Упр.24.9 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) cos(3a)-cos(4a)-cos(5a)+cos(6a)=-4sin(a/2)sin(a)cos(9a/2);
2) cos(3?/2+4a)+sin(3?-8a)-sin(4?-12a)=4cos(2a)cos(4a)sin(6a);
3) cos^2(a)-cos^2(?)=sin(a+?)sin(?-a);
4) (sin(a)+sin(3a)+sin(5a)+sin(7a))/(cos(a)+cos(3a)+cos(5a)+cos(7a))=tg(4a);
5) 2(sin(2a)+2cos^2(a)-1)/(cos(a)-sin(a)-cos(3a)+sin(3a))=1/sin(a);
6) (1+cos(a)+cos(2a)+cos(3a))/(cos(a)+2cos^2(a)-1)=2cos(a);
7) (cos(a)-cos(?))^2+(sin(a)-sin(?))^2=4sin^2((a-?)/2);
8) ((sin(a)-cos(a))^2-1+sin(4a))/(cos(2a)+cos(4a))=tg(a);
9) (sin(a)/sin(2a)-cos(a)/cos(2a))·(cos(a)-cos(7a))/sin(a)=-4sin(3a);
10) ((cos(a)-cos(3a))(sin(a)+sin(3a))/(1-cos(4a))=sin(2a).

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

2. $$\cos\left(\frac{3\pi}{2}+4a\right)+\sin(3\pi-8a)-\sin(4\pi-12a)$$
   $$=\sin 4a+\sin 8a+\sin 12a$$
   $$=2\sin 6a\cos 2a+\sin 12a$$
   $$=2\sin 6a\cos 2a+2\sin 6a\cos 6a$$
   $$=2\sin 6a(\cos 2a+\cos 6a)$$
   $$=2\sin 6a\cdot 2\cos 4a\cos 2a=4\cos 2a\cos 4a\sin 6a.$$

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

4. $$\frac{\sin a+\sin 3a+\sin 5a+\sin 7a}{\cos a+\cos 3a+\cos 5a+\cos 7a}$$
   $$=\frac{2\sin 2a\cos a+2\sin 6a\cos a}{2\cos 2a\cos a+2\cos 6a\cos a}$$
   $$=\frac{\sin 2a+\sin 6a}{\cos 2a+\cos 6a}$$
   $$=\frac{2\sin 4a\cos 2a}{2\cos 4a\cos 2a}=\tg 4a.$$

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

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

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

8. $$\frac{(\sin a-\cos a)^2-1+\sin 4a}{\cos 2a+\cos 4a}$$
   $$=\frac{\sin^2 a+\cos^2 a-2\sin a\cos a-1+\sin 4a}{\cos 2a+\cos 4a}$$
   $$=\frac{\sin 4a-2\sin a\cos a}{\cos 2a+\cos 4a}$$
   $$=\frac{2\sin 2a\cos 2a-2\sin a\cos a}{\cos 2a+\cos 4a}$$
   $$=\tg a.$$

9. $$\left(\frac{\sin a}{\sin 2a}-\frac{\cos a}{\cos 2a}\right)\cdot \frac{\cos a-\cos 7a}{\sin a}$$
   $$=\frac{\sin a\cos 2a-\sin 2a\cos a}{\sin 2a\cos 2a}\cdot \frac{-2\sin 4a\sin(-3a)}{\sin a}$$
   $$=\frac{\sin(a-2a)}{\frac12\sin 4a}\cdot \frac{2\sin 4a\sin 3a}{\sin a}$$
   $$=\frac{-\sin a}{\frac12\sin 4a}\cdot \frac{2\sin 4a\sin 3a}{\sin a}=-4\sin 3a.$$

10. $$\frac{(\cos a-\cos 3a)(\sin a+\sin 3a)}{1-\cos 4a}$$
    $$=\frac{-2\sin 2a\sin(-a)\cdot 2\sin 2a\cos a}{1-(1-2\sin^2 2a)}$$
    $$=\frac{4\sin a\cos a\sin^2 2a}{2\sin^2 2a}=2\sin a\cos a=\sin 2a.$$

Ответ

1) $$-4\sin \frac{a}{2}\sin a\cos \frac{9a}{2}$$;
2) $$4\cos 2a\cos 4a\sin 6a$$;
3) $$\sin(a+\beta)\sin(\beta-a)$$;
4) $$\tg 4a$$;
5) $$\frac{1}{\sin a}$$;
6) $$2\cos a$$;
7) $$4\sin^2 \frac{a-\beta}{2}$$;
8) $$\tg a$$;
9) $$-4\sin 3a$$;
10) $$\sin 2a$$.