1. Muammo: Agar $\tan \alpha + \cot \alpha = 4$ bo'lsa, $\sin 2\alpha$ ni toping. 2. Formulalar va qoidalar: - $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ - $\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$ - $\sin 2\alpha = 2 \sin \alpha \cos \alpha$ 3. Hisoblash: $$\tan \alpha + \cot \alpha = \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} = \frac{1}{\sin \alpha \cos \alpha} = 4$$ 4. Demak, $$\sin \alpha \cos \alpha = \frac{1}{4}$$ 5. $\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \times \frac{1}{4} = \frac{1}{2}$ Javob: C) $\frac{1}{2}$ 6. Muammo: $\sin^4 \alpha + \cos^4 \alpha$ ning eng kichik qiymatini toping. 7. Formulalar: $$\sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha)^2 + (\cos^2 \alpha)^2$$ 8. $a = \sin^2 \alpha$, $b = \cos^2 \alpha$, $a + b = 1$ 9. $a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab$ 10. $ab = \sin^2 \alpha \cos^2 \alpha = (\sin \alpha \cos \alpha)^2 = \left(\frac{\sin 2\alpha}{2}\right)^2 = \frac{\sin^2 2\alpha}{4}$ 11. Shunday qilib, $$\sin^4 \alpha + \cos^4 \alpha = 1 - 2 \times \frac{\sin^2 2\alpha}{4} = 1 - \frac{\sin^2 2\alpha}{2}$$ 12. Eng kichik qiymat $\sin^2 2\alpha$ maksimal bo'lganda olinadi, ya'ni $\sin^2 2\alpha = 1$ 13. Demak, $$\min (\sin^4 \alpha + \cos^4 \alpha) = 1 - \frac{1}{2} = \frac{1}{2}$$ Javob: D) $\frac{1}{2}$ 14. Muammo: $\frac{2}{\tan \alpha + \cot \alpha}$ ni soddalashtiring. 15. $\tan \alpha + \cot \alpha = \frac{1}{\sin \alpha \cos \alpha}$ (oldingi masaladan) 16. Shunday qilib, $$\frac{2}{\tan \alpha + \cot \alpha} = 2 \sin \alpha \cos \alpha = \sin 2\alpha$$ Javob: E) $\sin 2\alpha$ 17. Muammo: $\frac{2}{\cot \alpha - \tan \alpha}$ ni soddalashtiring. 18. $\cot \alpha - \tan \alpha = \frac{\cos \alpha}{\sin \alpha} - \frac{\sin \alpha}{\cos \alpha} = \frac{\cos^2 \alpha - \sin^2 \alpha}{\sin \alpha \cos \alpha} = \frac{\cos 2\alpha}{\sin \alpha \cos \alpha}$ 19. Shunday qilib, $$\frac{2}{\cot \alpha - \tan \alpha} = \frac{2}{\frac{\cos 2\alpha}{\sin \alpha \cos \alpha}} = 2 \times \frac{\sin \alpha \cos \alpha}{\cos 2\alpha}$$ 20. $2 \sin \alpha \cos \alpha = \sin 2\alpha$ 21. Demak, $$\frac{2}{\cot \alpha - \tan \alpha} = \frac{\sin 2\alpha}{\cos 2\alpha} = \tan 2\alpha$$ Javob: C) $\tan 2\alpha$ 22. Muammo: $\frac{\cos 2\alpha + \cos(\frac{\pi}{2} - \alpha) \sin \alpha}{\sin(\frac{\pi}{2} + \alpha)}$ ni soddalashtiring. 23. $\cos(\frac{\pi}{2} - \alpha) = \sin \alpha$, $\sin(\frac{\pi}{2} + \alpha) = \cos \alpha$ 24. Ifoda: $$\frac{\cos 2\alpha + \sin \alpha \sin \alpha}{\cos \alpha} = \frac{\cos 2\alpha + \sin^2 \alpha}{\cos \alpha}$$ 25. $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ 26. Shunday qilib, $$\frac{\cos^2 \alpha - \sin^2 \alpha + \sin^2 \alpha}{\cos \alpha} = \frac{\cos^2 \alpha}{\cos \alpha} = \cos \alpha$$ Javob: D) $\cos \alpha$ 27. Muammo: $\frac{\sin 2\alpha + \cos(\pi - \alpha) \sin \alpha}{\sin(\frac{\pi}{2} - \alpha)}$ ni soddalashtiring. 28. $\cos(\pi - \alpha) = -\cos \alpha$, $\sin(\frac{\pi}{2} - \alpha) = \cos \alpha$ 29. Ifoda: $$\frac{\sin 2\alpha - \cos \alpha \sin \alpha}{\cos \alpha}$$ 30. $\sin 2\alpha = 2 \sin \alpha \cos \alpha$ 31. Shunday qilib, $$\frac{2 \sin \alpha \cos \alpha - \cos \alpha \sin \alpha}{\cos \alpha} = \frac{\sin \alpha \cos \alpha}{\cos \alpha} = \sin \alpha$$ Javob: B) $\sin \alpha$ 32. Muammo: $\sin(\frac{\pi}{8}) \cos^3(\frac{\pi}{8}) - \sin^3(\frac{\pi}{8}) \cos(\frac{\pi}{8})$ ni hisoblang. 33. Formulalar: $$a = \sin \frac{\pi}{8}, b = \cos \frac{\pi}{8}$$ 34. Ifoda: $$a b^3 - a^3 b = ab(b^2 - a^2) = ab(b - a)(b + a)$$ 35. $b^2 - a^2 = \cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ 36. $a b = \sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}$ 37. Shunday qilib, $$a b^3 - a^3 b = ab(b^2 - a^2) = \frac{\sqrt{2}}{4} \times \frac{\sqrt{2}}{2} = \frac{1}{4}$$ Javob: A) 0.25 38. Muammo: $\sin(\frac{\pi}{16}) \cos^3(\frac{\pi}{16}) - \sin^3(\frac{\pi}{16}) \cos(\frac{\pi}{16})$ ni hisoblang. 39. Xuddi oldingi kabi, $$a = \sin \frac{\pi}{16}, b = \cos \frac{\pi}{16}$$ 40. Ifoda: $$a b^3 - a^3 b = ab(b^2 - a^2) = ab \cos \frac{\pi}{8}$$ 41. $ab = \frac{1}{2} \sin \frac{\pi}{8} = \frac{1}{2} \times \frac{\sqrt{2 - \sqrt{2}}}{2} = \frac{\sqrt{2 - \sqrt{2}}}{4}$ 42. $\cos \frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ 43. Shunday qilib, $$a b^3 - a^3 b = \frac{\sqrt{2 - \sqrt{2}}}{4} \times \frac{\sqrt{2 + \sqrt{2}}}{2} = \frac{\sqrt{(2 - \sqrt{2})(2 + \sqrt{2})}}{8} = \frac{\sqrt{4 - 2}}{8} = \frac{\sqrt{2}}{8}$$ Javob: A) $\frac{\sqrt{2}}{8}$ 44. Muammo: $14 \sqrt{2} (\sin^4(\frac{3\pi}{8}) - \cos^4(\frac{3\pi}{8}))$ ni hisoblang. 45. $a = \sin^2 \frac{3\pi}{8}$, $b = \cos^2 \frac{3\pi}{8}$ 46. $\sin^4 x - \cos^4 x = (a^2 - b^2) = (a - b)(a + b) = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = \cos 2x \times 1 = \cos 2x$ 47. Shunday qilib, $$\sin^4 \frac{3\pi}{8} - \cos^4 \frac{3\pi}{8} = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$ 48. Hisoblash: $$14 \sqrt{2} \times \left(-\frac{\sqrt{2}}{2}\right) = 14 \sqrt{2} \times -\frac{\sqrt{2}}{2} = 14 \times -1 = -14$$ Javob: B) -14 49. Muammo: $\sin^4(\frac{23\pi}{12}) - \cos^4(\frac{13\pi}{12})$ ni hisoblang. 50. $\sin^4 x - \cos^4 y = (\sin^2 x - \cos^2 y)(\sin^2 x + \cos^2 y)$ 51. $\sin^2 \frac{23\pi}{12} + \cos^2 \frac{13\pi}{12} = 1$ (chunki $\frac{23\pi}{12} = 2\pi - \frac{\pi}{12}$ va $\sin^2 \theta = \sin^2 (2\pi - \theta)$) 52. $\sin^2 \frac{23\pi}{12} - \cos^2 \frac{13\pi}{12} = \sin^2 \frac{\pi}{12} - \cos^2 \frac{13\pi}{12} = \sin^2 \frac{\pi}{12} - \cos^2 \frac{\pi}{12} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ 53. Shunday qilib, $$\sin^4 \frac{23\pi}{12} - \cos^4 \frac{13\pi}{12} = -\frac{\sqrt{3}}{2} \times 1 = -\frac{\sqrt{3}}{2}$$ Javob: A) $-\frac{\sqrt{3}}{2}$