1. Muammo: $29(0-3-53). \cos \alpha = \frac{1}{2} (\sqrt{2} + \sqrt{3})$ tenglik uchun $\alpha$ ning o'tkir burchak qiymatini toping. 2. Formulalar va qoidalar: - $\cos \alpha$ qiymati $[-1,1]$ oraliqda bo'lishi kerak. - O'tkir burchaklar $0^\circ < \alpha < 90^\circ$ oraliqda. 3. Hisoblash: - $\frac{1}{2} (\sqrt{2} + \sqrt{3}) \approx \frac{1}{2} (1.414 + 1.732) = \frac{3.146}{2} = 1.573$. - $\cos \alpha$ uchun qiymat $1.573$ bo'lib, bu $[-1,1]$ oraliqdan tashqarida. - Demak, berilgan ifoda noto'g'ri yoki xato yozilgan. 4. Ehtimol, ifoda $\cos \alpha = \frac{1}{2} \sqrt{2 + \sqrt{3}}$ bo'lishi kerak. - Hisoblaymiz: $\sqrt{2 + \sqrt{3}} \approx \sqrt{2 + 1.732} = \sqrt{3.732} \approx 1.932$. - $\frac{1}{2} \times 1.932 = 0.966$. - $\cos \alpha = 0.966$. 5. $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$ yaqin. - Shunday qilib, $\alpha = 15^\circ$. 6. Javob: B) 15°. "slug": "cos_alpha_value","subject": "trigonometry","desmos": {"latex": "y=\cos\alpha","features": {"intercepts": true,"extrema": true}},"q_count": 1