Упр.63 Повторение ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)
Докажите тождество:
- $$\cos(3?)-\cos(4?)-\cos(5?)+\cos(6?)=-4\sin\left(\frac{?}{2}\right)\sin(?)\cos\left(\frac{9?}{2}\right).$$
- $$2\frac{\sin(2?)+2\cos^2(?)-1}{\cos(?)-\sin(?)-\cos(3?)+\sin(3?)}=\frac{1}{\sin(?)}.$$
- $$(\cos(?)-\cos(?))^2+(\sin(?)-\sin(?))^2=4\sin^2\left(\frac{?-?}{2}\right).$$
- $$\sin(?)+\sin(?)+\sin(?-?)=4\sin\left(\frac{?}{2}\right)\cos\left(\frac{?}{2}\right)\cos\left(\frac{?-?}{2}\right).$$
- $$\frac{\sin(2?)\cos(4?)(1+\cos(2?))}{(\sin(3?)+\sin(?))(\cos(3?)+\cos(5?))}=\frac{1}{2}.$$
- $$\sin^2\left(\frac{15?}{8}-2?\right)-\cos^2\left(\frac{17?}{8}-2?\right)=-\frac{\cos(4?)}{\sqrt{2}}.$$
1) $$\cos 3\alpha-\cos 4\alpha-\cos 5\alpha+\cos 6\alpha$$
$$=(\cos 3\alpha+\cos 6\alpha)-(\cos 4\alpha+\cos 5\alpha)$$
$$=2\cos \frac{9\alpha}{2}\cos \frac{3\alpha}{2}-2\cos \frac{9\alpha}{2}\cos \frac{\alpha}{2}$$
$$=2\cos \frac{9\alpha}{2}\left(\cos \frac{3\alpha}{2}-\cos \frac{\alpha}{2}\right)$$
$$=2\cos \frac{9\alpha}{2}\cdot\left(-2\sin \frac{\alpha}{2}\sin \alpha\right)$$
$$=-4\sin \frac{\alpha}{2}\sin \alpha \cos \frac{9\alpha}{2}.$$
Тождество доказано.
2) $$\frac{2(\sin 2\alpha+2\cos^2\alpha-1)}{\cos \alpha-\sin \alpha-\cos 3\alpha+\sin 3\alpha}$$
$$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{\cos \alpha-\sin \alpha-\cos 3\alpha+\sin 3\alpha}$$
$$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{-2\sin 2\alpha\sin(-\alpha)+\sin \alpha\cos 2\alpha}$$
$$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{2\sin \alpha(\sin 2\alpha+\cos 2\alpha)}$$
$$=\frac{1}{\sin \alpha}.$$
Тождество доказано.
3) $$\left(\cos \alpha-\cos \beta\right)^2+\left(\sin \alpha-\sin \beta\right)^2$$
$$=\cos^2\alpha-2\cos \alpha\cos \beta+\cos^2\beta+\sin^2\alpha-2\sin \alpha\sin \beta+\sin^2\beta$$
$$=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2(\cos \alpha\cos \beta+\sin \alpha\sin \beta)$$
$$=2-2\cos(\alpha-\beta)$$
$$=4\sin^2\frac{\alpha-\beta}{2}.$$
Тождество доказано.
4) $$\sin \alpha+\sin \beta+\sin(\alpha-\beta)$$
$$=2\sin \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2}+2\sin \frac{\alpha-\beta}{2}\cos \frac{\alpha-\beta}{2}$$
$$=2\cos \frac{\alpha-\beta}{2}\left(\sin \frac{\alpha+\beta}{2}+\sin \frac{\alpha-\beta}{2}\right)$$
$$=2\cos \frac{\alpha-\beta}{2}\cdot 2\sin \frac{\alpha}{2}\cos \frac{\beta}{2}$$
$$=4\sin \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\alpha-\beta}{2}.$$
Тождество доказано.
5) $$\frac{\sin 2\alpha\cos 4\alpha(1+\cos 2\alpha)}{(\sin 3\alpha+\sin \alpha)(\cos 3\alpha+\cos 5\alpha)}$$
$$=\frac{\sin 2\alpha\cos 4\alpha(1+\cos 2\alpha)}{(2\sin 2\alpha\cos \alpha)(2\cos 4\alpha\cos \alpha)}$$
$$=\frac{1+\cos 2\alpha}{4\cos^2\alpha}$$
$$=\frac{2\cos^2\alpha}{4\cos^2\alpha}=\frac12.$$
Тождество доказано.
6) $$\sin^2\left(\frac{15\pi}{8}-2\alpha\right)-\cos^2\left(\frac{17\pi}{8}-2\alpha\right)$$
$$=\sin^2\left(-\frac{\pi}{8}-2\alpha\right)-\cos^2\left(\frac{\pi}{8}-2\alpha\right)$$
$$=\sin^2\left(\frac{\pi}{8}+2\alpha\right)-\cos^2\left(\frac{\pi}{8}-2\alpha\right)$$
$$=\frac{1-\cos\left(\frac{\pi}{4}+4\alpha\right)}{2}-\frac{1+\cos\left(\frac{\pi}{4}-4\alpha\right)}{2}$$
$$=-\frac12\left(\cos\left(\frac{\pi}{4}+4\alpha\right)+\cos\left(\frac{\pi}{4}-4\alpha\right)\right)$$
$$=-\cos \frac{\pi}{4}\cos 4\alpha=-\frac{\cos 4\alpha}{\sqrt2}.$$
Тождество доказано.
