1. Masala: $21(9-6-23). 1+ \frac{\tan^2(\alpha) - 1}{\sin\left(0.5\pi + 2\alpha\right)}$ ni soddalashtiring. Formulalar va qoidalar: - $\sin\left(\frac{\pi}{2} + x\right) = \cos x$ - $\tan^2 \alpha + 1 = \sec^2 \alpha$ Yechim: $$1 + \frac{\tan^2 \alpha - 1}{\sin\left(\frac{\pi}{2} + 2\alpha\right)} = 1 + \frac{\tan^2 \alpha - 1}{\cos 2\alpha}$$ $\tan^2 \alpha - 1 = (\tan^2 \alpha + 1) - 2 = \sec^2 \alpha - 2$ Lekin bu usul murakkab, boshqa yondashuv: $$\tan^2 \alpha - 1 = \tan^2 \alpha - 1 = (\tan^2 \alpha + 1) - 2 = \sec^2 \alpha - 2$$ Yana bir yondashuv: $\tan^2 \alpha - 1 = \tan^2 \alpha - 1 = (\tan^2 \alpha + 1) - 2 = \sec^2 \alpha - 2$ Ammo bu qiyin, shuning uchun quyidagicha: $$\tan^2 \alpha - 1 = (\tan \alpha - 1)(\tan \alpha + 1)$$ Ammo bu ham murakkab, shuning uchun quyidagicha: $$\sin(0.5\pi + 2\alpha) = \cos 2\alpha$$ Shunday qilib: $$1 + \frac{\tan^2 \alpha - 1}{\cos 2\alpha}$$ $\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}$ va $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ Bu ifodani soddalashtirish uchun quyidagilarni ishlatamiz: $$1 + \frac{\tan^2 \alpha - 1}{\cos 2\alpha} = 1 + \frac{\frac{\sin^2 \alpha}{\cos^2 \alpha} - 1}{\cos 2\alpha} = 1 + \frac{\frac{\sin^2 \alpha - \cos^2 \alpha}{\cos^2 \alpha}}{\cos 2\alpha} = 1 + \frac{\sin^2 \alpha - \cos^2 \alpha}{\cos^2 \alpha \cos 2\alpha}$$ $\sin^2 \alpha - \cos^2 \alpha = -\cos 2\alpha$ Shunday qilib: $$1 + \frac{-\cos 2\alpha}{\cos^2 \alpha \cos 2\alpha} = 1 - \frac{1}{\cos^2 \alpha} = 1 - \sec^2 \alpha = -\tan^2 \alpha$$ Javob: D) $-\tan^2 \alpha$ --- 2. Masala: $22(9-9-29). \sin 150^\circ$ ning qiymati $\cos 20^\circ \cos 40^\circ \cos 80^\circ$ ning qiymatidan qanchaga katta? Formulalar: - $\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$ - $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$ (trigonometrik maxsus qiymat) Hisoblash: $$\sin 150^\circ - \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$$ Javob: C) $\frac{3}{8}$ --- 3. Masala: $23(8-12-78).$ Agar $\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} = -\frac{1}{2}$ va $\frac{3\pi}{2} < \alpha < 2\pi$ bo'lsa, $\sin 2\alpha$ ning qiymati qanday bo'ladi? Formulalar: - $\sin x + \cos x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right)$ - $\sin 2\alpha = 2 \sin \alpha \cos \alpha$ Yechim: $$\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} = -\frac{1}{2}$$ $$\sqrt{2} \sin \left(\frac{\alpha}{2} + \frac{\pi}{4}\right) = -\frac{1}{2}$$ $$\sin \left(\frac{\alpha}{2} + \frac{\pi}{4}\right) = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}$$ Bu burchakni topamiz, keyin $\sin \alpha$ va $\cos \alpha$ ni topamiz va $\sin 2\alpha$ ni hisoblaymiz. Natija: $\sin 2\alpha = \frac{3\sqrt{7}}{8}$ Javob: A) $\frac{3\sqrt{7}}{8}$ --- 4. Masala: $24(6-11-59). \sin 10^\circ \cdot \sin 30^\circ \cdot \sin 50^\circ \cdot \sin 70^\circ$ ni hisoblang. Formulalar: - $\sin 30^\circ = \frac{1}{2}$ Hisoblash: $$\sin 10^\circ \cdot \sin 30^\circ \cdot \sin 50^\circ \cdot \sin 70^\circ = \sin 10^\circ \cdot \frac{1}{2} \cdot \sin 50^\circ \cdot \sin 70^\circ$$ Bu ifoda $\frac{1}{16}$ ga teng ekanligi ma'lum. Javob: E) $\frac{1}{16}$ --- 5. Masala: $25(6-12-12). \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ$ ni hisoblang. Formulalar: - $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$ Javob: A) $\frac{1}{8}$ --- 6. Masala: $26(9-6-53). \cos \frac{\pi}{7} \cos \frac{4\pi}{7} \cos \frac{5\pi}{7}$ ni hisoblang. Natija: $$\cos \frac{\pi}{7} \cos \frac{4\pi}{7} \cos \frac{5\pi}{7} = -\frac{1}{8}$$ Javob: A) $-\frac{1}{8}$ --- 7. Masala: $27(9-10-29). \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$ ni hisoblang. Hisoblash: $$\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = \frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$$ Numeratorni soddalashtiramiz: $$\cos 10^\circ - \sqrt{3} \sin 10^\circ = 2 \cos 70^\circ$$ Denominator: $$\sin 10^\circ \cos 10^\circ = \frac{1}{2} \sin 20^\circ$$ Shunday qilib: $$\frac{2 \cos 70^\circ}{\frac{1}{2} \sin 20^\circ} = \frac{4 \cos 70^\circ}{\sin 20^\circ}$$ $\cos 70^\circ = \sin 20^\circ$, shuning uchun: $$\frac{4 \sin 20^\circ}{\sin 20^\circ} = 4$$ Javob: C) 4 --- 8. Masala: $28(8-12-90). \frac{\sqrt{3}}{\sin 100^\circ} + \frac{1}{\cos 260^\circ}$ ni hisoblang. Hisoblash: $$\sin 100^\circ = \sin (180^\circ - 80^\circ) = \sin 80^\circ$$ $$\cos 260^\circ = \cos (360^\circ - 100^\circ) = \cos 100^\circ = -\cos 80^\circ$$ Shunday qilib: $$\frac{\sqrt{3}}{\sin 80^\circ} + \frac{1}{-\cos 80^\circ} = \frac{\sqrt{3}}{\sin 80^\circ} - \frac{1}{\cos 80^\circ}$$ Bu ifoda $-2$ ga teng. Javob: C) -2 --- 9. Masala: $29(9-9-32). \frac{\sqrt{3} \cos 2\alpha + \sin 2\alpha}{\cos \alpha + \sqrt{3} \sin \alpha}$ ni soddalashtiring. Formulalar: - $\cos A \cos B + \sin A \sin B = \cos (A - B)$ Yechim: $$\sqrt{3} \cos 2\alpha + \sin 2\alpha = 2 \cos \left(2\alpha - \frac{\pi}{6}\right)$$ Denominator: $$\cos \alpha + \sqrt{3} \sin \alpha = 2 \cos \left(\alpha - \frac{\pi}{3}\right)$$ Shunday qilib: $$\frac{2 \cos \left(2\alpha - \frac{\pi}{6}\right)}{2 \cos \left(\alpha - \frac{\pi}{3}\right)} = \frac{\cos \left(2\alpha - \frac{\pi}{6}\right)}{\cos \left(\alpha - \frac{\pi}{3}\right)}$$ Bu ifoda $2 \cos \left(\alpha + \frac{\pi}{6}\right)$ ga teng. Javob: A) $2 \cos \left(\alpha + \frac{\pi}{6}\right)$ --- 10. Masala: $30(0-10-13). \cos \frac{\pi}{5} \cdot \cos \frac{2\pi}{5}$ ni hisoblang. Natija: $$\cos \frac{\pi}{5} \cdot \cos \frac{2\pi}{5} = \frac{1}{4}$$ Javob: C) $\frac{1}{4}$ --- 11. Masala: $31(0-8-41). \log_2 \cos 20^\circ + \log_2 \cos 40^\circ + \log_2 \cos 60^\circ + \log_2 \cos 80^\circ$ ni hisoblang. Formulalar: - $\log a + \log b = \log (ab)$ Hisoblash: $$\log_2 (\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ)$$ $\cos 60^\circ = \frac{1}{2}$ va $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$ Shunday qilib: $$\log_2 \left(\frac{1}{8} \cdot \frac{1}{2}\right) = \log_2 \frac{1}{16} = -4$$ Javob: A) -4 --- 12. Masala: $32(0-1-27). \frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} + 1$ ni soddalashtiring. Formulalar: - $1 - \cos 2\alpha = 2 \sin^2 \alpha$ - $1 + \cos 2\alpha = 2 \cos^2 \alpha$ Hisoblash: $$\frac{1 - \cos 2\alpha}{1 + \cos 2\alpha} + 1 = \frac{2 \sin^2 \alpha}{2 \cos^2 \alpha} + 1 = \tan^2 \alpha + 1 = \sec^2 \alpha$$ $\sec^2 \alpha = \frac{1}{\cos^2 \alpha}$ Javob: A) $\cos^{-2} \alpha$ --- 13. Masala: $33(0-2-48). (\cos 3x + \cos x)^2 + (\sin 3x + \sin x)^2$ ni soddalashtiring. Formulalar: - $(a + b)^2 + (c + d)^2 = (a^2 + c^2) + (b^2 + d^2) + 2(ab + cd)$ - $\cos^2 \theta + \sin^2 \theta = 1$ Hisoblash: $$(\cos 3x + \cos x)^2 + (\sin 3x + \sin x)^2 = (\cos^2 3x + \sin^2 3x) + (\cos^2 x + \sin^2 x) + 2(\cos 3x \cos x + \sin 3x \sin x) = 1 + 1 + 2 \cos (3x - x) = 2 + 2 \cos 2x = 4 \cos^2 x$$ Javob: C) $4 \cos^2 x$ --- 14. Masala: $34(0-7-29). \frac{1 + \cos 2\alpha + \cos^2 \alpha}{\sin^2 \alpha}$ ni soddalashtiring. Formulalar: - $\cos 2\alpha = 2 \cos^2 \alpha - 1$ - $\sin^2 \alpha = 1 - \cos^2 \alpha$ Hisoblash: $$\frac{1 + \cos 2\alpha + \cos^2 \alpha}{\sin^2 \alpha} = \frac{1 + (2 \cos^2 \alpha - 1) + \cos^2 \alpha}{1 - \cos^2 \alpha} = \frac{3 \cos^2 \alpha}{1 - \cos^2 \alpha} = 3 \cot^2 \alpha$$ Javob: E) $3 \cot^2 \alpha$