1. Muammo: cos\alpha = \frac{1}{2} \sqrt{2} + \sqrt{3} tenglik qaysi \alpha o‘tkir burchak uchun to‘g‘ri?\n\n2. Formulalar va qoidalar: cos\alpha qiymati -1 dan 1 gacha bo‘ladi. \n\n3. Hisoblash: \frac{1}{2} \sqrt{2} + \sqrt{3} ni hisoblaymiz:\n$$\frac{1}{2} \sqrt{2} + \sqrt{3} \approx 0.707 + 1.732 = 2.439$$\nBu qiymat cos\alpha uchun mumkin emas, chunki cos\alpha \in [-1,1].\n\n4. Shuning uchun berilgan ifoda xato yoki noto‘g‘ri yozilgan.\n\n5. Ikkinchi masala: cos^{2}22^{\circ}30' - sin^{2}22^{\circ}30' ni toping.\n\n6. Formuladan foydalanamiz: cos^{2}x - sin^{2}x = cos 2x\n\n7. Hisoblash: cos 2 \times 22^{\circ}30' = cos 45^{\circ} = \frac{\sqrt{2}}{2} \approx 0.707\n\n8. Javob: B) \frac{\sqrt{2}}{4} emas, chunki \frac{\sqrt{2}}{2} = 0.707, \frac{\sqrt{2}}{4} = 0.353. To‘g‘ri javob A) \frac{\sqrt{2}}{8} emas, chunki u ham kichik.\n\n9. To‘g‘ri javob: cos 45^{\circ} = \frac{\sqrt{2}}{2} ya'ni 0.707, bu variantlarda yo‘q.\n\n10. Uchinchi masala: cos^{2}5 + cos^{2}1 - cos6 \cdot cos4 ni hisoblang.\n\n11. Formulalar: cos^{2}x = \frac{1 + cos 2x}{2}\n\n12. Hisoblash: cos^{2}5 = \frac{1 + cos 10}{2}, cos^{2}1 = \frac{1 + cos 2}{2}\n\n13. Shunday qilib,\n$$cos^{2}5 + cos^{2}1 - cos6 \cdot cos4 = \frac{1 + cos 10}{2} + \frac{1 + cos 2}{2} - cos6 \cdot cos4$$\n\n14. cos6 \cdot cos4 ni ifodalash uchun cos A cos B = \frac{cos(A+B) + cos(A-B)}{2} formulasidan foydalanamiz:\n$$cos6 \cdot cos4 = \frac{cos(10) + cos(2)}{2}$$\n\n15. Endi ifodani yozamiz:\n$$\frac{1 + cos 10}{2} + \frac{1 + cos 2}{2} - \frac{cos 10 + cos 2}{2} = \frac{1 + cos 10 + 1 + cos 2 - cos 10 - cos 2}{2} = \frac{2}{2} = 1$$\n\n16. Javob: D) 1\n\n17. To‘rtinchi masala: sin^{4}15^{\circ} + cos^{4}15^{\circ} ni hisoblang.\n\n18. Formulalar: a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2a^{2}b^{2}\n\n19. sin^{2}15^{\circ} + cos^{2}15^{\circ} = 1\n\n20. sin^{2}15^{\circ} \cdot cos^{2}15^{\circ} ni hisoblaymiz:\n$$sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}, cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$$\n$$sin^{2}15^{\circ} = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^{2} = \frac{6 - 2\sqrt{12} + 2}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$$\n$$cos^{2}15^{\circ} = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^{2} = \frac{6 + 2\sqrt{12} + 2}{16} = \frac{8 + 4\sqrt{3}}{16} = \frac{2 + \sqrt{3}}{4}$$\n\n21. Shunday qilib,\n$$sin^{2}15^{\circ} \cdot cos^{2}15^{\circ} = \frac{2 - \sqrt{3}}{4} \times \frac{2 + \sqrt{3}}{4} = \frac{4 - 3}{16} = \frac{1}{16}$$\n\n22. Endi sin^{4}15^{\circ} + cos^{4}15^{\circ} ni hisoblaymiz:\n$$= (sin^{2}15^{\circ} + cos^{2}15^{\circ})^{2} - 2 sin^{2}15^{\circ} cos^{2}15^{\circ} = 1^{2} - 2 \times \frac{1}{16} = 1 - \frac{1}{8} = \frac{7}{8}$$\n\n23. Javob: E) \frac{7}{8}