Упр.1477 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

Используем формулы приведения к произведению:

$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$
$$\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$
$$\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$
$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$

1. $$\sin\frac{\pi}{5}+\sin\frac{3\pi}{5}=2\sin\frac{\frac{\pi}{5}+\frac{3\pi}{5}}{2}\cos\frac{\frac{\pi}{5}-\frac{3\pi}{5}}{2}$$
   $$=2\sin\frac{2\pi}{5}\cos\frac{-\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{\pi}{5}$$

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

3. $$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=2\cos\frac{\frac{3\pi}{10}+\frac{\pi}{10}}{2}\sin\frac{\frac{3\pi}{10}-\frac{\pi}{10}}{2}$$
   $$=2\cos\frac{\pi}{5}\sin\frac{\pi}{10}$$

4. $$\cos\frac{\pi}{4}-\cos\frac{3\pi}{4}=-2\sin\frac{\frac{\pi}{4}+\frac{3\pi}{4}}{2}\sin\frac{\frac{\pi}{4}-\frac{3\pi}{4}}{2}$$
   $$=-2\sin\frac{\pi}{2}\sin\frac{-\pi}{4}=2\sin\frac{\pi}{2}\sin\frac{\pi}{4}$$

5. $$\sin\left(\frac{\pi}{6}+x\right)+\sin\left(\frac{\pi}{6}-x\right)=2\sin\frac{\left(\frac{\pi}{6}+x\right)+\left(\frac{\pi}{6}-x\right)}{2}\cos\frac{\left(\frac{\pi}{6}+x\right)-\left(\frac{\pi}{6}-x\right)}{2}$$
   $$=2\sin\frac{\pi}{6}\cos x$$

6. $$\sin\left(\frac{\pi}{3}+a\right)-\sin\left(\frac{\pi}{3}-a\right)=2\cos\frac{\left(\frac{\pi}{3}+a\right)+\left(\frac{\pi}{3}-a\right)}{2}\sin\frac{\left(\frac{\pi}{3}+a\right)-\left(\frac{\pi}{3}-a\right)}{2}$$
   $$=2\cos\frac{\pi}{3}\sin a$$

7. $$\cos\left(\frac{\pi}{3}-y\right)+\cos\left(\frac{\pi}{3}+y\right)=2\cos\frac{\left(\frac{\pi}{3}-y\right)+\left(\frac{\pi}{3}+y\right)}{2}\cos\frac{\left(\frac{\pi}{3}-y\right)-\left(\frac{\pi}{3}+y\right)}{2}$$
   $$=2\cos\frac{\pi}{3}\cos y$$

8. $$\cos\left(\frac{\pi}{6}-\beta\right)-\cos\left(\frac{\pi}{6}+\beta\right)=-2\sin\frac{\left(\frac{\pi}{6}-\beta\right)+\left(\frac{\pi}{6}+\beta\right)}{2}\sin\frac{\left(\frac{\pi}{6}-\beta\right)-\left(\frac{\pi}{6}+\beta\right)}{2}$$
   $$=-2\sin\frac{\pi}{6}\sin(-\beta)=2\sin\frac{\pi}{6}\sin\beta$$

Ответ

а) $$2\sin\frac{2\pi}{5}\cos\frac{\pi}{5}$$;
б) $$2\cos\frac{5\pi}{12}\cos\frac{\pi}{4}$$;
в) $$2\cos\frac{\pi}{5}\sin\frac{\pi}{10}$$;
г) $$2\sin\frac{\pi}{2}\sin\frac{\pi}{4}$$;
д) $$2\sin\frac{\pi}{6}\cos x$$;
е) $$2\cos\frac{\pi}{3}\sin a$$;
ж) $$2\cos\frac{\pi}{3}\cos y$$;
з) $$2\sin\frac{\pi}{6}\sin\beta$$.