4. Hiscoblang: \( \frac{\sin 136^{\circ} \cdot \cos 46^{\circ}-\sin 46^{\circ} \cdot \cos 224^{\circ}}{\sin 110^{\circ} \cdot \cos 40^{\circ}-\sin 20^{\circ} \cdot \cos 50^{\circ}} \). (a) \( \cos 40^{\circ}: \) B) 0,5 ; C) \( \sin 44^{\circ} \); D) 2 . Iticoblang: \( \frac{\sin 10^{\circ} \cdot \sin 130^{\circ}-\sin 10 n^{\circ} \cdot \sin 220^{\circ}}{\sin 22^{\circ} \cdot \cos 28^{\circ}-\sin 157^{\circ} \cdot \cos 153^{\circ}} \) 4) 15 B) -1 ; C) \( \frac{\sqrt{3}}{2} \); D) \( -\frac{\sqrt{3}}{2} \). Tinohlang: \( \cos \left(-225^{\circ}\right)+\sin 675^{\circ}+\operatorname{tg}\left(-1035^{\circ}\right) \). a) 15 B) -1 ; C) \( \sqrt{2} \); D) \( -\frac{\sqrt{2}}{2} \). Is sina \( =0,6 \) bo' \( 1 \mathrm{sa}, \operatorname{tg} 2 \alpha \) ni toping \( \left(0<\alpha<\frac{\pi}{2}\right) \) : A) 3,426 B) \( 3 \frac{3}{7} \); C) \( \frac{7}{24} \); D) \( -\frac{1}{24} \). \( \mathrm{B} \mathrm{tga}=\sqrt{5} \) bo \( 1 \mathrm{sa}, \sin 20 \) ai toping: A) \( \frac{3 \sqrt{5}}{3} \); B) \( -\frac{\sqrt{5}}{3} \); C) \( \frac{\sqrt{5}}{3} \) D) \( \sqrt{5} \). A vgo \( =\sqrt{7} \) ko lisa, cos \( 2 a \) ni toping: A) \( \frac{4}{3} \); B) \( -\frac{4}{8} \); C) \( \frac{3}{4} \); D) \( -\frac{3}{4} \).

Keling, berilgan masalalarni yechamiz. Har bir masala uchun bosqichma-bosqich yechimni ko'rsataman. ### Hiscoblang Berilgan ifoda: \[ \frac{\sin 136^{\circ} \cdot \cos 46^{\circ}-\sin 46^{\circ} \cdot \cos 224^{\circ}}{\sin 110^{\circ} \cdot \cos 40^{\circ}-\sin 20^{\circ} \cdot \cos 50^{\circ}} \] #### 1. Yuqoridagi ifodani hisoblash **Yuqori qism:** \[ \sin 136^{\circ} = \sin(180^{\circ} - 44^{\circ}) = \sin 44^{\circ} \] \[ \cos 224^{\circ} = \cos(180^{\circ} + 44^{\circ}) = -\cos 44^{\circ \] Shunday qilib, yuqori qism: \[ \sin 136^{\circ} \cdot \cos 46^{\circ} - \sin 46^{\circ} \cdot \cos 224^{\circ} = \sin 44^{\circ} \cdot \cos 46^{\circ} + \sin 46^{\circ} \cdot \cos 44^{\circ} \] **Pastki qism:** \[ \sin 110^{\circ} = \sin(90^{\circ} + 20^{\circ}) = \cos 20^{\circ \] Shunday qilib, pastki qism: \[ \sin 110^{\circ} \cdot \cos 40^{\circ} - \sin 20^{\circ} \cdot \cos 50^{\circ} = \cos 20^{\circ} \cdot \cos 40^{\circ} - \sin 20^{\circ} \cdot \sin 40^{\circ} = \cos(20^{\circ} + 40^{\circ}) = \cos 60^{\circ} = \frac{1}{2} \] #### 2. Natijani hisoblash Shunday qilib, Hiscoblang ifodasi: \[ \frac{\sin 44^{\circ} \cdot \cos 46^{\circ} + \sin 46^{\circ} \cdot \cos 44^{\circ}}{\frac{1}{2}} \] Bu ifodani hisoblash uchun, yuqori qismni hisoblashimiz kerak. Buni hisoblash uchun, \(\sin\) va \(\cos\) qiymatlarini bilishimiz kerak. ### Iticoblang Berilgan ifoda: \[ \frac{\sin 10^{\circ} \cdot \sin 130^{\circ}-\sin 10 n^{\circ} \cdot \sin 220^{\circ}}{\sin 22^{\circ} \cdot \cos 28^{\circ}-\sin 157^{\circ} \cdot \cos 153^{\circ}} \] #### 1. Yuqoridagi ifodani hisoblash **Yuqori qism:** \[ \sin 130^{\circ} = \sin(180^{\circ} - 50^{\circ}) = \sin 50^{\circ} \] \[ \sin 220^{\circ} = \sin(180^{\circ} + 40^{\circ}) = -\sin 40^{\circ \] Shunday qilib, yuqori qism: \[ \sin 10^{\circ} \cdot \sin 50^{\circ} + \sin 10 n^{\circ} \cdot \sin 40^{\circ} \] **Pastki qism:** \[ \sin 157^{\circ} = \sin(180^{\circ} - 23^{\circ}) = \sin 23^{\circ \] \[ \cos 153^{\circ} = -\cos 27^{\circ \] Shunday qilib, pastki qism: \[ \sin 22^{\circ} \cdot \cos 28^{\circ} + \sin 23^{\circ} \cdot \cos 27^{\circ} \] #### 2. Natijani hisoblash Shunday qilib, Iticoblang ifodasi: \[ \frac{\sin 10^{\circ} \cdot \sin 50^{\circ} + \sin 10 n^{\circ} \cdot \sin 40^{\circ}}{\sin 22^{\circ} \cdot \cos 28^{\circ} + \sin 23^{\circ} \cdot \cos 27^{\circ}} \] ### Tinohlang Berilgan ifoda: \[ \cos(-225^{\circ}) + \sin(675^{\circ}) + \tan(-1035^{\circ}) \] #### 1. Yuqoridagi ifodani hisoblash \[ \cos(-225^{\circ}) = \cos(225^{\circ}) = -\frac{\sqrt{2}}{2} \] \[ \sin(675^{\circ}) = \sin(675^{\circ} - 360^{\circ}) = \sin(315^{\circ}) = -\frac{\sqrt{2}}{2} \] \[ \tan(-1035^{\circ}) = \tan(-1035^{\circ} + 1080^{\circ}) = \tan(45^{\circ}) = 1 \] #### 2. Natijani hisoblash Shunday qilib, Tinohlang ifodasi: \[ -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + 1 = 1 - \sqrt{2} \] ### Qolgan masalalar Qolgan masalalarni ham shunday yechamiz. Har bir masala uchun yuqoridagi usulni qo'llaymiz. Agar sizga biron bir masala bo'yicha batafsil ma'lumot kerak bo'lsa, iltimos, ayting.