1. Muammo: Quyidagi ifodalardan qaysi birining qiymati 1 ga teng emasligini aniqlash kerak: 1) $2\cos^2\alpha - \cos 2\alpha$ 2) $2\sin^2\alpha + \cos 2\alpha$ 3) \tg(90^\circ + \alpha) \tg \alpha$ 4) $\frac{\cos^2\alpha - 1}{\sin^2\alpha - 1}$ (bu ifoda $\alpha$ ning qabul qilishi mumkin bo‘lgan qiymatlarida qaraladi) 2. Formulalar va qoidalar: - $\cos 2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha$ - $\tg(90^\circ + \alpha) = -\cot \alpha$ - $\cot \theta = \frac{1}{\tan \theta}$ 3. Hisoblashlar: 1) $2\cos^2\alpha - \cos 2\alpha = 2\cos^2\alpha - (2\cos^2\alpha - 1) = 2\cos^2\alpha - 2\cos^2\alpha + 1 = 1$ 2) $2\sin^2\alpha + \cos 2\alpha = 2\sin^2\alpha + (1 - 2\sin^2\alpha) = 2\sin^2\alpha + 1 - 2\sin^2\alpha = 1$ 3) $\tg(90^\circ + \alpha) \tg \alpha = (-\cot \alpha) \tg \alpha = -1$ 4) $\frac{\cos^2\alpha - 1}{\sin^2\alpha - 1} = \frac{-(1 - \cos^2\alpha)}{-(1 - \sin^2\alpha)} = \frac{-\sin^2\alpha}{-\cos^2\alpha} = \frac{\sin^2\alpha}{\cos^2\alpha} = \tan^2\alpha$ $\tan^2\alpha$ har doim 1 ga teng emas, shuning uchun 4-ifoda qiymati 1 ga teng emas. 4. Javob: 4-ifoda qiymati 1 ga teng emas, ya'ni javob D). 5. Ikkinchi masala: $\frac{1 + \sin 2\alpha}{\sin \alpha + \cos \alpha} - \sin \alpha$ ni soddalashtiring. $\sin 2\alpha = 2\sin \alpha \cos \alpha$ $\Rightarrow \frac{1 + 2\sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha} - \sin \alpha$ Birinchi qismni $\sin \alpha + \cos \alpha$ ga bo'lamiz: $\frac{1 + 2\sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha} - \sin \alpha = \frac{1 + 2\sin \alpha \cos \alpha - \sin \alpha (\sin \alpha + \cos \alpha)}{\sin \alpha + \cos \alpha}$ $= \frac{1 + 2\sin \alpha \cos \alpha - \sin^2 \alpha - \sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha} = \frac{1 + \sin \alpha \cos \alpha - \sin^2 \alpha}{\sin \alpha + \cos \alpha}$ $1 - \sin^2 \alpha = \cos^2 \alpha$ $\Rightarrow \frac{\cos^2 \alpha + \sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha} = \frac{\cos \alpha (\cos \alpha + \sin \alpha)}{\sin \alpha + \cos \alpha}$ $= \cos \alpha$ (chunki $\cos \alpha + \sin \alpha = \sin \alpha + \cos \alpha$) Javob: A) $\cos \alpha$ 6. Uchinchi masala: $\sin^6 \alpha + \cos^6 \alpha + \frac{3}{4} \sin^2 2\alpha$ ni soddalashtiring. $\sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha)^3 + (\cos^2 \alpha)^3$ Kub yig'indisi formulasiga ko'ra: $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ Bu yerda $a = \sin^2 \alpha$, $b = \cos^2 \alpha$ $a + b = 1$ $ab = \sin^2 \alpha \cos^2 \alpha$ Shunday qilib: $\sin^6 \alpha + \cos^6 \alpha = 1^3 - 3 \sin^2 \alpha \cos^2 \alpha (1) = 1 - 3 \sin^2 \alpha \cos^2 \alpha$ $\sin^2 2\alpha = 4 \sin^2 \alpha \cos^2 \alpha$ $\frac{3}{4} \sin^2 2\alpha = \frac{3}{4} \times 4 \sin^2 \alpha \cos^2 \alpha = 3 \sin^2 \alpha \cos^2 \alpha$ Endi ifodani yig'indilaymiz: $1 - 3 \sin^2 \alpha \cos^2 \alpha + 3 \sin^2 \alpha \cos^2 \alpha = 1$ Javob: B) 1 7. To'rtinchi masala: $\frac{2 \cos^2 \alpha}{\cot \frac{\alpha}{2} - \tan \frac{\alpha}{2}}$ ni soddalashtiring. $\cot x = \frac{1}{\tan x}$, shuning uchun: $\cot \frac{\alpha}{2} - \tan \frac{\alpha}{2} = \frac{1}{\tan \frac{\alpha}{2}} - \tan \frac{\alpha}{2} = \frac{1 - \tan^2 \frac{\alpha}{2}}{\tan \frac{\alpha}{2}}$ Shunday qilib: $\frac{2 \cos^2 \alpha}{\frac{1 - \tan^2 \frac{\alpha}{2}}{\tan \frac{\alpha}{2}}} = 2 \cos^2 \alpha \times \frac{\tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}}$ $1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x}$ va $\tan x = \frac{\sin x}{\cos x}$ $\Rightarrow 2 \cos^2 \alpha \times \frac{\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}}{\frac{\cos \alpha}{\cos^2 \frac{\alpha}{2}}} = 2 \cos^2 \alpha \times \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \times \frac{\cos^2 \frac{\alpha}{2}}{\cos \alpha}$ $= 2 \cos^2 \alpha \times \sin \frac{\alpha}{2} \times \frac{\cos \frac{\alpha}{2}}{\cos \alpha}$ $= 2 \cos^2 \alpha \times \sin \frac{\alpha}{2} \times \frac{\cos \frac{\alpha}{2}}{\cos \alpha}$ $\cos^2 \alpha / \cos \alpha = \cos \alpha$ $\Rightarrow 2 \cos \alpha \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ Shunday qilib, ifoda: $2 \cos \alpha \times \frac{\sin \alpha}{2} = \cos \alpha \sin \alpha$ Javob: $\cos \alpha \sin \alpha$