12. \( \sin \alpha=0,6 \) bo'lsa, \( \operatorname{tg} 2 \alpha \) ni toping \( \left(0<\alpha<\frac{\pi}{2}\right) \) : \( \begin{array}{llll}\text { A) } 3,42 ; & \text { B) } 3 \frac{3}{7} ; & \text { C) } \frac{7}{24} ; & \text { D) } \\ \text { 13. } \operatorname{tg} \alpha=\sqrt{5} \text { bo'lsa, } \sin 2 \alpha \text { ni toping: } \\ \begin{array}{llll}\text { A) } \frac{3 \sqrt{5}}{5} ; & \text { B) }-\frac{\sqrt{5}}{3} ; & \text { C) } \frac{\sqrt{5}}{3} ; & \text { D) } \\ \text { 14. } \operatorname{tg} \alpha=\sqrt{7} \text { bo'lsa, } \cos 2 \alpha \text { ni toping: } \\ \begin{array}{llll}\text { A) } \frac{4}{3} ; & \text { B) }-\frac{4}{3} ; & \text { C) } \frac{3}{4}\end{array}\end{array}\end{array} \)

Keling, har bir masalani alohida ko'rib chiqamiz. ### Masala 12: \( \sin \alpha = 0.6 \) bo'lsa, \( \operatorname{tg} 2\alpha \) ni toping. 1. **Birinchi qadam:** \( \sin \alpha \) dan \( \cos \alpha \) ni topamiz. \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (0.6)^2 = 1 - 0.36 = 0.64 \] \[ \cos \alpha = \sqrt{0.64} = 0.8 \] 2. **Ikkinchi qadam:** \( \operatorname{tg} \alpha \) ni hisoblaymiz. \[ \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{0.6}{0.8} = 0.75 \] 3. **Uchinchi qadam:** \( \operatorname{tg} 2\alpha \) formulasi: \[ \operatorname{tg} 2\alpha = \frac{2 \operatorname{tg} \alpha}{1 - \operatorname{tg}^2 \alpha} \] \[ \operatorname{tg} 2\alpha = \frac{2 \cdot 0.75}{1 - (0.75)^2} = \frac{1.5}{1 - 0.5625} = \frac{1.5}{0.4375} = \frac{1.5 \cdot 16}{7} = \frac{24}{7} \approx 3.42857 \] Natija: \( \operatorname{tg} 2\alpha \approx 3.42 \) (A). ### Masala 13: \( \operatorname{tg} \alpha = \sqrt{5} \) bo'lsa, \( \sin 2\alpha \) ni toping. 1. **Birinchi qadam:** \( \sin 2\alpha \) formulasi: \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \] 2. **Ikkinchi qadam:** \( \sin \alpha \) va \( \cos \alpha \) ni topamiz: \[ \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha = \sqrt{5} \cos \alpha \] \[ \sin^2 \alpha + \cos^2 \alpha = 1 \Rightarrow (\sqrt{5} \cos \alpha)^2 + \cos^2 \alpha = 1 \] \[ 5 \cos^2 \alpha + \cos^2 \alpha = 1 \Rightarrow 6 \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{6} \Rightarrow \cos \alpha = \frac{1}{\sqrt{6}} \] \[ \sin \alpha = \sqrt{5} \cdot \frac{1}{\sqrt{6}} = \frac{\sqrt{5}}{\sqrt{6}} \] 3. **Uchinchi qadam:** \( \sin 2\alpha \) ni hisoblaymiz: \[ \sin 2\alpha = 2 \cdot \frac{\sqrt{5}}{\sqrt{6}} \cdot \frac{1}{\sqrt{6}} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \] Natija: \( \sin 2\alpha = \frac{\sqrt{5}}{3} \) (C). ### Masala 14: \( \operatorname{tg} \alpha = \sqrt{7} \) bo'lsa, \( \cos 2\alpha \) ni toping. 1. **Birinchi qadam:** \( \cos 2\alpha \) formulasi: \[ \cos 2\alpha = \frac{1 - \operatorname{tg}^2 \alpha}{1 + \operatorname{tg}^2 \alpha} \] 2. **Ikkinchi qadam:** \( \operatorname{tg}^2 \alpha \) ni hisoblaymiz: \[ \operatorname{tg}^2 \alpha = (\sqrt{7})^2 = 7 \] 3. **Uchinchi qadam:** \( \cos 2\alpha \) ni hisoblaymiz: \[ \cos 2\alpha = \frac{1 - 7}{1 + 7} = \frac{-6}{8} = -\frac{3}{4} \] Natija: \( \cos 2\alpha = -\frac{3}{4} \) (bu variantlar orasida yo'q, lekin hisoblash to'g'ri). Shunday qilib, natijalar: - 12. A) 3,42 - 13. C) \( \frac{\sqrt{5}}{3} \) - 14. \( \cos 2\alpha = -\frac{3}{4} \) (variantlar orasida yo'q).