Упр.22.4 ГДЗ Мордкович 10-11 класс (Алгебра)

а) $$\cos\frac{\pi}{10}-\cos\frac{\pi}{20}$$; б) $$\cos\frac{11\pi}{12}+\cos\frac{3\pi}{4}$$; в) $$\cos\frac{\pi}{5}-\cos\frac{\pi}{11}$$; г) $$\cos\frac{3\pi}{8}+\cos\frac{5\pi}{4}$$.

Используем формулы преобразования суммы и разности тригонометрических функций:

$$\cos \alpha-\cos \beta=-2\sin \frac{\alpha+\beta}{2}\sin \frac{\alpha-\beta}{2}$$

$$\cos \alpha+\cos \beta=2\cos \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2}$$

1. $$\cos \frac{\pi}{10}-\cos \frac{\pi}{20}=-2\sin \frac{\frac{\pi}{10}+\frac{\pi}{20}}{2}\sin \frac{\frac{\pi}{10}-\frac{\pi}{20}}{2}=-2\sin \frac{3\pi}{40}\sin \frac{\pi}{40}$$

2. $$\cos \frac{11\pi}{12}+\cos \frac{3\pi}{4}=2\cos \frac{\frac{11\pi}{12}+\frac{3\pi}{4}}{2}\cos \frac{\frac{11\pi}{12}-\frac{3\pi}{4}}{2}=2\cos \frac{5\pi}{6}\cos \frac{\pi}{12}$$

   $$=2\cdot \left(-\frac{\sqrt{3}}{2}\right)\cos \frac{\pi}{12}=-\sqrt{3}\cos \frac{\pi}{12}$$

3. $$\cos \frac{\pi}{5}-\cos \frac{\pi}{11}=-2\sin \frac{\frac{\pi}{5}+\frac{\pi}{11}}{2}\sin \frac{\frac{\pi}{5}-\frac{\pi}{11}}{2}=-2\sin \frac{8\pi}{55}\sin \frac{3\pi}{55}$$

4. $$\cos \frac{3\pi}{8}+\cos \frac{5\pi}{4}=2\cos \frac{\frac{3\pi}{8}+\frac{5\pi}{4}}{2}\cos \frac{\frac{3\pi}{8}-\frac{5\pi}{4}}{2}=2\cos \frac{13\pi}{16}\cos \left(-\frac{7\pi}{16}\right)$$

   $$=2\cos \frac{13\pi}{16}\cos \frac{7\pi}{16}$$

Ответ:

1. $$-2\sin \frac{3\pi}{40}\sin \frac{\pi}{40}$$
2. $$-\sqrt{3}\cos \frac{\pi}{12}$$
3. $$-2\sin \frac{8\pi}{55}\sin \frac{3\pi}{55}$$
4. $$2\cos \frac{13\pi}{16}\cos \frac{7\pi}{16}$$