Упр.26.18 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

Задача

1) 1+2cos(2?)+cos(4?)=4cos^2(?)cos(2?);
2) (1+sin(2?)-cos(2?))/(1+sin(2?)+cos(2?))=tg(?);
3) (sin^2(2?)+4sin^2(?)-4)/(1-8sin^2(?)-cos(4?))=1/2ctg^4(?);
4) cos(4?-?/2)sin(5?/2+2?)/((1+cos(2?))(1+cos4?)))=tg(?);
5) (cos(4?)+1)/(ctg(?)-tg(?))=1/2sin(4?);
6) (2cos(2?)-sin(4?))/(2cos(2?)+sin(4?))=tg^2(45°-?).

Подробный ответ

$$1+2\cos 2\alpha+\cos 4\alpha=1+2\cos 2\alpha+(2\cos^2 2\alpha-1)$$
$$=2\cos 2\alpha+2\cos^2 2\alpha=2\cos 2\alpha(1+\cos 2\alpha)$$
$$=2\cos 2\alpha\cdot 2\cos^2\alpha=4\cos^2\alpha\cos 2\alpha.$$
$$\frac{1+\sin 2\alpha-\cos 2\alpha}{1+\sin 2\alpha+\cos 2\alpha} =\frac{2\sin\alpha\cos\alpha+1-(1-2\sin^2\alpha)}{2\sin\alpha\cos\alpha+1+(2\cos^2\alpha-1)}$$
$$=\frac{2\sin\alpha(\cos\alpha+\sin\alpha)}{2\cos\alpha(\sin\alpha+\cos\alpha)} =\frac{\sin\alpha}{\cos\alpha}=\tg\alpha.$$
$$\frac{\sin^2 2\alpha+4\sin^2\alpha-4}{1-8\sin^2\alpha-\cos 4\alpha} =\frac{4\sin^2\alpha\cos^2\alpha-4\cos^2\alpha}{2\sin^2 2\alpha-8\sin^2\alpha}$$
$$=\frac{4\cos^2\alpha(\sin^2\alpha-1)}{8\sin^2\alpha(\cos^2\alpha-1)} =\frac{4\cos^2\alpha(-\cos^2\alpha)}{8\sin^2\alpha(-\sin^2\alpha)}$$
$$=\frac12\cdot\frac{\cos^4\alpha}{\sin^4\alpha}=\frac12\ctg^4\alpha.$$
$$\frac{\cos\left(4\alpha-\frac{\pi}{2}\right)\sin\left(\frac{5\pi}{2}+2\alpha\right)}{(1+\cos 2\alpha)(1+\cos 4\alpha)} =\frac{\sin 4\alpha}{(1+\cos 2\alpha)(1+2\cos^2 2\alpha-1)}$$
$$=\frac{2\sin 2\alpha\cos 2\alpha\cdot \cos 2\alpha}{(1+\cos 2\alpha)\cdot 2\cos^2 2\alpha} =\frac{\sin 2\alpha}{1+\cos 2\alpha}$$
$$=\frac{2\sin\alpha\cos\alpha}{1+2\cos^2\alpha-1} =\frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} =\tg\alpha.$$
$$\frac{\cos 4\alpha+1}{\ctg\alpha-\tg\alpha} =\frac{2\cos^2 2\alpha}{\ctg\alpha-\tg\alpha}$$
$$=\frac{2\cos^2 2\alpha\cdot\tg\alpha}{1-\tg^2\alpha} =\frac{2\cos^2 2\alpha\cdot\tg\alpha}{\frac{\cos 2\alpha}{\cos^2\alpha}}$$
$$=\cos^2 2\alpha\cdot\tg 2\alpha=\cos 2\alpha\sin 2\alpha=\frac12\sin 4\alpha.$$
$$\frac{2\cos 2\alpha-\sin 4\alpha}{2\cos 2\alpha+\sin 4\alpha} =\frac{2\cos 2\alpha-2\sin 2\alpha\cos 2\alpha}{2\cos 2\alpha+2\sin 2\alpha\cos 2\alpha}$$
$$=\frac{2\cos 2\alpha(1-\sin 2\alpha)}{2\cos 2\alpha(1+\sin 2\alpha)} =\frac{1-\sin 2\alpha}{1+\sin 2\alpha}$$
$$=\frac{1-\cos(90^\circ-2\alpha)}{1+\cos(90^\circ-2\alpha)} =\tg^2\left(45^\circ-\alpha\right).$$