1. Masala: $0 < \alpha < \frac{\pi}{2}$ va $\cos \alpha = \frac{1}{2} \sqrt{2} + \sqrt{2}$ bo'lsa, $\alpha$ ning qiymatini toping. 2. Masala: $\frac{\sin^4 \alpha + 2 \cos \alpha \sin \alpha - \cos^4 \alpha}{2 \cos^2 \alpha - 1}$ ifodasini soddalashtiring. 3. Masala: $\frac{\cos^2 x + \cos x}{2 \cos^2 \frac{x}{2}} + 1$ ifodasini soddalashtiring. 4. Masala: $8 \sin^2 \left(\frac{15 \pi}{16}\right) \cdot \cos^2 \left(\frac{17 \pi}{16}\right) - 1$ ni hisoblang. 5. Masala: $\frac{\sin^2 2.5 \alpha - \sin^2 1.5 \alpha}{\sin 4 \alpha \cdot \sin \alpha + \cos 3 \alpha \cdot \cos 2 \alpha}$ ifodasini soddalashtiring. 6. Masala: $\sin 112.5^\circ$ ni hisoblang. 7. Masala: $\sin 195^\circ$ ning qiymatini aniqlang. 8. Masala: $\cos 2227^\circ 30'$ ni hisoblang. 9. Masala: $\tan 105^\circ$ ning qiymatini hisoblang. 10. Masala: $\frac{4 \cos^2 2 \alpha - 4 \cos^2 \alpha + 3 \sin^2 \alpha}{4 \cos^2 \left(\frac{5 \pi}{2} - \alpha\right) - \sin^2 2 (\alpha - \pi)}$ ifodasini soddalashtiring. --- 1. $\cos \alpha = \frac{1}{2} \sqrt{2} + \sqrt{2} = \frac{3}{2} \sqrt{2}$, bu $>1$ emas, ehtimol $\cos \alpha = \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2}$ noto'g'ri, to'g'ri ifoda $\cos \alpha = \frac{1}{2} (\sqrt{2} + \sqrt{2}) = \sqrt{2}$ ham noto'g'ri. Ehtimol $\cos \alpha = \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2}$ noto'g'ri. To'g'ri ifoda $\cos \alpha = \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2}$ emas. Agar $\cos \alpha = \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2}$ bo'lsa, $\cos \alpha > 1$ bo'ladi, bu mumkin emas. Ehtimol $\cos \alpha = \frac{1}{2} (\sqrt{2} + 1)$ bo'lishi kerak. Agar $\cos \alpha = \frac{1}{2} (\sqrt{2} + 1)$ bo'lsa, $\alpha = \frac{\pi}{8}$. 2. $\sin^4 \alpha + 2 \cos \alpha \sin \alpha - \cos^4 \alpha = (\sin^2 \alpha)^2 - (\cos^2 \alpha)^2 + 2 \cos \alpha \sin \alpha = (\sin^2 \alpha - \cos^2 \alpha)(\sin^2 \alpha + \cos^2 \alpha) + 2 \cos \alpha \sin \alpha = -\cos 2 \alpha + 2 \sin \alpha \cos \alpha$. Denominator: $2 \cos^2 \alpha - 1 = \cos 2 \alpha$. Ifodani soddalashtiramiz: $$\frac{-\cos 2 \alpha + 2 \sin \alpha \cos \alpha}{\cos 2 \alpha} = \frac{-\cos 2 \alpha}{\cos 2 \alpha} + \frac{2 \sin \alpha \cos \alpha}{\cos 2 \alpha} = -1 + \frac{\sin 2 \alpha}{\cos 2 \alpha} = -1 + \tan 2 \alpha$$ Javob: $\tan 2 \alpha - 1$. 3. $\frac{\cos^2 x + \cos x}{2 \cos^2 \frac{x}{2}} + 1$. $\cos^2 x + \cos x = \cos x (\cos x + 1)$. $2 \cos^2 \frac{x}{2} = 1 + \cos x$. Ifodani yozamiz: $$\frac{\cos x (\cos x + 1)}{1 + \cos x} + 1 = \cos x + 1$$ $\cos x + 1 = 2 \cos^2 \frac{x}{2}$. Javob: $2 \cos^2 \frac{x}{2}$. 4. $8 \sin^2 \left(\frac{15 \pi}{16}\right) \cdot \cos^2 \left(\frac{17 \pi}{16}\right) - 1$. $\sin \left(\frac{15 \pi}{16}\right) = \cos \left(\frac{\pi}{16}\right)$, $\cos \left(\frac{17 \pi}{16}\right) = -\cos \left(\frac{\pi}{16}\right)$. Shunday qilib: $$8 \sin^2 \left(\frac{15 \pi}{16}\right) \cos^2 \left(\frac{17 \pi}{16}\right) - 1 = 8 \cos^2 \left(\frac{\pi}{16}\right) \cdot \cos^2 \left(\frac{\pi}{16}\right) - 1 = 8 \cos^4 \left(\frac{\pi}{16}\right) - 1$$ $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$, shuning uchun hisoblash davom ettiriladi. Natija: $-\frac{1}{2}$. 5. $\frac{\sin^2 2.5 \alpha - \sin^2 1.5 \alpha}{\sin 4 \alpha \sin \alpha + \cos 3 \alpha \cos 2 \alpha}$. $\sin^2 A - \sin^2 B = \sin (A-B) \sin (A+B)$. $\sin 4 \alpha \sin \alpha + \cos 3 \alpha \cos 2 \alpha = \cos (3 \alpha - 2 \alpha) - \cos (3 \alpha + 2 \alpha) + \cos 3 \alpha \cos 2 \alpha$. Soddalashtirish natijasi: $2 \tan 2 \alpha$. 6. $\sin 112.5^\circ = \sin \left(90^\circ + 22.5^\circ\right) = \cos 22.5^\circ = \frac{1}{2} \sqrt{2 + \sqrt{2}}$. 7. $\sin 195^\circ = \sin (180^\circ + 15^\circ) = -\sin 15^\circ = -\frac{\sqrt{6} - \sqrt{2}}{4} = -\frac{\sqrt{3} - 1}{2 \sqrt{2}}$. Javob: $-\frac{\sqrt{3} - \sqrt{2}}{2}$. 8. $2227^\circ 30' = 2227.5^\circ$. $2227.5^\circ - 6 \times 360^\circ = 2227.5^\circ - 2160^\circ = 67.5^\circ$. $\cos 67.5^\circ = \cos (45^\circ + 22.5^\circ) = \frac{\sqrt{2} - \sqrt{6}}{4}$. 9. $\tan 105^\circ = \tan (60^\circ + 45^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = 2 + \sqrt{3}$. 10. $\frac{4 \cos^2 2 \alpha - 4 \cos^2 \alpha + 3 \sin^2 \alpha}{4 \cos^2 \left(\frac{5 \pi}{2} - \alpha\right) - \sin^2 2 (\alpha - \pi)}$. $\cos \left(\frac{5 \pi}{2} - \alpha\right) = \sin \alpha$. $\sin 2 (\alpha - \pi) = \sin 2 \alpha$. Soddalashtirish natijasi: $4 \cos 2 \alpha - 1$. --- Javoblar: 1) $\frac{\pi}{8}$ 2) $\tan 2 \alpha - 1$ 3) $2 \cos^2 \frac{x}{2}$ 4) $-\frac{1}{2}$ 5) $2 \tan 2 \alpha$ 6) $\frac{1}{2} \sqrt{2 + \sqrt{2}}$ 7) $-\frac{\sqrt{3} - \sqrt{2}}{2}$ 8) $\frac{\sqrt{2} - \sqrt{6}}{4}$ 9) $2 + \sqrt{3}$ 10) $4 \cos 2 \alpha - 1$