Упр.63 Повторение ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) cos(3?)-cos(4?)-cos(5?)+cos(6?)=-4sin(?/2)sin(?)cos(9?/2);
2) 2(sin(2?)+2cos^2(?)-1)/(cos(?)-sin(?)-cos(3?)+sin(3?))=1/sin(?);
3) (cos(?)-cos(?))^2+(sin(?)-sin(?))^2=4sin^2((?-?)/2);
4) sin(?)+sin(?)+sin(?-?)=4sin(?/2)cos(?/2)cos((?-?)/2);
5) sin(2?)cos(4?)(1+cos(2?))/((sin(3?)+sin(?))(cos(3?)+cos(5?))=1/2;
6) sin^2(15?/8-2?)-cos^2(17?/8-2?)=-cos(4?)/v2.

1. $$\cos 3\alpha-\cos 4\alpha-\cos 5\alpha+\cos 6\alpha$$
   $$=(\cos 3\alpha+\cos 6\alpha)-(\cos 4\alpha+\cos 5\alpha)$$
   $$=2\cos \frac{9\alpha}{2}\cos \frac{3\alpha}{2}-2\cos \frac{9\alpha}{2}\cos \frac{\alpha}{2}$$
   $$=2\cos \frac{9\alpha}{2}\left(\cos \frac{3\alpha}{2}-\cos \frac{\alpha}{2}\right)$$
   $$=-4\sin \frac{\alpha}{2}\sin \alpha \cos \frac{9\alpha}{2}.$$

2. $$\frac{2(\sin 2\alpha+2\cos^2\alpha-1)}{\cos\alpha-\sin\alpha-\cos 3\alpha+\sin 3\alpha}$$
   $$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{(\cos\alpha-\cos 3\alpha)+(\sin 3\alpha-\sin\alpha)}$$
   $$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{2\sin 2\alpha\sin\alpha+2\cos 2\alpha\sin\alpha}$$
   $$=\frac{2(\sin 2\alpha+\cos 2\alpha)}{2\sin\alpha(\sin 2\alpha+\cos 2\alpha)}$$
   $$=\frac{1}{\sin\alpha}.$$

3. $$ (\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2 $$
   $$=\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta$$
   $$=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)$$
   $$=2-2\cos(\alpha-\beta)$$
   $$=4\sin^2\frac{\alpha-\beta}{2}.$$

4. $$\sin\alpha+\sin\beta+\sin(\alpha-\beta)$$
   $$=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+\sin(\alpha-\beta)$$
   $$=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha-\beta}{2}$$
   $$=2\cos\frac{\alpha-\beta}{2}\left(\sin\frac{\alpha+\beta}{2}+\sin\frac{\alpha-\beta}{2}\right)$$
   $$=4\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\alpha-\beta}{2}.$$

5. $$\frac{\sin 2\alpha\cos 4\alpha(1+\cos 2\alpha)}{(\sin 3\alpha+\sin\alpha)(\cos 3\alpha+\cos 5\alpha)}$$
   $$=\frac{\sin 2\alpha\cos 4\alpha(1+\cos 2\alpha)}{(2\sin 2\alpha\cos\alpha)(2\cos 4\alpha\cos\alpha)}$$
   $$=\frac{1+\cos 2\alpha}{4\cos^2\alpha}$$
   $$=\frac{2\cos^2\alpha}{4\cos^2\alpha}=\frac12.$$

6. $$\sin^2\left(\frac{15\pi}{8}-2\alpha\right)-\cos^2\left(\frac{17\pi}{8}-2\alpha\right)$$
   $$=\sin^2\left(-\frac{\pi}{8}-2\alpha\right)-\cos^2\left(\frac{\pi}{8}-2\alpha\right)$$
   $$=\sin^2\left(\frac{\pi}{8}+2\alpha\right)-\cos^2\left(\frac{\pi}{8}-2\alpha\right)$$
   $$=\frac{1-\cos\left(\frac{\pi}{4}+4\alpha\right)}{2}-\frac{1+\cos\left(\frac{\pi}{4}-4\alpha\right)}{2}$$
   $$=-\frac12\left(\cos\left(\frac{\pi}{4}+4\alpha\right)+\cos\left(\frac{\pi}{4}-4\alpha\right)\right)$$
   $$=-\cos\frac{\pi}{4}\cos 4\alpha=-\frac{\cos 4\alpha}{\sqrt2}.$$

Ответ

1) $$\cos 3\alpha-\cos 4\alpha-\cos 5\alpha+\cos 6\alpha=-4\sin\frac{\alpha}{2}\sin\alpha\cos\frac{9\alpha}{2}$$

2) $$\frac{2(\sin 2\alpha+2\cos^2\alpha-1)}{\cos\alpha-\sin\alpha-\cos 3\alpha+\sin 3\alpha}=\frac1{\sin\alpha}$$

3) $$(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=4\sin^2\frac{\alpha-\beta}{2}$$

4) $$\sin\alpha+\sin\beta+\sin(\alpha-\beta)=4\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\alpha-\beta}{2}$$

5) $$\frac{\sin 2\alpha\cos 4\alpha(1+\cos 2\alpha)}{(\sin 3\alpha+\sin\alpha)(\cos 3\alpha+\cos 5\alpha)}=\frac12$$

6) $$\sin^2\left(\frac{15\pi}{8}-2\alpha\right)-\cos^2\left(\frac{17\pi}{8}-2\alpha\right)=-\frac{\cos 4\alpha}{\sqrt2}$$