Упр.24.1 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
1) cos(50°)+cos(20°); 5) cos(?/18)+cos(?/12);
2) cos(2a)-cos(4a); 6) sin(x+a)+sin(x-a);
3) sin(?)+sin(4?); 7) sin(2a-?/6)-sin(2a-?/3);
4) sin(5°)-sin(3°); 8) cos(?/3+a)+cos(?/3-a).
$$\cos 50^\circ+\cos 20^\circ=2\cos\frac{50^\circ+20^\circ}{2}\cdot\cos\frac{50^\circ-20^\circ}{2}=2\cos35^\circ\cdot\cos15^\circ.$$
$$\cos 2a-\cos 4a=-2\sin\frac{2a+4a}{2}\cdot\sin\frac{2a-4a}{2}=-2\sin3a\cdot\sin(-a)=2\sin3a\cdot\sin a.$$
$$\sin\beta+\sin4\beta=2\sin\frac{\beta+4\beta}{2}\cdot\cos\frac{\beta-4\beta}{2}=2\sin\frac{5\beta}{2}\cdot\cos\frac{3\beta}{2}.$$
$$\sin5^\circ-\sin3^\circ=2\sin\frac{5^\circ-3^\circ}{2}\cdot\cos\frac{5^\circ+3^\circ}{2}=2\sin1^\circ\cdot\cos4^\circ.$$
$$\cos\frac{\pi}{18}+\cos\frac{\pi}{12}=2\cos\frac{\frac{\pi}{18}+\frac{\pi}{12}}{2}\cdot\cos\frac{\frac{\pi}{18}-\frac{\pi}{12}}{2}$$
$$=2\cos\frac{5\pi}{72}\cdot\cos\frac{\pi}{72}.$$
$$\sin(x+a)+\sin(x-a)=2\sin\frac{(x+a)+(x-a)}{2}\cdot\cos\frac{(x+a)-(x-a)}{2}$$
$$=2\sin x\cdot\cos a.$$
$$\sin\left(2a-\frac{\pi}{6}\right)-\sin\left(2a-\frac{\pi}{3}\right)=2\sin\frac{\left(2a-\frac{\pi}{6}\right)-\left(2a-\frac{\pi}{3}\right)}{2}\cdot\cos\frac{\left(2a-\frac{\pi}{6}\right)+\left(2a-\frac{\pi}{3}\right)}{2}$$
$$=2\sin\frac{\pi}{12}\cdot\cos\left(2a-\frac{\pi}{4}\right).$$
$$\cos\left(\frac{\pi}{3}+a\right)+\cos\left(\frac{\pi}{3}-a\right)=2\cos\frac{\left(\frac{\pi}{3}+a\right)+\left(\frac{\pi}{3}-a\right)}{2}\cdot\cos\frac{\left(\frac{\pi}{3}+a\right)-\left(\frac{\pi}{3}-a\right)}{2}$$
$$=2\cos\frac{\pi}{3}\cdot\cos a.$$
Ответ
- $$2\cos35^\circ\cdot\cos15^\circ$$
- $$2\sin3a\cdot\sin a$$
- $$2\sin\frac{5\beta}{2}\cdot\cos\frac{3\beta}{2}$$
- $$2\sin1^\circ\cdot\cos4^\circ$$
- $$2\cos\frac{5\pi}{72}\cdot\cos\frac{\pi}{72}$$
- $$2\sin x\cdot\cos a$$
- $$2\sin\frac{\pi}{12}\cdot\cos\left(2a-\frac{\pi}{4}\right)$$
- $$2\cos\frac{\pi}{3}\cdot\cos a$$