consider triangle $BCD$ with sides $BC = 2.4$ km, $CD = 3.5$ km, and angle $C = 80^\circ$. 34. **Step 2:** Use Law of Cosines to find diagonal $BD$: $$BD^2 = BC^2 + CD^2 - 2 \times BC \times CD \times \cos 80^\circ$$ $$= 2.4^2 + 3.5^2 - 2 \times 2.4 \times 3.5 \times \cos 80^\circ$$ 35. **Step 3:** Calculate values: $$2.4^2 = 5.76$$ $$3.5^2 = 12.25$$ $$\cos 80^\circ \approx 0.1736$$ 36. **Step 4:** Substitute: $$BD^2 = 5.76 + 12.25 - 2 \times 2.4 \times 3.5 \times 0.1736 = 18.01 - 2.918 = 15.092$$ 37. **Step 5:** Find $BD$: $$BD = \sqrt{15.092} \approx 3.885$$ 38. **Step 6:** In triangle $BDA$, $BA = 2$ km, $BD \approx 3.885$ km, and $\angle CBA = 120^\circ$ is at $B$ between $BA$ and $BC$. 39. **Step 7:** To find $\angle DBC$, note that $\angle CBA = 120^\circ$ is split into $\angle DBA + \angle DBC$. 40. **Step 8:** Use Law of Cosines in triangle $BDA$ or use vector/angle subtraction. Since $\angle CBA = 120^\circ$ and $\angle DBA$ is angle between $BD$ and $BA$, find $\angle DBA$ first. 41. **Step 9:** Use Law of Cosines in triangle $BDA$ to find $\angle DBA$: $$BD^2 = BA^2 + AD^2 - 2 \times BA \times AD \times \cos \angle DBA$$ 42. **Step 10:** We lack $AD$ length, so use given info: $DA$ is vertical side, but no length given, so assume $AD$ is vertical and use given angles. 43. **Step 11:** Since insufficient data for $AD$, use angle sum at $B$: $$\angle DBC = 120^\circ - \angle DBA$$ 44. **Step 12:** Use Law of Cosines in triangle $BCD$ to find $\angle DBC$: $$\cos \angle DBC = \frac{BC^2 + BD^2 - CD^2}{2 \times BC \times BD}$$ 45. **Step 13:** Substitute values: $$\cos \angle DBC = \frac{2.4^2 + 3.885^2 - 3.5^2}{2 \times 2.4 \times 3.885} = \frac{5.76 + 15.092 - 12.25}{18.648} = \frac{8.602}{18.648} = 0.4615$$ 46. **Step 14:** Calculate angle: $$\angle DBC = \arccos(0.4615) \approx 62.5^\circ$$ --- 47. **Answer 1.4:** $\boxed{62.5^\circ}$