Упр.22.7 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) (sin^2(315°)cos(300°)+tg(-315°))/(sin(-120°)cos(150°));
2) sin(5?/4)cos(5?/6)tg(-2?/3)ctg(4?/3);
3) (sin(20°)cos(10°)+cos(160°)cos(100°))/(sin(21°)cos(9°)+cos(159°)cos(99°));
4) (cos(66°)cos(5°)+cos(84°)cos(24°))/(cos(65°)cos(5°)+cos(85°)cos(25°)).

1. $$\frac{\sin^2 315^\circ \cdot \cos 300^\circ + \tg(-315^\circ)}{\sin(-120^\circ)\cdot \cos 150^\circ}$$

   $$\sin 315^\circ=-\frac{\sqrt2}{2},\quad \cos 300^\circ=\frac12,\quad \tg(-315^\circ)=\tg 45^\circ=1,$$

   $$\sin(-120^\circ)=-\sin 120^\circ=-\frac{\sqrt3}{2},\quad \cos 150^\circ=-\cos 30^\circ=-\frac{\sqrt3}{2}.$$

   Тогда
   $$\frac{\left(-\frac{\sqrt2}{2}\right)^2\cdot \frac12+1}{\left(-\frac{\sqrt3}{2}\right)\left(-\frac{\sqrt3}{2}\right)} =\frac{\frac12\cdot \frac12+1}{\frac34} =\frac{\frac14+1}{\frac34} =\frac{5}{3}.$$

2. $$\sin\frac{5\pi}{4}\cdot \cos\frac{5\pi}{6}\cdot \tg\left(-\frac{2\pi}{3}\right)\cdot \ctg\frac{4\pi}{3}$$

   $$\sin\frac{5\pi}{4}=-\frac{\sqrt2}{2},\quad \cos\frac{5\pi}{6}=-\frac{\sqrt3}{2},\quad \tg\left(-\frac{2\pi}{3}\right)=\tg\frac{\pi}{3}=\sqrt3,\quad \ctg\frac{4\pi}{3}=\ctg\frac{\pi}{3}=\frac{1}{\sqrt3}.$$

   Тогда
   $$\left(-\frac{\sqrt2}{2}\right)\left(-\frac{\sqrt3}{2}\right)\cdot \sqrt3 \cdot \frac{1}{\sqrt3} =\frac{\sqrt6}{4}.$$

3. $$\frac{\sin 20^\circ\cdot \cos 10^\circ+\cos 160^\circ\cdot \cos 100^\circ}{\sin 21^\circ\cdot \cos 9^\circ+\cos 159^\circ\cdot \cos 99^\circ}$$

   $$\cos 160^\circ=\cos(180^\circ-20^\circ)=-\cos 20^\circ,\quad \cos 100^\circ=\cos(90^\circ+10^\circ)=-\sin 10^\circ,$$

   $$\cos 159^\circ=\cos(180^\circ-21^\circ)=-\cos 21^\circ,\quad \cos 99^\circ=\cos(90^\circ+9^\circ)=-\sin 9^\circ.$$

   Тогда
   $$\frac{\sin 20^\circ\cos 10^\circ+\cos 20^\circ\sin 10^\circ}{\sin 21^\circ\cos 9^\circ+\cos 21^\circ\sin 9^\circ} =\frac{\sin(20^\circ+10^\circ)}{\sin(21^\circ+9^\circ)} =\frac{\sin 30^\circ}{\sin 30^\circ}=1.$$

4. $$\frac{\cos 66^\circ\cdot \cos 5^\circ+\cos 84^\circ\cdot \cos 24^\circ}{\cos 65^\circ\cdot \cos 5^\circ+\cos 85^\circ\cdot \cos 25^\circ}$$

   $$\cos 84^\circ=\sin 6^\circ,\quad \cos 24^\circ=\sin 66^\circ,\quad \cos 85^\circ=\sin 5^\circ,\quad \cos 25^\circ=\sin 65^\circ.$$

   Тогда
   $$\frac{\cos 66^\circ\cos 5^\circ+\sin 6^\circ\sin 66^\circ}{\cos 65^\circ\cos 5^\circ+\sin 5^\circ\sin 25^\circ} =\frac{\cos(66^\circ-6^\circ)}{\cos(65^\circ-5^\circ)} =\frac{\cos 60^\circ}{\cos 60^\circ}=1.$$

Ответ

1) $$\frac{5}{3}$$; 2) $$\frac{\sqrt6}{4}$$; 3) $$1$$; 4) $$1$$.