Упр.28.324 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)

1) 4sin(a)cos^3(a)-4sin^3(a)cos(a);
2) sin(a/4)cos(a/4)cos(a/2);
3) sin(6a)/sin(2a)+cos(6a)/cos(2a);
4) (ctg(a)+tg(a))/(ctg(a)-tg(a));
5) cos(a)/(tg^2(a/2)-ctg^2(a/2));
6) (1-2sin^2(2a))/(2tg(3п/4-2a)cos^2(3п/4+2a));
7) (2cos^2(a)-1)/(2ctg(п/4-a)sin^2(п/4-a));
8) cos^4(a-п)/(cos^4(a-3п/2)+sin^4(a+3п/2)-1).

1. $$4\sin a\cos^3 a-4\sin^3 a\cos a=4\sin a\cos a(\cos^2 a-\sin^2 a)$$
   $$=2\sin 2a\cdot \cos 2a=\sin 4a.$$
2. $$\sin \frac a4\cos \frac a4\cos \frac a2=\frac12\sin \frac a2\cos \frac a2=\frac14\sin a.$$
3. $$\frac{\sin 6a}{\sin 2a}+\frac{\cos 6a}{\cos 2a}
   =\frac{\sin 6a\cos 2a+\cos 6a\sin 2a}{\sin 2a\cos 2a}$$
   $$=\frac{\sin(6a+2a)}{\sin 2a\cos 2a}
   =\frac{\sin 8a}{\sin 2a\cos 2a}
   =\frac{2\sin 4a\cos 4a}{\frac12\sin 4a}=4\cos 4a.$$
4. $$\frac{\ctg a+\tg a}{\ctg a-\tg a}
   =\frac{\frac{1+\tg^2 a}{\tg a}}{\frac{1-\tg^2 a}{\tg a}}
   =\frac{1+\tg^2 a}{1-\tg^2 a}$$
   $$=\frac{1}{\cos^2 a}\left(2-\frac{1}{\cos^2 a}\right)
   =\frac{2\cos^2 a-1}{\cos^2 a}
   =\frac{1}{\cos 2a}.$$
5. $$\frac{\cos a}{\tg^2 \frac a2-\ctg^2 \frac a2}
   =\frac{\cos a}{\frac{1-\cos a}{1+\cos a}-\frac{1+\cos a}{1-\cos a}}$$
   $$=\cos a\cdot \frac{(1-\cos a)(1+\cos a)}{(1-\cos a)^2-(1+\cos a)^2}$$
   $$=\cos a\cdot \frac{1-\cos^2 a}{-4\cos a}
   =-\frac14\sin^2 a.$$
6. $$\frac{1-2\sin^2 2a}{2\tg\left(\frac{3\pi}{4}-2a\right)\cos^2\left(\frac{3\pi}{4}+2a\right)}$$
   $$=\frac{\cos 4a}{2\tg\left(\frac{3\pi}{4}-2a\right)\sin^2\left(\frac{3\pi}{2}-\left(\frac{3\pi}{4}-2a\right)\right)}$$
   $$=\frac{\cos 4a}{2\tg\left(\frac{3\pi}{4}-2a\right)\cos^2\left(\frac{3\pi}{4}-2a\right)}
   =\frac{\cos 4a}{2\sin\left(\frac{3\pi}{4}-2a\right)\cos\left(\frac{3\pi}{4}-2a\right)}$$
   $$=\frac{\cos 4a}{\sin\left(\frac{3\pi}{2}-4a\right)}
   =\frac{\cos 4a}{-\cos 4a}=-1.$$
7. $$\frac{2\cos^2 a-1}{2\ctg\left(\frac{\pi}{4}-a\right)\sin^2\left(\frac{\pi}{4}-a\right)}
   =\frac{\cos 2a}{2\frac{\cos\left(\frac{\pi}{4}-a\right)}{\sin\left(\frac{\pi}{4}-a\right)}\sin^2\left(\frac{\pi}{4}-a\right)}$$
   $$=\frac{\cos 2a}{2\sin\left(\frac{\pi}{4}-a\right)\cos\left(\frac{\pi}{4}-a\right)}
   =\frac{\cos 2a}{\sin\left(\frac{\pi}{2}-2a\right)}
   =\frac{\cos 2a}{\cos 2a}=1.$$
8. $$\frac{\cos^4(a-\pi)}{\cos^4\left(a-\frac{3\pi}{2}\right)+\sin^4\left(a+\frac{3\pi}{2}\right)-1}
   =\frac{\cos^4 a}{\sin^4 a+\cos^4 a-1}$$
   $$=\frac{\cos^4 a}{(\sin^2 a+\cos^2 a)^2-2\sin^2 a\cos^2 a-1}
   =\frac{\cos^4 a}{-2\sin^2 a\cos^2 a}$$
   $$=-\frac12\ctg^2 a.$$

Ответ

1) $$\sin 4a$$; 2) $$\frac14\sin a$$; 3) $$4\cos 4a$$; 4) $$\frac{1}{\cos 2a}$$; 5) $$-\frac14\sin^2 a$$; 6) $$-1$$; 7) $$1$$; 8) $$-\frac12\ctg^2 a$$.