Упр.1422 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) cos(8?/15)cos(?/5)+sin(8?/15)sin(?/5);
б) cos(?/10)cos(2?/5)-sin(?/10)sin(2?/5);
в) sin(?/6)cos(?/12)+cos(?/6)sin(?/12);
г) sin(1/9?)cos(4/9?)-cos(1/9?)sin(4/9?).

Используем формулы:

$$\cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha-\beta),$$

$$\cos \alpha \cos \beta — \sin \alpha \sin \beta = \cos(\alpha+\beta),$$

$$\sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin(\alpha+\beta),$$

$$\sin \alpha \cos \beta — \cos \alpha \sin \beta = \sin(\alpha-\beta).$$

а)

$$\cos \frac{8\pi}{15}\cos \frac{\pi}{5}+\sin \frac{8\pi}{15}\sin \frac{\pi}{5}=\cos\left(\frac{8\pi}{15}-\frac{\pi}{5}\right)=\cos\left(\frac{8\pi}{15}-\frac{3\pi}{15}\right)=\cos \frac{\pi}{3}=\frac12.$$

б)

$$\cos \frac{\pi}{10}\cos \frac{2\pi}{5}-\sin \frac{\pi}{10}\sin \frac{2\pi}{5}=\cos\left(\frac{\pi}{10}+\frac{2\pi}{5}\right)=\cos\left(\frac{\pi}{10}+\frac{4\pi}{10}\right)=\cos \frac{\pi}{2}=0.$$

в)

$$\sin \frac{\pi}{6}\cos \frac{\pi}{12}+\cos \frac{\pi}{6}\sin \frac{\pi}{12}=\sin\left(\frac{\pi}{6}+\frac{\pi}{12}\right)=\sin \frac{\pi}{4}=\frac{\sqrt2}{2}.$$

г)

$$\sin \frac{\pi}{9}\cos \frac{4\pi}{9}-\cos \frac{\pi}{9}\sin \frac{4\pi}{9}=\sin\left(\frac{\pi}{9}-\frac{4\pi}{9}\right)=\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt3}{2}.$$

Ответ

а) $$\frac12$$; б) $$0$$; в) $$\frac{\sqrt2}{2}$$; г) $$-\frac{\sqrt3}{2}$$.