Упр.23.33 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) cos^4(a)-6sin^2(a)cos^2(a)+sin^4(a);
2) cos(2a)/(ctg(a)-sin(2a));
3) (2sin(4a)(1-tg^2(2a)))/(1+ctg^2(?/2+2a));
4) (2cos^2(a)-1)/(2ctg(?/4-a)sin^2(?/4-a)).

1. $$\cos^4 a-6\sin^2 a\cos^2 a+\sin^4 a$$
   $$=(\cos^4 a-2\sin^2 a\cos^2 a+\sin^4 a)-4\sin^2 a\cos^2 a$$
   $$=(\cos^2 a-\sin^2 a)^2-(2\sin a\cos a)^2$$
   $$=\cos^2 2a-\sin^2 2a=\cos 4a.$$

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

3. $$\frac{2\sin 4a\,(1-\tg^2 2a)}{1+\ctg^2\left(\frac{\pi}{2}+2a\right)}$$
   $$=\frac{2\sin 4a\,(1-\tg^2 2a)}{1+\tg^2 2a}$$
   $$=2\sin 4a\cdot \cos 4a=\sin 8a.$$

4. $$\frac{2\cos^2 a-1}{2\ctg\left(\frac{\pi}{4}-a\right)\sin^2\left(\frac{\pi}{4}-a\right)}$$
   $$=\frac{\cos 2a}{2\cdot \frac{\cos\left(\frac{\pi}{4}-a\right)}{\sin\left(\frac{\pi}{4}-a\right)}\cdot \sin^2\left(\frac{\pi}{4}-a\right)}$$
   $$=\frac{\cos 2a}{2\sin\left(\frac{\pi}{4}-a\right)\cos\left(\frac{\pi}{4}-a\right)}$$
   $$=\frac{\cos 2a}{\sin\left(\frac{\pi}{2}-2a\right)}=\frac{\cos 2a}{\cos 2a}=1.$$

Ответ

1) $$\cos 4a$$; 2) $$\tg a$$; 3) $$\sin 8a$$; 4) $$1$$.