1. **Masalani bayon qilish:** Berilgan masalalar trigonometrik ifodalar va ularning qiymatlarini topish bilan bog'liq. 2. **Formulalar va qoidalar:** - Tangens va kotangensning asosiy qiymatlari. - Trigonometrik ko'paytmalar va yig'indilar formulalari. - Ikki burchak formulasidan foydalanish: $$\tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2 \alpha}$$ - Arktangensning xossalari. 3. **Masala 7:** $$\tan 15^0 = 2 - \sqrt{3}$$ va $$\cot 15^0 = \frac{1}{\tan 15^0} = 2 + \sqrt{3}$$ Hisoblaymiz: $$\tan 15^0 - \cot 15^0 = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}$$ Javob: B) $-2\sqrt{3}$ 4. **Masala 8:** $$\alpha = 15^0$$ Hisoblaymiz: $$1 + \cos 2\alpha = 1 + \cos 30^0 = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$$ $$\tan \alpha = \tan 15^0 = 2 - \sqrt{3}$$ Shunday qilib: $$ (1 + \cos 2\alpha) \tan \alpha = \frac{2 + \sqrt{3}}{2} (2 - \sqrt{3}) = \frac{(2 + \sqrt{3})(2 - \sqrt{3})}{2} = \frac{4 - 3}{2} = \frac{1}{2}$$ Endi $\frac{1}{2}$ ni $\frac{1}{8}$ bilan solishtiramiz: $\frac{1}{2} = 4 \times \frac{1}{8}$ Javob: D) u $\frac{1}{8}$ dan 4 marta katta 5. **Masala 9:** $$\tan \alpha = \frac{1}{2}$$ Formuladan foydalanamiz: $$\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$ Javob: B) $\frac{4}{3}$ 6. **Masala 10:** $$\tan 22.5^0 + \tan^{-1} 22.5^0$$ Bu yerda $\tan^{-1} 22.5^0$ burchak emas, balki arktangens funksiyasi, lekin ehtimol savolda $\tan^{-1} 22.5^0$ $\cot 22.5^0$ degan ma'noda ishlatilgan. $$\tan 22.5^0 = \sqrt{2} - 1$$ $$\cot 22.5^0 = \frac{1}{\tan 22.5^0} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$$ Yig'indisi: $$ (\sqrt{2} - 1) + (\sqrt{2} + 1) = 2\sqrt{2}$$ Javob: E) $2\sqrt{2}$ 7. **Masala 11:** Berilgan: $$\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} = -\frac{1}{2}$$ va $$\frac{3\pi}{2} < \alpha < 2\pi$$ Formuladan foydalanamiz: $$\left(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}\right)^2 = \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} + 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = 1 + \sin \alpha$$ Chap tomon kvadrati: $$\left(-\frac{1}{2}\right)^2 = \frac{1}{4}$$ Shunday qilib: $$1 + \sin \alpha = \frac{1}{4} \Rightarrow \sin \alpha = -\frac{3}{4}$$ Endi $$\sin 2\alpha = 2 \sin \alpha \cos \alpha$$ $$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left(-\frac{3}{4}\right)^2} = -\sqrt{1 - \frac{9}{16}} = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}$$ $$\sin 2\alpha = 2 \cdot \left(-\frac{3}{4}\right) \cdot \left(-\frac{\sqrt{7}}{4}\right) = \frac{3\sqrt{7}}{8}$$ Ammo $\alpha$ 3$\pi/2$ dan katta, 2$\pi$ dan kichik, shuning uchun $\sin 2\alpha$ manfiy: $$\sin 2\alpha = -\frac{3\sqrt{7}}{8}$$ Javob: A) $-\frac{3\sqrt{7}}{8}$ 8. **Masala 12:** Hisoblang: $$\cos 92^0 \cdot \cos 2^0 + 0.5 \cdot \sin 4^0 + 1$$ $$\cos 92^0 = \cos (90^0 + 2^0) = -\sin 2^0 \approx -0.0349$$ $$\cos 2^0 \approx 0.9994$$ $$\sin 4^0 \approx 0.0698$$ Hisoblaymiz: $$-0.0349 \times 0.9994 + 0.5 \times 0.0698 + 1 = -0.0349 + 0.0349 + 1 = 1$$ Javob: B) 1