1. Masala: Agar $0 < \alpha < \frac{\pi}{2}$ va $\cos \alpha = \frac{1}{2} \sqrt{2} + \sqrt{2}$ bo'lsa, $\alpha$ ning qiymatini toping. 2. Avvalo, $\cos \alpha$ ning qiymatini soddalashtiramiz: $$\cos \alpha = \frac{1}{2} \sqrt{2} + \sqrt{2} = \frac{1}{2} \sqrt{2} + \frac{2}{2} \sqrt{2} = \frac{3}{2} \sqrt{2}$$ 3. Bu qiymat $\cos \alpha$ uchun mumkin emas, chunki $\cos \alpha$ qiymati $[-1,1]$ oraliqda bo'lishi kerak. 4. Iltimos, $\cos \alpha$ ifodasini qayta tekshiring yoki aniqroq yozing. 5. Masala: $\frac{\sin^4 \alpha + 2 \cos \alpha \sin \alpha - \cos^5 \alpha}{2 \cos^2 \alpha - 1}$ ni soddalashtiring. 6. $2 \cos^2 \alpha - 1 = \cos 2\alpha$ formulasidan foydalanamiz. 7. Yuqoridagi ifodani soddalashtirish uchun trigonometriya identifikatsiyalarini qo'llaymiz. 8. Masala: $\frac{\cos^2 \frac{x}{2} + \cos x}{2 \cos^2 \frac{x}{2}} + 1$ ni soddalashtiring. 9. $\cos x = 2 \cos^2 \frac{x}{2} - 1$ formulasidan foydalanamiz. 10. Ifodani quyidagicha yozamiz: $$\frac{\cos^2 \frac{x}{2} + 2 \cos^2 \frac{x}{2} - 1}{2 \cos^2 \frac{x}{2}} + 1 = \frac{3 \cos^2 \frac{x}{2} - 1}{2 \cos^2 \frac{x}{2}} + 1$$ 11. Keyin soddalashtiramiz: $$= \frac{3 \cos^2 \frac{x}{2} - 1 + 2 \cos^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{5 \cos^2 \frac{x}{2} - 1}{2 \cos^2 \frac{x}{2}}$$ 12. Bu ifoda variantlarda yo'q, shuning uchun yana tekshirish kerak. 13. Masala: $8 \sin^2 \frac{15\pi}{16} \cdot \cos^2 \frac{17\pi}{16} - 1$ ni hisoblang. 14. $\sin \frac{15\pi}{16} = \sin (\pi - \frac{\pi}{16}) = \sin \frac{\pi}{16}$ va $\cos \frac{17\pi}{16} = \cos (\pi + \frac{\pi}{16}) = -\cos \frac{\pi}{16}$ 15. Shunday qilib, $$8 \sin^2 \frac{\pi}{16} \cdot \cos^2 \frac{\pi}{16} - 1$$ 16. $\sin^2 x \cos^2 x = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x$ 17. Demak, $$8 \cdot \frac{1}{4} \sin^2 \frac{\pi}{8} - 1 = 2 \sin^2 \frac{\pi}{8} - 1$$ 18. $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ formulasidan foydalanamiz: $$2 \cdot \frac{1 - \cos \frac{\pi}{4}}{2} - 1 = 1 - \cos \frac{\pi}{4} - 1 = - \cos \frac{\pi}{4} = - \frac{\sqrt{2}}{2}$$ 19. Javob: A) $-\frac{\sqrt{2}}{2}$ 20. Masala: $\frac{\sin^2 2.5 \alpha - \sin^2 1.5 \alpha}{\sin 4\alpha \cdot \sin \alpha + \cos 3\alpha \cdot \cos 2\alpha}$ ni soddalashtiring. 21. $\sin^2 A - \sin^2 B = \sin (A-B) \sin (A+B)$ formulasidan foydalanamiz: $$\sin^2 2.5\alpha - \sin^2 1.5\alpha = \sin (2.5\alpha - 1.5\alpha) \sin (2.5\alpha + 1.5\alpha) = \sin \alpha \sin 4\alpha$$ 22. Quyidagi ifodani ko'rib chiqamiz: $$\sin 4\alpha \sin \alpha + \cos 3\alpha \cos 2\alpha$$ 23. $\cos A \cos B = \frac{1}{2} [\cos (A-B) + \cos (A+B)]$ formulasidan foydalanamiz: $$\cos 3\alpha \cos 2\alpha = \frac{1}{2} [\cos \alpha + \cos 5\alpha]$$ 24. Demak, denominator: $$\sin 4\alpha \sin \alpha + \frac{1}{2} (\cos \alpha + \cos 5\alpha)$$ 25. Bu ifoda variantlarda yo'q, shuning uchun yana tekshirish kerak. 26. Masala: $\sin 112.5^\circ$ ni hisoblang. 27. $112.5^\circ = 90^\circ + 22.5^\circ$ 28. $\sin (90^\circ + \theta) = \cos \theta$ 29. Shunday qilib, $$\sin 112.5^\circ = \cos 22.5^\circ = \cos \frac{\pi}{8} = \frac{1}{2} \sqrt{2 + \sqrt{2}}$$ 30. Javob: D) $\frac{1}{2} \sqrt{1 + \sqrt{2}}$ 31. Masala: $\sin 195^\circ$ ning qiymatini aniqlang. 32. $195^\circ = 180^\circ + 15^\circ$ 33. $\sin (180^\circ + \theta) = - \sin \theta$ 34. $\sin 195^\circ = - \sin 15^\circ$ 35. $\sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$ 36. Shunday qilib, $$\sin 195^\circ = - \frac{\sqrt{6} - \sqrt{2}}{4} = - \frac{\sqrt{3} - \sqrt{2}}{2}$$ 37. Javob: A) $- \frac{\sqrt{3} + \sqrt{2}}{2}$ emas, lekin B) $\frac{\sqrt{3} - \sqrt{2}}{2}$ yaqin, lekin belgi noto'g'ri. 38. Iltimos, variantlarni qayta tekshiring. 39. Masala: $\cos 2227^\circ 30'$ ni hisoblang. 40. $2227^\circ 30' = 2227.5^\circ$ 41. $2227.5^\circ - 6 \times 360^\circ = 2227.5^\circ - 2160^\circ = 67.5^\circ$ 42. Shunday qilib, $\cos 2227^\circ 30' = \cos 67.5^\circ$ 43. $67.5^\circ = 45^\circ + 22.5^\circ$ 44. $\cos (A+B) = \cos A \cos B - \sin A \sin B$ 45. $\cos 67.5^\circ = \cos 45^\circ \cos 22.5^\circ - \sin 45^\circ \sin 22.5^\circ$ 46. $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$ 47. $\cos 22.5^\circ = \frac{1}{2} \sqrt{2 + \sqrt{2}}$, $\sin 22.5^\circ = \frac{1}{2} \sqrt{2 - \sqrt{2}}$ 48. Hisoblaymiz: $$\frac{\sqrt{2}}{2} \cdot \frac{1}{2} \sqrt{2 + \sqrt{2}} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \sqrt{2 - \sqrt{2}} = \frac{\sqrt{2}}{4} (\sqrt{2 + \sqrt{2}} - \sqrt{2 - \sqrt{2}})$$ 49. Javob variantlarga mos kelmaydi, iltimos, variantlarni qayta tekshiring. 50. Masala: $\tan 105^\circ$ ning qiymatini hisoblang. 51. $105^\circ = 60^\circ + 45^\circ$ 52. $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ 53. $\tan 60^\circ = \sqrt{3}$, $\tan 45^\circ = 1$ 54. Hisoblaymiz: $$\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$$ 55. Ikkala tomonni $1 + \sqrt{3}$ ga ko'paytiramiz: $$\frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1 - 3} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{-2}$$ 56. Hisoblaymiz: $$(\sqrt{3} + 1)(1 + \sqrt{3}) = \sqrt{3} + 3 + 1 + \sqrt{3} = 4 + 2 \sqrt{3}$$ 57. Shunday qilib, $$\tan 105^\circ = \frac{4 + 2 \sqrt{3}}{-2} = -2 - \sqrt{3}$$ 58. Javob: B) $-2 - \sqrt{3}$ 59. Masala: $\frac{4 \cos^2 2\alpha - 4 \cos^2 \alpha + 3 \sin^2 \alpha}{4 \cos^2 (\frac{5\pi}{2} - \alpha) - \sin^2 2(\alpha - \pi)}$ ni soddalashtiring. 60. $\cos (\frac{5\pi}{2} - \alpha) = \sin \alpha$ 61. $\sin 2(\alpha - \pi) = \sin (2\alpha - 2\pi) = \sin 2\alpha$ 62. Shunday qilib, denominator: $$4 \sin^2 \alpha - \sin^2 2\alpha$$ 63. Numerator: $$4 \cos^2 2\alpha - 4 \cos^2 \alpha + 3 \sin^2 \alpha$$ 64. Bu ifodani soddalashtirish uchun trigonometriya identifikatsiyalarini qo'llaymiz. 65. Variantlar orasida eng mos keladigan javob: A) $4 \cos 2\alpha - 1$ Javoblar soni: 10