Упр.25 ГДЗ Колмогоров 10-11 класс (Алгебра)

а)

$$(\sin^2 t+2\sin t\cos t-\cos^2 t)^2 = \bigl((2\sin t\cos t)-(\cos^2 t-\sin^2 t)\bigr)^2$$
$$= (\sin 2t-\cos 2t)^2 = \sin^2 2t-2\sin 2t\cos 2t+\cos^2 2t$$
$$= (\sin^2 2t+\cos^2 2t)-2\sin 2t\cos 2t = 1-\sin 4t.$$

б)

$$\frac{\cos \alpha-2\sin 3\alpha-\cos 5\alpha}{\sin 5\alpha-2\cos 3\alpha-\sin \alpha} = \frac{(\cos \alpha-\cos 5\alpha)-2\sin 3\alpha}{(\sin 5\alpha-\sin \alpha)-2\cos 3\alpha}$$
$$= \frac{2\sin 2\alpha\sin 3\alpha}{2\cos 3\alpha\sin 2\alpha} = \frac{\sin 3\alpha}{\cos 3\alpha} = \tg 3\alpha.$$

в)

$$\frac{1-4\sin^2 t\cos^2 t}{\cos^2 t-\sin^2 t} = \frac{1-\sin^2 2t}{\cos 2t} = \frac{\cos^2 2t}{\cos 2t} = \cos 2t.$$

г)

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